Stat Final

a) 0.12 / 2 = 0.6 Enter (2nd)+(VARS)+(3InvNormal)+(%0.06) = 1.55 (two decimals)

Find zα/2 for α=0.12.

U b. Complete the hypotheses below. H0​: μ = 820 Ha​: μ ≠ 820 c. A Type I error means the researcher concludes the mean is different from 820 but in fact the mean is equal to 820 A Type II error means the researcher concludes the mean is equal to 820 but in fact the mean is different from 820

A certain national lunch program mandates that for a high school to receive reimbursement for school​ lunches, the number of calories served at lunch must be no more than 820 calories. Suppose a nutritionist believes that the true mean number of calories served at lunch at all schools in the country is is not 820 calories. a. Identify the parameter of interest. b. Specify the null and alternative hypotheses for testing this claim. c. Describe a Type I error in the words of the problem. d. Describe a Type II error in the words of the problem.

a. What is ​p(22​)? (just find 22 and find the value on the y axis for probability b. What is the probability that x equals 21 or 23​? (add the probabilities up) \ c. What is ​ P(x≤21​)? (add all probabilities up)

A discrete random variable x can assume five possible​ values, 19​, 20​, 21​, 22​, and 23. The following histogram shows the likelihood of each value. Complete parts a through c below.

H0​: p =.45 Ha​: p <.45 if more >

A media research company conducted an online survey of adults in a certain country to determine their favorite sport. A friend believes professional football​ (with revenue of about $10 billion per​ year) is the favorite sport of less than 45% of adults in the country. Specify the null and alternative hypotheses for testing this belief. Be sure to identify the parameter of interest.

6! 6(5) ------ = ------------ 4! (6-4) 4(3)(2)(1) {2(1)}

Compute the following.

No, because np<15 and nq≥15.

For the binomial sample information summarized​ below, indicate whether the sample size is large enough to use the large sample approximation to construct a confidence interval for p. n=40​, p=0.25

The test does not prove the alternative hypothesis​ correct, because of the possibility of rejecting the null hypothesis when the null hypothesis is true.

If a hypothesis test is performed and the null hypothesis is rejected in favor of the alternative​ hypothesis, does the test prove that the alternative hypothesis is​ correct? Explain

ENTER STAT+(EDIT)+(ENTER DATA) ENTER STAT +2 SAMP T TEST + (DATA) +POOLED (ENTER X,Q, N) ENTER +STAT +(9) 2-SAMP T INT)+(DATA) +Pooled (ENTER INFO)

Independent random samples from approximately normal populations produced the results shown below. Complete parts a and b.

a) just plug in the 95% CI ( 1.61058, 2.24342) b) ​No, because we can be​ 95% confident that the mean surface roughness lies between the lower and upper bounds of the​ interval, and 2.5 lies outside the interval. (because it doesn't lie in between interval)

Oil field pipes are internally coated in order to prevent corrosion. Researchers at a university investigated the influence that coating may have on the surface roughness of oil field pipes. A scanning probe instrument was used to measure the surface roughness of each section in a sample of 10 sections of coated interior pipe. The data​ (in micrometers) are shown in the table to the right. A Minitab analysis of the data is given below. Complete parts a and b. a. Locate a​ 95% confidence interval for the mean surface roughness of coated interior pipe on the accompanying Minitab printout. b. Would you expect the average surface roughness to be as high as 2.5​ micrometers? Explain

Ho: p=0.060 Ha​: p>0.060

Several years​ ago, a government agency reported the default rate​ (the proportion of borrowers who default on their​ loans) on a certain type of loan at 0.060. Set up the null and alternative hypotheses to determine if the default rate this year is greater than 0.060.

the answer is 1 dunno The answer is 2 dunno L1 L2 high light the top of l3 and the l1-l2 μd=μ1−μ2 ENTER + STAT + 8TINTRVAL +(DATA) ENTER + STAT + T-Test + DATA

The data for a random sample of six paired observations are shown in the table. a. Calculate d and s2d. b. Express μd in terms of μ1 and μ2. c. Form a 90​% confidence interval for μd. d. Test the null hypothesis H0​: μd=0 against the alternative hypothesis Ha​: μd≠0. Use α=0.10.

yes , because np≥15 and nq≥15.

A random sample of size n=200 yielded p=0.70. a. Is the sample size large enough to use the large sample approximation to construct a confidence interval for​ p? Explain. b. Construct a 99​% confidence interval for p. c. Interpret the 99​% confidence interval. d. Explain what is meant by the phrase ​"99​% confidence​ interval."

The mean of the sampling distribution of p is p. The standard deviation of the sampling distribution of p is SQRT (pq/n) The sampling distribution of p is approximately normal.

Describe the sampling distribution of p based on large samples of size n—that ​is, given the​ mean, the standard​ deviation, and the​ (approximate) shape of the distribution of p when large samples of size n are​ (repeatedly) selected from the binomial distribution with a probability of success p. Let q=1−p.

ux = u q= Q/(Srt n )

Suppose a random sample of n=100 measurements is selected from a population with mean μ and standard deviation σ. For each of the following values of μ and σ​, give the values of μx and σx. a. μ=5​, σ=3 b. μ=25​, σ=100 c. μ=10​, σ=80 d. μ=5​, σ=

Reject the null hypothesis since the​p-value is less than the value of α.

For the α and observed significance level​ (p-value) pair, indicate whether the null hypothesis would be rejected. α=0.10​, ​p-value = 0.05

x-u/ o

Suppose the random variable x is best described by a normal distribution with μ=25 and σ=5. Find the​ z-score that corresponds to each of the following​ x-values. a. x=15 b. x=25 c. x=10 d. x=13 e. x=19 f. x=

a. Locate the values of​ SSE, s2​, and s. SSE=382722271 (SS +SND LINE) s2= 1,234,588 (MS +SND LINE) s=1111.12 (top row under s) The value s is an estimator for σ. Approximately​ 95% of the observed values of y will fall within 2s of their respective least squares predicted values.

Use the output to the right to complete parts a and b below.

The level of​ significance, α​, is the probability of committing a Type I​ error, rejecting the null hypothesis when it​ is, in​ fact, true

What is the level of significance of a test of​ hypothesis?

a) enter (Stat) +(1edit)+(plug in numbers) enter (Stat) + (test) +(8Tinterrval)+(data)+(plug in CL%) (282.26 , 433.3) b) The confidence interval means that we are 99​% confident that the true mean skidding distance for the whole road is in the interval. c) The sample must be randomly selected from the population. The population distribution must be approximately normal. d) ​Yes, because the sample distribution is approximately normal. e) ​Yes, because the problem statement indicates that it is. f) The condition does not apply. g) The condition does not apply. h) No, because the value 456 lies outside the confidence interval.

A group of researchers wants to estimate the true mean skidding distance along a new road in a certain forest. The skidding distances​ (in meters) were measured at 20 randomly selected road sites. These values are given in the accompanying table. Complete parts a through d. 489 347 454 195 289 407 575 437 547 384 a Estimate the true mean skidding distance for the road with a 99​% confidence interval. b. Give a practical interpretation of the​ interval, part a. c. What conditions are required for the​ inference, part​ b, to be​ valid? Are these conditions reasonably​ satisfied? Check all correct statements below. d) Is it reasonable to assume that the population distribution is approximately​ normal? e) Is it reasonable to assume that the sample is​ random? f) Is it reasonable to assume that the sample size is much smaller than the population​ size? g) Is it reasonable to assume that the sample was selected carefully to represent the whole​ population? h) A logger working on the road claims the mean skidding distance is at least 456 meters. Do you​ agree?

State the null and alternative hypotheses. H0​: u = 3 Ha​: u > 3 Enter (STAT)+(TEST)+(1Z-TEST)+(STAT)+(Plug in U, Q,xbar ,n, and >) Do not rejectH0.There is insufficient evidence to conclude that the true mean response for all school children is greater than 3. The conclusion is still valid because the sample size is large enough that the Central Limit Theorem applies.

A health journal conducted a study to see if packaging a healthy food product like junk food would influence​ children's desire to consume the product. A fictitious brand of a healthy food product—sliced apples—was packaged to appeal to children. The researchers showed the packaging to a sample of 352 school children and asked each whether he or she was willing to eat the product. Willingness to eat was measured on a​ 5-point scale, with 1=​"not willing at​ all" and 5=​"very ​willing." The data are summarized as x=3.24 and s=2.47. Suppose the researchers knew that the mean willingness to eat an actual brand of sliced apples​ (which is not packaged for​ children) is μ=3.Complete parts a and b below. a. Conduct a test to determine whether the true mean willingness to eat the brand of sliced apples packaged for children exceeded 3. Use α=0.01 to make your conclusion. State the null and alternative hypotheses b. The data​ (willingness to eat​ values) are not normally distributed. How does this impact​ (if at​ all) the validity of your conclusion in part a​?

There are 13 identical trials. For each​ trial, there are two possible​ outcome(s). The probability of each possible outcome is the same for each​ trial, and the trials are all independent. b. p=nothing ​(Round to four decimal places as​ needed.) p = .67 *.69 take both percentages and multiply Snd+ VARS+(BINOMIALPDF) +input (n, probity, x )

A hotel guest satisfaction study revealed that 69​% of hotel guests were aware of the​ hotel's "green" conservation program. Among these​ guests, 67​% actually participate in the program by reusing towels and bed linens. In a random sample of 13 hotel​ guests, consider the number​ (x) of guests who are aware and participate in the​ hotel's conservation efforts. a. Explain why x is​ (approximately) a binomial random variable. b. Use the rules of probability to determine the value of p for this binomial experiment. c. Assume p=0.47. Find the probability that at least 8 of the 13 hotel guests are aware of and participate in the​ hotel's conservation efforts.

No​, because the population​mean, μM​, is within the confidence interval. H0: μM=58,000 Ha​: μM≠58,000 Enter (Stat) +(Test) +(1Z-Test)+(Stat)+(put in u0, xbar, stdevU > ) z<−1.96 or z>1.96 = .95 or .05 Do not reject H0. There is insufficient evidence to conclude that the true mean salary of males with postgraduate degrees differs from ​$58,000. c. Do the inferences in parts a and b​agree? Explain why or why not. The inferences agree because the confidence expressed in the confidence interval is being expressed as the error rate for the hypothesis test. yes​, because the population​mean, μF​, is not within the confidence interval. H0​: μF=38,000 Ha​: μF≠38,000 Enter (Stat) +(Test) +(1Z-Test)+(Stat)+(put in u0, xbar, stdevU not equal ) A. z<−1.96 or z>1.96 Reject H0. There is sufficient evidence to conclude that the true mean salary of females with postgraduate degrees differs from ​$38,000. The inferences agree because the confidence expressed in the confidence interval is being expressed as the error rate for the hypothesis test.

A magazine published a paper on the relationship between education level and earnings. The data for the research were obtained from a national survey of over 36,000 respondents. The survey revealed that males with a postgraduate degree had a mean salary of ​$61,284 (with standard error sx=​$2,195​), while females with a postgraduate degree had a mean of ​$32,236 (with standard error sx=​$925​). Complete parts a through f. a. The article reports that a 95​% confidence interval for μM​, the population mean salary of all males with postgraduate​degrees, is ​($56,982​,$65,586​). Based on this​interval, is there evidence to say that μMdiffers from ​$58,000​? Explain. b. Use the summary information to test the hypothesis that the true mean salary of males with postgraduate degrees differs from $58,000. Use α=0.05. (Note: sx=s/n​.) d. The article reports that a 95​% confidence interval for μF​, the population mean salary of all females with postgraduate​degrees, is ​($30,423​, $34,049​). Based on this​interval, is there evidence to say that μF differs from $38,000​? Explain. f. Do the inferences in parts d and e​agree? Explain why or why not.

enter (Stat) + (test) +(8Tinterrval)+(STAT)+(plug in CL%, Xbar, SXdev) The new formula is better than the​ handbook's formula because μ is greater than the confidence interval's upper limit.

A new formula for estimating the water evaporation from occupied swimming pools was proposed and analyzed in an engineering journal. The key components of the new formula are number of pool​ occupants, area of​ pool's water​ surface, and the density difference between room air temperature and the air at the​ pool's surface. Data were collected from a wide range of pools for which the evaporation level was known. The new formula was applied to each pool in the​ sample, yielding an estimated evaporation level. The absolute value of the deviation between the actual and estimated evaporation level was then recorded as a percentage. The researchers reported the following summary statistics for absolute deviation​ percentage: x=15.3​,s=19.Assume that the sample contained n=13 swimming pools. Answer parts a and b below. b. The handbook of a certain engineering society also provides a formula for estimating pool evaporation. Suppose the​ handbook's mean absolute deviation percentage is μ=34%. On​ average, is the new formula better than the​ handbook's formula? Explain.

have to find x 1000*.65 = 650 1000*.17 = 170 ENTER+(STAT)+(TEST) +(A) 1-propZint) +(enter x,n,cl%) b. If the true percentage of adults who would choose to sleep when they are at home sick is 73​%, would you be​ surprised? Yes

A newspaper reported on the results of an opinion poll in which adults were asked what one thing they are most likely to do when they are home sick with a cold or the flu. In the​ survey, 65​% said that they are most likely to sleep and 17​% said that they would watch television. Although the sample size was not​ reported, typically opinion polls include approximately​ 1,000 randomly selected respondents. a. Assuming a sample size of​ 1,000 for this​ poll, construct a 95​% confidence interval for the true percentage of all adults who would choose to sleep when they are at home sick.

a. Identify the population of interest. Choose the correct answer below. adults b. Identify the sample. Choose the correct answer below. 100 adults Identify the parameter of interest. Choose the correct answer below. p, the population proportion of adults who say that some coffee shops are overpriced ENTER+(STAT)+(TEST) +(A) 1-propZint) +(enter x,n,cl%) We are 95​% confident that the parameter of interest lies in the confidence interval

A newspaper reported that 20​% of people say that some coffee shops are overpriced. The source of this information was a telephone survey of 100 adults. a. Identify the population of interest in this study. b. Identify the sample for the study. c. Identify the parameter of interest in the study. d. Find and interpret a 95​% confidence interval for the parameter of interest. a. Identify the population of interest. Choose the correct answer below.

Adults not working during summer vacation B. The experiment consists of identical​ trials, there are only two possible outcomes on each trial​ (works or does not​ work), and the trials are independent p = the percent in the beginning of question. = .25 Snd+ VARS+(BINOMIALPDF) +input (n, probity, x )

A poll found that 25​% of adults do not work at all while on summer vacation. In a random sample of 9 adults, let x represent the number who do not work during summer vacation. Complete parts a through e.

ALL ANSWERS ARE ON POP UP The proportion would be . 50 H0: p=0.50 vs. Ha: p≠0.50 because its two tailed The​ p-value is 0.000. (on pop up) Reject the null​hypothesis, because the​p-value is less than α. There is sufficient evidence to conclude that color and flavor are related.

A professor at a local university designed an experiment to see if someone could identify the color of a candy based on taste alone. Students were blindfolded and then given a​ red-colored or​ yellow-colored candy to chew.​ (Half the students were assigned to receive the red candy and half to receive the yellow candy. The students could not see what color candy they were​ given.) After​ chewing, the students were asked to guess the color of the candy based on the flavor. Of the 124 students who participated in the​ study, 96 correctly identified the color of the candy. The results are shown in the accompanying technology printout. Complete parts a through c below.

H0​: μ = 2.3 ​Ha: μ ≠ 2.3 Enter (STAT)+(TEST)+(1Z-TEST)+(STAT)+(Plug in U, Q,xbar ,n, and not =) reject H0. There is sufficient evidence to conclude that the true mean facial WHR is not 2.3.

A psychology student conducted a study on using a chief executive​ officer's facial structure to predict a​ firm's financial performance. The facial​ width-to-height ratio​ (WHR) for each in a sample of 59 CEOs at publicly traded firms was determined. The sample resulted x=1.58 and s=2.14.The student wants to predict the financial performance of a firm based on the value of the true mean facial WHR of CEOs. The student wants to use the value of μ=2.3.Do you recommend he use this​ value? Conduct a test of hypothesis for μ to help you answer the question. Specify all the elements of the​ test, including H0​, Ha​, test​ statistic, p-value, and your conclusion. Test at α=0.05. What is the value of the test​ statistic?

ux = mean Q = Q/sqr n The shape is that of a normal distribution and depends on the sample size. c. Calculate the standard normal​ z-score corresponding to a value of x=21.8. z= 21.8- 22 /Q/Sqtn ​(Type an integer or a​ decimal.

A random sample of n=100 observations is drawn from a population with a mean equal to 22 and a standard deviation equal to 20. Complete parts a through g below.

a) 100-90 = 10 10/2 = 5 5/100 = 0.05 Use calculator for z score (2nd)+(VARS)+(3InvNorm)+(.percentage) (2nd)+(VARS)+(3InvNorm)+(0.05) =1.64 Use Q/Sqrt(n) * Z Score + xbar (upper Bound) 8.5/Sqrt(100) * 1.64 + 81.2 (upper Bound) =82.59 Use 8.5/Sqrt(100) * 1.64 -81.2 (lower bound) = 79.81 Use absolute value no negative b) A confidence coefficient of 0.90 means that there is a probability of 0.90 that an interval estimator constructed using this coefficient will enclose the population parameter. c) Use calculator for z score (2nd)+(VARS)+(3InvNorm)+(.percentage) (2nd)+(VARS)+(3InvNorm)+(0.025) =1.96 Use Q/Sqrt(n) * Z Score + xbar (upper Bound) 8.5/Sqrt(100) * 1.96 + 81.2 (upper Bound) =82.87 Use 8.5/Sqrt(100) * 1.96 -81.2 (lower bound) = 79.53 Use absolute value no negative d) an increase , critical value. increase. e) ​Yes, since the sample sizes are large (n≥​30) and randomly selected from the target​ population, the condition guarantees that the sampling distribution of x is approximately normal.

A random sample of 100 observations from a normally distributed population possesses a mean equal to 81.2 and a standard deviation equal to 8.5. Use this information to complete parts a through e below. a. Find a 90​% confidence interval for μ. b. What do you mean when you say that a confidence coefficient is 0.90​? c. Find a 95​% confidence interval for μ. d. What happens to the width of a confidence interval as the value of the confidence coefficient is increased while the sample size is held​ fixed? Increasing the confidence coefficient while keeping the sample size fixed will cause an increase in the critical value. This means that the width of the confidence interval will increase. e. Would your confidence intervals of parts a and c be valid if the distribution of the original population were not​ normal? Explain.

( = not equal too ) Enter (STAT)+(TEST)+(1Z-TEST)+(STAT)+(Plug in U, Q,xbar ,n, and not =) There is sufficient evidence to reject H0 for α>0.11. Do not reject H0. There is insufficient evidence to indicate that the true population mean μ is not equal to 100 at α=0.05 The results differ because the alternative hypothesis in part a is more specific than the one in part b

A random sample of 100 observations from a population with standard deviation 43 yielded a sample mean of 108. Complete parts a through c below. a. Test the null hypothesis that μ=100 against the alternative hypothesis that μ>​100, using α=0.05. Interpret the results of the test. What is the value of the test​ statistic?

a) Enter (Stat) +(Test) +(1Z-Test)+(Stat)+(put in u0, xbar, stdev not > uo) z=1.77 p-value=.038 Reject H0. There is sufficient evidence to indicate that the true population mean is greater than 100 at α=0.05.(>.038 p value) b) Enter (Stat) +(Test) +(1Z-Test)+(Stat)+(put in u0, xbar, stdev not equal uo) z=1.77 p-value=0.076 Do not reject H0. There is insufficient evidence to indicate that the true population mean μ is not equal to 100 at α=0.05. (<0.076) A. The results differ because the alternative hypothesis in part a is more specific than the one in part b.

A random sample of 100 observations from a population with standard deviation 62 yielded a sample mean of 111. Complete parts a through c below.

USE list Then use 51propztest then invnorm

A random sample of 50consumers​ taste-tested a new snack food. Their responses​ (where 0=do not​ like; 1=​like; 2=​indifferent) are reproduced below. Complete parts a and b below.

find Square root of standard dev. Enter (Stat) +(Test) +(1Z-Test)+(Stat)+(put in u0, xbar, stdev <uo) Reject H0. There is sufficient evidence to indicate that the true population mean is less than 0.43 at α=0.10. Enter (Stat) +(Test) +(1Z-Test)+(Stat)+(put in u0, xbar, stdev not equal uo) Do not reject H0. There is insufficient evidence to indicate that the true population mean μ is not equal to 100 at α=0.05. (<0.076)

A random sample of 68 produced the summary statistics x=0.391 and s^2=0.045. Complete parts a and b.

100-95 =5 100-90 =10 100-99 = 1 5/2 =2.5 10/2 =5 1/2 = .5 =2.5/100 = 0.025 5/100 = 0.05 .5/100 = .005 Use calculator for z score (2nd)+(VARS)+(3InvNorm)+(.percentage) (2nd)+(VARS)+(3InvNorm)+(split percent) =1.96 1.64 = 2.58 Use Q/Sqrt(n) * Z Score + xbar (upper Bound) Use Q/Sqrt(n) * Z Score -xbar (lower bound) use absolute value * keep changing z score

A random sample of 89 observations produced a mean x=25.7 and a standard deviation s=2.7. a. Find a​ 95% confidence interval for μ. b. Find a​ 90% confidence interval for μ. c. Find a​ 99% confidence interval for μ.

100-(percentage) = Alpha Alpha / 2 = divided alpha 100-95 = 5 5/2 = 2.5 2.5/100= .025 Use calculator for z score (2nd)+(VARS)+(3InvNorm)+(.percentage) (2nd)+(VARS)+(3InvNorm)+(.025) Use Q/Sqrt(n) * Z Score + xbar (upper Bound) 30/Sqrt(60) * 1.28 + 32 (upper Bound) = 36.96 Use Q/Sqrt(n) * Z Score -xbar (lower bound) 30/Sqrt(60) * 1.28 + 32 (upper Bound) No, since the sample sizes are large (n≥​30) and randomly selected from the target​ population, the condition guarantees that the sampling distribution of x is approximately normal.

A random sample of n measurements was selected from a population with unknown mean μ and standard deviation σ=30 for each of the situations in parts a through d. Calculate a 90​% confidence interval for μ for each of these situations. a. n=60​, xbar=32 b. n=250​, xbar=104 c. n=120​, xbar=17 d. n=120​, xbar=5.29 e. Is the assumption that the underlying population of measurements is normally distributed necessary to ensure the validity of the confidence intervals in parts a through​ d? Explain.

Steps convert n = df by reducing by 1 Step 2 if its two tailed divide a by 2 then use student t scale. no equal = two tailed >left tailed t is greater t> <right tailed t is less that a -t< (DONT FORGET NEGATIVE SIGN)

A random sample of n observations is selected from a normal population to test the null hypothesis that μ=10. Specify the rejection region for each of the following combinations of Ha​, α​, and n. a. Ha​: μ≠​10; α=0.01​; n=14 b. Ha​: μ>​10; α=0.05​; n=23 c. Ha​: μ>​10; α=0.10​; n=9 d. Ha​: μ<​10; α=0.05​; n=13 e. Ha​: μ≠​10; α =0.10​; n=19 f. Ha​: μ<​10; α=0.01​; n=5

step 1 ) p* n = x (.70*100) = x=70 ENTER+(STAT)+(TEST) +(A) 1-propZint) +(enter x,n,cl%) We are 99​% confident that p lies in the confidence interval. ​99% of similarly constructed intervals would contain the true value of the parameter.

A random sample of size n=100 yielded p=0.70. a. Is the sample size large enough to use the large sample approximation to construct a confidence interval for​ p? Explain. b. Construct a 99​% confidence interval for p. c. Interpret the 99​% confidence interval. d. Explain what is meant by the phrase ​"99​% confidence​ interval."

H0​: μ=425 Ha​: μ<425 Enter (STAT)+(TEST)+(2T-TEST)+(STAT)+(Plug in U, Q, xbar ,n, and < Reject H0. There is sufficient evidence at the α=0.05 level of significance to conclude that the true mean skidding distance is less than 425 meters.​Thus, there is enough evidence to refute the claim.

A recent study investigated tractor skidding distances along a road in a forest. The skidding distances​ (in meters) were measured at 20 randomly selected road sites. The data are given in the accompanying table. A logger working on the road claims that the mean skidding distance is at least 425 meters. Is there sufficient evidence to refute this​ claim? Use α=0.05. Click the icon to view the table.

a) just plug in xbar = u = 14.7 b) 100-(percentage) = Alpha Alpha / 2 = divided alpha 100-95 = 5 5/2 = 2.5 2.5/100= .025 Use calculator for z score (2nd)+(VARS)+(3InvNorm)+(.percentage) (2nd)+(VARS)+(3InvNorm)+(.025) = 1.96 Use Q/Sqrt(n) * Z Score + xbar (upper Bound) 10.4/Sqrt(2816) * 1.96 + 14.7 (upper Bound) = 15.08 Use 10.4/Sqrt(2816) * 1.96 - 14.7 (lower bound) = 14.32 Use absolute value (no negative) c) The statement is incorrect. A correct statement would be​ "One can be​ 95% confident that the true mean number of people named per person will fall in the interval computed in part b.​" d) It does not impact the validity of the interpretation because the sampling space of the sample mean is approximately normal according to the Central Limit Theorem.

A research program used a representative random sample of men and women to gauge the size of the personal network of older adults. Each adult in the sample was asked to​ "please name the people you have frequent contact with and who are also important to​ you." The responses of 2,816 adults in this sample yielded statistics on network​ size, that​ is, the mean number of people named per person was x=14.7​, with a standard deviation of s=10.4. Complete parts a through d. a. Give a point estimate for μ. b. Give an interval estimate for μ. Use a confidence coefficient of 0.95. c. Comment on the validity of the following​statement: "95% of the​time, the true mean number of people named per person will fall in the interval computed in part b​." Choose the correct answer below. d. It is unlikely that the personal network sizes of adults are normally distributed. In​ fact, it is likely that the distribution is highly skewed. If​ so, what​ impact, if​ any, does this have on the validity of inferences derived from the confidence​ interval?

Steps convert n = df by reducing by 1 Step 2 if its two tailed divide a by 2 then use student t scale. no equal = two tailed >left tailed t is greater t> <right tailed t is less that a -t< (DONT FORGET NEGATIVE SIGN) find t enter (Stat) + (test) +(8Tinterrval)+(STAT)+(plug in CL%, Xbar, SXdev) Do not reject H0. There is insufficient evidence at the α=0.05 level of significance to conclude that the true mean of the population is less than 6 get df from original n

A sample of five​ measurements, randomly selected from a normally distributed​ population, resulted in the following summary​ statistics: x=4.8​, s=1.3. Complete parts a through c. a. Test the null hypothesis that the mean of the population is 6 against the alternative​ hypothesis, μ<6. Use α=0.05. If α=0.05​, find the rejection region for the test. Choose the correct answer below.

​H0: μ=69 Ha​: μ≠69 b. For this​ problem, what is a Type I​ error? A Type II​ error? What is a Type I error in this​ problem? A Type I error would be to conclude that the true mean level of support for sustainability is not 69 ​when, in​ fact, the mean is equal to 69. What is a Type II error in this​ problem? A Type II error would be to conclude that the true mean level of support for sustainability is 69 ​when, in​ fact, the mean is not equal to 69. c. The provided printout of the analysis gives the results of the test. Locate the test statistic and the​ p-value on the printout. (just copy) test​ statistic: negative −1.4104 ​p-value: .1584 d. At α=0.05​, give the appropriate conclusion. Choose the correct answer below. Do not reject the null hypothesis. There is insufficient evidence at the α=0.05 level of significance to conclude that the true mean level of support for sustainability is not equal to 69 e. What​ assumptions, if​ any, about the distribution of support levels must hold true in order for the inference derived from the test to be​ valid? Explain. Select all that apply. No assumptions are necessary because the sample size is very large.

A study analyzed the sustainability behaviors of CPA corporations. The level of support for corporate sustainability​ (measured on a quantitative scale ranging from 0 to 160​ points) was obtained for each in a sample of 975 senior managers at CPA firms. The CEO of a CPA firm claims that the true mean level of support for sustainability is 69. Complete parts a through e.

H0​: μ=0 Ha​: μ>0 . z>1.645 (0.05 level) Reject the null hypothesis that the true mean rate of return of​ round-trip trades is zero if the​ p-value is less than the level of significance α=0.05. plug in numbers for p and t Reject H0. There is sufficient evidence to indicate that the true population mean is less than 0.05

A study analyzed the trading skills of investors. The study focused on​ "round-trip" trades, that​ is, trades in which the same stock was both bought and sold in the same quarter. Consider a random sample of 212 ​round-trip trades made by investors. Suppose the sample mean rate of return is 2.99​% and the sample standard deviation is 8.98​%. If the true mean rate of return of​ round-trip trades is​ positive, then the population of investors is considered to have performed successfully. Complete parts a through e below.

H0​: β1=​0, Ha​: β1≠0

A study of the effect of massage on boxing performance measured a​ boxer's blood lactate concentration​ (in mM) and perceived recovery​ (on a​ 28-point scale). On the basis of the information provided by the​ study, the data shown in the accompanying table were obtained for 16​ five-round boxing performances in which a massage was given to the boxer between rounds. Conduct a test to determine whether blood lactate level​ (y) is linearly related to perceived recovery​ (x). Use α=0.10. Click the icon to view the table of boxer blood data.

point estimate is 577/884 = .65 H0​: p=0.7 Ha​: p>0.7 Enter (STAT)+(TEST)+(51-PropZTEST) +(TYPE x,n,>) Enter (2nd)+(VARS)+(INVNORM) =alpha = rejection space. Do not reject the null hypothesis because the test statistic is not in the rejection region.​Therefore, there is insufficient evidence at the 0.05 level of significance to indicate that the true proportion of all​Internet-using adults who have paid to download music exceeds 0.7. Make the appropriate conclusion using the​ p-value. Do not reject the null hypothesis because the​ p-value is greater than α=0.05. ​Therefore, there is insufficient evidence at the 0.05 level of significance to indicate that the true proportion of all​ Internet-using adults who have paid to download music exceeds 0.7.

A study was conducted to see if people who use the Internet have also paid to download music. In a representative sample of 884 adults who use the​ Internet, 577 admitted that they have paid to download music. Let p represent the true proportion of all​ Internet-using adults who have paid to download music. Complete parts a through g below.

H0​: μ1−μ2=​0, Ha: μ1−μ2>0 ENTER+(STAT) +(EDIT) (ENTER DATA on L1+L2) ENTER+(STAT) +(4) (2-SAMPTTEST) +(POOLED) (DATA) > Reject H0. There is sufficient evidence the true mean performance level for students in the rudeness condition is lower than the true mean performance level for students in the control group.

A study was done to determine the effect of rudeness on a​ victim's task performance. Students were randomly assigned to one of two experimental​ conditions, a rudeness condition and a control group. Each student was asked to write down as many uses for a brick as possible in five minutes. The number of different uses for a brick was recorded for each of the 60 students and the data is shown in the accompanying table. Conduct a statistical analysis​ (at α=0.01​) to determine if the true mean performance level for students in the rudeness condition is lower than the true mean performance level for students in the control group. Click the icon to view the data table.

H0​:μ=0.338 Ha​:μ≠0.338 Reject H0. There is sufficient evidence at the α=0.05 level of significance to conclude that the true mean bubble rising velocity is not 0.338 m/s. Enter (STAT)+(TEST)+(2T-TEST)+(STAT)+(Plug in U, Q, xbar ,n, and not =) not generate , small

A study was performed on a method of purifying nuclear fuel waste. The process involves oxidation in molten salt and tends to produce oxygen bubbles with a rising velocity. To monitor the​ process, the researchers collected data on bubble velocity​ (measured in meters per​ second) for a random sample of 18 photographic bubble images. These data​ (simulated) are reproduced in the accompanying table. When oxygen is inserted into the molten salt at a rate​ (called the sparging​ rate) of 3.33×10−6​, the researchers discovered that the true mean bubble rising velocity is μ=0.338. Complete parts a and b.

s=0.2192 LOOK UNDER S on pop up b. Give a practical interpretation of s. Select the correct choice below and fill in the answer box to complete your choice. ​(Round to four decimal places as​ needed.) Most​ (about 95%) of the errors of prediction will fall within 0.4384 mm of the least squares line.

A team of civil engineers used regression analysis to estimate y=the ratio of repair to replacement cost of commercial pipe. The independent variable was x=the diameter​ (in millimeters) of the pipe. Use the accompanying simple linear regression printout to complete parts a and b below. Click the icon to view the simple linear regression technology printout. a. Locate the value of s on the printout.

The parameter of interest is p the true proportion of fish fillets that are actually red snapper. It is inappropriate because either the number of successes in the sample or the number of failures in the sample is less than 15. With 90​% confidence, the interval found in part c contains the true value of p.

A team of researchers analyzed the meat from each in a sample of 29 "red snapper" fish fillets purchased from vendors across a region in an effort estimate the true proportion of fillets that are really red snapper. DNA tests revealed that 6 of the 29 fillets​ (or 21​%) were not red snapper but the cheaper​ look-alike variety of fish. Complete parts a through d. a. Identify the parameter of interest to the researchers.

a) Since geologists use the combination of the first and second holes to make an​ estimation, the observations are not independent—each set of twinned holes should be treated as a pair. Deduct the amounts given ( or find the difference) to find the interval find the xbar and stdev or difference in mean. ENTER + STAT + TEST+(8: TINTERVAL) +(DATA)+(ENTER DATA 0, sdev, xbar use diffence, c-level) e) With 95​% confidence, the true mean difference in chemical compound percentages falls within the interval. f) The geologists can conclude that there is no evidence of a difference in the true means of all original holes and their twin holes drilled at the mine because 0 is contained within the interval.

A traditional method of verifying mineralization grades in mining is to drill twinned​ holes, i.e., the drilling of a new​ hole, or​ "twin," next to an earlier drillhole. Geologists use data collected at both holes to estimate the total amount of a particular chemical compound present at the drilling site. The data in the associated table represent the​ chemicals' percentages for a sample of 15 twinned holes drilled at a diamond mine. The geologists want to know if there is any evidence of a difference in the true chemical compound means of all original holes and their twin holes drilled at the mine. Complete parts a through e below. A a. Explain why the data should be analyzed as paired differences. Choose the correct answer below. e. Interpret the interval from part d. Can the geologists conclude that there is no evidence of a difference in the true means of all original holes and their twin holes drilled at the​ mine? Choose the correct interpretation below.

H0​: μ=865​, Ha​: μ<865

A university economist conducted a study of elementary school lunch menus. During the​ state-mandated testing​ period, school lunches averaged 865 calories. The economist claimed that after the testing period​ ended, the average caloric content of the school lunches dropped significantly. Set up the null and alternative hypothesis to test the​ economist's claim.

H0​: u = 882 Ha​: u > 882

A university economist conducted a study of elementary school lunch menus. During the​ state-mandated testing​ period, school lunches averaged 882 calories. The economist claimed that after the testing period​ ended, the average caloric content of the school lunches increased significantly. Set up the null and alternative hypothesis to test the​ economist's claim.

The experiment consists of n identical trials. There are only two possible outcomes on each trial. The probability of success remains the same from trial to trial. The trials are independent. B. The sample size and the percentage of that particular kind of golf ball that meets all the requirements would need to be known. 5 x 12 dozen = n p = .20 is given in question =u = p*n 0 = sqrt (pnq)

According to a certain golf​ association, the weight of the golf ball shall not be greater than 1.620 ounces​ (45.93 grams). The diameter of the ball shall not be less than 1.680 inches. The velocity of the ball shall not be greater than 250 feet per second. The golf association periodically checks the specifications of golf balls using random sampling. Five dozen of each kind are​ sampled, and if more than three do not meet size or velocity​ requirements, that kind of ball is removed from the golf​ association's approved list. Complete parts a through c below.

use excel to find mean and standard deviation. Enter (STAT)+(EDIT) (plug in numbers) Enter (STAT)+(CALC)+(1-Var Stat) (enter) H0​:μ=6.6 Ha​:μ>6.6 Enter (Stat) +(Test) +(1Z-Test)+(Stat)+(put in u0, xbar, stdevU > ) H0 is not rejected. There is insufficient evidence at the α=0.05 level of significance to conclude that the average fee this year is higher than last year c. In conducting the​ test, was it necessary to assume that the population of average​ full-service fees was normally​ distributed? Justify your answer. ​No, because the sample size is greater than 30

According to a recent survey of the 20​,000 funeral homes of a certain​ nation, funeral homes collected an average of $6,600 per​ full-service funeral last year. A random sample of 36 funeral homes reported revenue data for the current year. Among other​ measures, each reported its average fee for a​ full-service funeral. These data​ (in thousands of​ dollars) are shown in the accompanying table. Complete parts a through c below

xbar = 18,289 /5,000 = 3.6578 n=5000 sxdev = 1.4 enter (Stat) + (test) +(8Tinterrval)+(STAT)+(plug in CL%, Xbar, SXdev)

According to​ scientists, the cockroach has had 300 million years to develop a resistance to destruction. In a study conducted by​ researchers, 5,000 roaches​ (the expected number in a​ roach-infested house) were released in the test kitchen. One week​ later, the kitchen was fumigated and 18,289 dead roaches were​ counted, a gain of 13,289 roaches for the​ 1-week period. Assume that none of the original roaches died during the​ 1-week period and that the standard deviation of​ x, the number of roaches produced per roach in a​ 1-week period, is 1.4. Use the number of roaches produced by the sample of 5,000 roaches to find a 95​% confidence interval for the mean number of roaches produced per week for each roach in a typical​ roach-infested house. Find a 95​% confidence interval for the mean number of roaches produced per week for each roach in a typical​ roach-infested house.

ENTER+(STAT)+(TEST) +(A) 1-propZint) +(enter x,n,cl%)] four decimals

An accounting firm annually monitors a certain mailing​ service's performance. One parameter of interest is the percentage of mail delivered on time. In a sample of 311,000 items mailed between Dec. 10 and Mar. 3—the most difficult delivery season due to bad weather and holidays—the accounting firm determined that 262,100 items were delivered on time. Use this information to make a statement about the likelihood of an item being delivered on time by that mailing service

a. The null hypothesis is H0​: There is no intrusion b. The alternative hypothesis is Ha​: There is an intrusion 8/800 = a 0.01 2000 - 500 = b 1,500 1,500/2000 = .75

An intrusion detection system​ (IDS) is designed to provide an alarm whenever unauthorized access to a computer system occurs. The probability of the system giving a false alarm is defined by the symbol α​, while the probability of a missed detection is defined by the symbol β. These symbols are used to represent Type I and Type II error​ rates, respectively, in a hypothesis testing scenario. a. What is the null​ hypothesis, H0​? b. What is the alternative​ hypothesis, Ha​? c. According to a college research​ laboratory, only 8 in 8,000 computer sessions with no intrusions resulted in a false alarm. For the same​ system, the laboratory found that only 500 of 2,000 intrusions were actually detected. Use this information to estimate the values of α and β.

a) u ( looking for the mean) b) enter (Stat) +(1edit)+(plug in numbers) enter (Stat) + (test) +(8Tinterrval)+(data)+(plug in CL%) plug in mean ( 89.29) c) The sample is small and the trap spacing population has unknown distribution and standard deviation d) enter (Stat) + (test) +(8Tinterrval)+(data)+(change CL%) (78., 100.5) e) One can be​ 95% confident the true mean trap spacing distance lies within the above interval. f) The sample is randomly selected from the population. The population has a relative frequency distribution that is approximately normal

An observational study of teams fishing for the red spiny lobster in a certain body of water was conducted and the results published in a science magazine. One of the variables of interest was the average distance separating traps−called "trap spacing"−deployed by the same team of fishermen. Trap spacing measurements​ (in meters) for a sample of seven teams of fishermen are shown in the accompanying table. Of interest is the mean trap spacing for the population of red spiny lobster fishermen fishing in this body of water. Complete parts a through f below. a. Identify the target parameter for this study. b. Compute a point estimate of the target parameter. c. What is the problem with using the normal​ (z) statistic to find a confidence interval for the target​ parameter? d. Find a​ 95% confidence interval for the target parameter e. Give a practical interpretation of the​ interval, part d f. What conditions must be satisfied for the​ interval, part d​, to be​ valid? Select all that apply.

H0​: p1−p2=0 Ha​: p1−p2>0 ENTER+STAT+6:2-PROPZTEST+ENTER INFO for probability 0.000 Reject H0. There is sufficient evidence to conclude that the proportion of MBA students from Group A who reported their employment status after graduation as​self-employed is significantly greater than the proportion of MBA students from Group B who reported their employment status after graduation as​self-employe

Are MBA students from Group A more likely to begin their careers as entrepreneurs than MBA students from Group​ B? This was a question of interest to a certain admission council who published the results of a survey of MBA alumni. Of the 1,327 students from Group A who responded to the​ survey, 209 reported their employment status after graduation as​ self-employed or a small business owner. Of the 7,122 students from Group B who responded to the​ survey, 368 reported their employment status after graduation as​ self-employed or a small business owner. Use this information to answer the research​ question, where α=0.05.

H0​: μ =​$2,300 Ha​: μ <​$2,300 The probability that the null hypothesis is rejected when the average gain is $2,300 is 0.05. z < -1.645

At many golf​ clubs, a teaching professional provides a free​ 10-minute lesson to new customers. A golf magazine reports that golf facilities that provide these free lessons​ gain, on​ average, ​$2,300 in green​ fees, lessons, or equipment expenditures. A teaching professional believes that the average gain is less than $2,300. Complete parts a through c below. b. Suppose you select α=0.05. Interpret this value in the words of the problem. c. For α=0.05​, specify the rejection region of a​ large-sample test. Choose the correct answer below.

type one error =0.01 H0​: p=0.50 Ha​: p>0.50 Do not reject H0. There is insufficient evidence to conclude that more than half of all Cola A drinkers selected Cola B in the blind taste test. Cola A is relieved because there is insufficient evidence to conclude that more than half of their customers prefer Cola B.

Cola A drinkers participated in a blind taste test where they were asked to taste unmarked cups of Cola B and Cola A and were asked to select their favorite. Suppose 118 Cola A drinkers took the test and 63 preferred the taste of Cola B. Determine if more than half of all Cola A drinkers selected Cola B in the blind taste test. Select α to minimize the probability of a Type I error. What were the consequences of the test results from Cola​ A's perspective? Determine the value of α that would minimize the probability of a Type I error. Choose the correct answer below.

Snd+ VARS+(BINOMIALPDF) +input (n, probity, x ) First find​ p(0). Substitute 0 for x. First find​ p(0). Substitute 0 for x. p(x) = 5x (0.5)x(0.5)5−x p(0) = (5/0) (0.5)^0 (0.5)^5−0 Calculate 50 . Recall that ​0!=1.

Consider the following binomial probability distribution. ​p(x)=(5/x) (0.48)x(0.52)5−x ​(x=​0, 1, 2,​ ..., 5​)

As x ​increases, y tends to increase.​ Thus, there appears to be a​ positive, linear relationship between x and y. c. Find the least squares estimates of β0 and β1. ENTER+STAT+EDIT+ENTER DATA ENTER +STAT+ CALC +4: LinREG(ax+b)+ENTER The line appears to fit the data quite well because the variation of the data points around the line is not very large.

Consider the following pairs of observations. Complete parts a through f below Use the method of least squares to fit a straight line to the six data points.

a) Since the sample size is sufficiently​ large, the sampling distribution is approximately normal b) Enter (Stat) +(Test) +(1Z-Test)+(Stat)+(put in u0, xbar, stdev not equal uo) d. Find the rejection region of the test for α=0.01 z>2.326 =0.01 Reject H0. There is sufficient evidence at the α=0.01 level of significance to conclude that μ>70. Reject H0.There is sufficient evidence at the α=0.01 level of significance to conclude that μ>70. Yes, because both conclusions reject H0.

Consider the test H0​: μ=70 versus Ha​: μ>70 using a large sample of size n=400. Assume σ=17. Complete parts a through g. a. Describe the sampling distribution of x. b. Find the value of the test statistic if x=72.2.

a) H0​: μd=0 Ha​: μd>0 b) Use a paired difference test because the assumption of independent samples is invalid. c) ENTER + STAT + TEST+(4:2-SAMPTTEST) +(STATS)+(ENTER DATA) +(NO POOLED) d) ENTER + STAT + TEST+(1: Z-TEST) +(STAT)+(ENTER DATA 0, sdev, xbar) e) The test statistic using the formula for paired difference test f) Reject H0 because the​ p-value is less than the significance level. There is sufficient evidence that the mean BMI at the end of camp is less than the mean BMI at the start of camp. g) No; the sample size is large enough that the Central Limit Theorem applies. ENTER + STAT + TEST+(7: ZINTERVAL) +(STAT)+(ENTER DATA 0, sdev, xbar use diffence, c-level) The true mean difference between the BMI at the start of camp and the BMI at the end of camp falls inside the limits of the​ interval, with 99​% confidence. Since 0 is not included in the​ interval, there is sufficient evidence that the mean BMI at the end of camp is less than the mean BMI at the start of camp.

Counselors at a summer camp for overweight and obese adolescents develop a​ weight-management program for each camper that centers on nutrition education and physical activity. In a recent​ study, the body mass index​ (BMI) was measured for each of 77 campers both at the start and end of camp. Summary statistics on BMI measurements are shown in the table to the right. Complete parts a through h. a. Give the null and alternative hypotheses for determining whether the mean BMI at the end of camp is less than the mean BMI at the start of camp. Let μ1 be the population mean BMI at the start of​ camp, μ2 be the population mean BMI at the end of​ camp, and μd=μ1−μ2. b. How should the data be​ analyzed, as an independent samples test or as a paired difference​ test? Explain. Choose the correct answer below. c. Calculate the test statistic using the formula for an independent samples test.​ (This may or may not be the correct method from part​ b.) d. Calculate the test statistic using the formula for paired difference test. ​ (This may or may not be the correct method from part​ b.) e. Compare the test​ statistics, parts c and d. Which test statistic provides more evidence in support of the alternative​ hypothesis? f. The​ p-value of the​ test, part d​, was reported as p<0.0001. Interpret this result assuming α=0.01. g. Do the differences in BMI values need to be normally distributed in order for the​ inference, part f​, to be​ valid? Explain. Choose the correct answer below. h. Find a​ 99% confidence interval for the true mean change in BMI for the campers.

H0​: μ=9 Ha​: μ< 9 A Type I error would be to conclude that the true mean number of solder joints inspected is less than 9 when, in​ fact, the mean is equal to 9. A Type II error would be to conclude that the true mean number of solder joints inspected is 9 when, in​ fact, the mean is less than 9. Do not reject H0. There is insufficient evidence at the α=0.05 level of significance to conclude that the true mean number of solder joints inspected is less than 9. (because p-value is greater than half of alpha)

Current technology uses​ high-resolution X-rays and lasers for inspection of​ solder-joint defects on printed circuit boards​ (PCBs). A particular manufacturer of​ laser-based inspection equipment claims that its product can inspect on average at least 9 solder joints per second when the joints are spaced 0.1 inch apart. The equipment was tested by a potential buyer on 48 different PCBs. In each​ case, the equipment was operated for exactly 1 second. The numbers of solder joints inspected on each run are shown in the accompanying data table. Complete parts a through c. b. In the context of this​ exercise, what is a Type I​ error? A Type II​ error? First identify what a Type I error is for this situation. Choose the correct answer below.

a) just plug in mean = u = 7.64 b) 100-90 = 10 10/2 = 5 5/100 = 0.05 Use calculator for z score (2nd)+(VARS)+(3InvNorm)+(.percentage) (2nd)+(VARS)+(3InvNorm)+(0.05) =1.64 Use Q/Sqrt(n) * Z Score + xbar (upper Bound) 8.95/Sqrt(64) * 1.64 + 7.64 (upper Bound) = 9.47 Use 8.95/Sqrt(64) * 1.64 - 7.64 (lower bound) = 5.81 Use absolute value (no negative) c) We are​ 90% confident that the mean sentence complexity score of all​ low-income children is between the endpoints of the confidence interval d) Yes

Each child in a sample of 64 low-income children was administered a language and communication exam. The sentence complexity scores had a mean of 7.64 and a standard deviation of 8.95. Complete parts a through d. a. From the​ sample, estimate the true mean sentence complexity score of all​ low-income children. b. Form a​ 90% confidence interval for the​estimate, part a c. Give a practical interpretation of the​ interval, part b. d. Suppose the true mean sentence complexity score of​middle-income children is known to be 15.55. Is there evidence that the true mean for​low-income children differs from​ 15.55?

Sample variations in the numbers of yards to the opposing goal line explain 16​% of the sample variation in the numbers of points scored using the least squares line.

Each week coaches in a certain football league face a decision during the game. On​ fourth-down, should the team punt the ball or go for a​ first-down? To aid in the​ decision-making process, statisticians at a particular university developed a regression model for predicting the number of points scored​ (y) by a team that has a​ first-down with a given number of yards​ (x) from the opposing goal line. One of the models fit to data collected on five league teams from a recent season was the simple linear regression​ model, E(y)=β0+β1x.The regression yielded the following​ results: y=4.66−0.46x​, r2=0.16. Complete parts a and b below.

ENTER+(STAT)+(TEST) +(A) 1-propZint) +(enter x,n,cl%) One can be​ 90% confident the true proportion of ice melt ponds in the region with​ first-year ice is within the above​ interval, though it is probably not 19​%

Environmental engineers are using data collected by weather data centers to learn how climate affects the sea ice. Of 502 ice melt ponds studied in a certain​ region, 94 were classified as having​ "first-year ice". The researchers estimated that about 19​% of melt ponds in the region have​ first-year ice.​ Estimate, with​ 90% confidence, the percentage of all​ ice-melt ponds in the region that have​ first-year ice. Give a practical interpretation of the results. Construct a​ 90% confidence interval around the sample proportion of ice melt ponds with​ first-year ice.

u = 500*.60 o = 500*(.60) * (1-.60) o = 500 * (0.60) * (0.40) o = squaroot (answer) No. Fewer than half of the food items observed lies outside the interval μ±2σ.

Every​ quarter, the Food and Drug Administration​ (FDA) produces a report called the Total Diet Study. The​ FDA's report covers a variety of food​ items, each of which is analyzed for potentially harmful chemical compounds. A Total Diet Study reported that no pesticides at all were found in 60​% of the domestically produced food samples. Consider a sample of 500 food items analyzed for the presence of pesticides. Complete parts a and b.

A single number calculated from the sample that estimates a target population parameter is called a point estimator. An interval estimator is a range of numbers that contain the target parameter with a high degree of confidence.

Explain the difference between an interval estimator and a point estimator for μ.

a) There are no​ differences, as both large and small samples will have normal sampling distributions. b) Large samples will have a normal sampling distribution due to the Central Limit​ Theorem, but the sampling distribution of small samples is unknown.

Explain the differences in the sampling distributions of x for large and small samples under the following assumptions. Complete parts a and b. b. Nothing is known about the distribution of the variable x. Choose the correct answer below.

The statement reflects the confidence in the estimation process rather than in the particular interval that is calculated from the sample data. It explains that over many repetitions of this application using the same​ procedure, 95% of the resulting intervals will contain μ.

Explain what is meant by the​ statement, "We are​ 95% confident that an interval estimate contains μ​."

Set up H0 and Ha for testing whether the true population mean intention score exceeds 2. Choose the correct answer below. H0​:μ=2 Ha​:μ>2 Enter (Stat) +(Test) +(1Z-Test)+(Stat)+(put in u0, xbar, stdev <uo) Calculate the value of the test statistic z. z=1.28 p-value=0.101 do not reject H0 There is insufficient evidence at the α=0.05 level of significance to conclude that the true population mean intention score exceeds 2

For a study of unethical corporate​ conduct, a sample of 87 accounting graduates was asked to perform an unethical task​ (to bribe a​ customer), and each​ subject's intention to comply with the unethical request score was measured.​ [Scores ranged from −1.5 ​(intention to resist the unethical​ request) to 2.5​ (intention to comply with the unethical​ request.] Summary statistics on the 87 scores show x=2.36 and s=2.63. One​ researcher, of the opinion that subjects will tend to perform the unethical​ task, believes the population mean intention score μ will exceed 2. Do the data support this​ belief? Test using α=0.05

H0​: p=0.5 Ha​: p≠0.5 n =149 x=88 ENTER+(STAT)+(TEST) +(A) 1-propZint) +(enter x,n,cl%)] Reject H0. There is sufficient evidence at the α=0.10 level of significance to conclude that the proportion of students who select a​three-grill display that is consistent with the compromise theory is not equal to 0.5.

For an experiment on influencing the choices of others by offering undesirable​ alternatives, each of 149 college students selected three portable grills from five to display on a showroom floor. The students were instructed to include Grill​ #2 (a​ smaller-sized grill) and select the remaining two grills in the display to maximize purchases of Grill​ #2. If the six possible grill display combinations​ (1-2-3, 1-2-4,​ 1-2-5, 2-3-4,​ 2-3-5, and​ 2-4-5) were selected at​ random, then the proportion of students selecting any display was ​1/6=0.167. One theory tested by the researcher was that the students would tend to choose the​ three-grill display so that Grill​ #2 was a compromise between a more desirable and a less desirable grill​ (that is, display​ 1-2-3, 1-2-4, or​ 1-2-5). Of the 149 ​students, 88 selected a​ three-grill display that was consistent with this theory. Use this information to test the theory proposed by the researcher at α=0.10.

H0​: p=0.85 Ha​: p≠0.85 ENTER+(STAT)+(TEST) +(A) 1-propZint) +(enter x,n,cl%)] Do not reject H0. Based on this​ sample, there is not sufficient evidence at the α=0.05 level of significance to conclude that the percentage of dairy farms in the continent that carry out calf dehorning is not equal to 85%. Therefore, the study supports the figure reported by the organization.

For safety​ reasons, calf dehorning has become a routine practice at dairy farms. A report from a particular organization stated that 85​% of dairy farms from a certain continent carry out calf dehorning. A later​ study, found that in a sample of 609 dairy farms from a region within the​ continent, 533 dehorned calves. Does the study support or refute the figure reported by the​ organization? Explain. Use α=0.05.

a. n=497​, p0=0.05 The sample size is large enough to use the normal approximation. b. n=100​, p0=0.99 The sample size is not large enough to use the normal approximation. c. n=48​, p0=0.2 The sample size is not large enough to use the normal approximation. d. n=20​, p0=0.2 The sample size is not large enough to use the normal approximation e. n=8​, p0=0.4 The sample size is not large enough to use the normal approximation.

For the binomial sample sizes and​ null-hypothesized values of p in each​ part, determine whether the sample size is large enough to meet the required conditions for using the normal approximation to conduct a valid​ large-sample hypothesis test of the null hypothesis H0​: p=p0. Complete parts a through e.

a )The companies on the​ magazine's list of the largest 216 private companies b) enter (Stat) +(1edit)+(plug in numbers) enter (Stat) + (test) +(8Tinterrval)+(data)+(plug in CL%) (two decimal numbers ) c) With 98​% confidence, the true mean value of the 216 private companies is contained within the confidence interval. d) The population must be approximately normally distributed as a criterion for using the​ t-distribution. A skewed distribution will be poorly approximated by the​ symmetric, mound shaped​ t-distribution. e) is , contained within the confidence interval.

IPOs—initial public offerings of stock—create billions of dollars of new wealth for​ owners, managers, and employees of companies that were previously privately owned.​ Nevertheless, hundreds of large and thousands of small companies remain privately owned. The revenues of a random sample of 15 firms from a​ magazine's 216 largest private companies list are given in the table to the right. Answer parts a through e below a. Describe the population from which the random sample was drawn. b. Use a 98​% confidence interval to estimate the mean revenue of the population of companies in question. Use the unit billions of dollars for the confidence interval limits c. Interpret the confidence interval in the context of the problem. Choose the correct answer below. d. What characteristic must the population possess to ensure the appropriateness of the estimation procedure in part b​? e. Suppose the same magazine reports that the true mean revenue of the 216 companies on the list is​ $5.0 billion. Is this claim​ believable? The claim is ​believable, since the claimed value is contained within the confidence interval.

What is the conclusion for a​ p-value of 0.02​? a) Do not reject the null hypothesis since the​p-value is not less than the value of α. b. What is the conclusion for a​p-value of 0.004​? reject the null hypothesis since the​ p-value is less than the value of α.

If a hypothesis test were conducted using α=0.01​, for which of the following​ p-values would the null hypothesis be​ rejected? a. 0.02 b. 0.004

a) the null hypothesis since the​ p-value is not less than the value of α. b) Reject the null hypothesis since the​p-value is less than the value of α.

If a hypothesis test were conducted using α=0.025​, for which of the following​ p-values would the null hypothesis be​ rejected? a. 0.026 b. 0.022 What is the conclusion for a​ p-value of 0.026​? b. What is the conclusion for a​p-value of 0.022​?

u = multiple n(p) = 29*0.5 o^2 = 29*(0.5) * (1-0.5) o^2 = 29*(0.5)*(0.5) q= take the square root of q^2

If x is a binomial random​ variable, calculate μ​, σ2​, and σ for each of the following values of n and p. Complete parts a through f. a. n=29​, p=0.5

Snd+ VARS+(BINOMIALPDF) +input (n, probity, x )

If x is a binomial random​ variable, compute​ p(x) for each of the cases below. a. n=3​, x=1​, p=0.8 b. n=4​, x=2​, q=0.4 c. n=5​, x=3​, p=0.3 d. n=3​, x=0​, p=0.7 e. n=4​, x=2​, q=0.6 f. n=5​, x=1​, p=

enter (Stat) + (test) +(8Tinterrval)+(STAT)+(plug in CL%, Xbar, SXdev) a) The 95​% confidence interval for mean facial WHR indicates that the true mean facial WHR for all CEOs at publicly traded Fortune 500 firms is between 1.886 and 1.974 with 95​% confidence b) ​no ,since the suggested value of μ does not lie within the confidence interval from part​a, it is not reasonable to assume that it is the true mean facial WHR of CEOs ( its not in interval)

In a psychology​ journal, researchers reported that a chief executive​ officer's facial structure can be used to predict a​ firm's financial performance. The study involved measuring the facial​ width-to-height ratio​ (WHR) for each in a sample of 68 CEOs at publicly traded Fortune 500 firms. These WHR values​ (determined by computer analyzing a photo of the​ CEO's face) had a mean of xbar=1.93 and a standard deviation of s=0.18. Use this information to complete parts a and b below. a. Find and interpret a 99​% confidence interval for μ​, the mean facial WHR for all CEOs at publicly traded Fortune 500 firms. b. The researchers found that CEOs with wider faces​ (relative to​ height) tended to be associated with firms that had greater financial performance. They based their inference on an equation that uses facial WHR to predict financial performance. Suppose an analyst wants to predict the financial performance of a Fortune 500 firm based on the value of the true mean facial WHR of CEOs. The analyst wants to use the value of μ=2.1. Do you recommend the use this​ value?

H0​: p=0.69 Ha: p<0.69 for x (add 431+100) = 531 ENTER+(STAT)+(TEST) +(A) 1-propZint) +(enter x,n,cl%)] Do not reject the null hypothesis because the​p-value is greater than the the probability of making a Type I error.​Therefore, there is insufficient evidence to indicate that fewer than 69% of the coffee growers in Country X are either organic certified or transitioning to become organic certified.

In a representative sample of 793 coffee growers from Country​ X, 431 growers were certified to sell to organic coffee markets while 100 growers were transitioning to become organic certified. In Country​ Y, 69​% of coffee growers are organic certified. Is there evidence to indicate that fewer than 69​% of the coffee growers in Country X are either organic certified or transitioning to become organic​ certified? State your conclusion so that there is only a 5​% chance of making a Type I error.

Enter (2nd)+(VARS)+(2normalcdf)+(z score, e^99) if The probability of observing a z-value>2.18 is equal to the​ p-value, if μ=90.

In a test of H0​: μ=90 against Ha​: μ>90​, the sample data yielded the test statistic z=1.80. Find and interpret the​ p-value for the test.

this is two tailed Enter (2nd)+(VARS)+(2normalcdf)+(z score, e^99) and multiply p value by 2.

In a test of H0​:μ=100 against Ha​:μ≠​100, the sample data yielded the test statistic z=2.38. Find the p​-value for the test.

Enter (Stat) +(Test) +(1Z-Test)+(Stat)+(put in u0, xbar, stdev not equal uo) There is sufficient evidence to reject H0for α>0.11

In a test of the hypothesis H0: μ=10 versus Ha: μ≠10​, a sample of n=50 observations possessed mean x=10.6 and standard deviation s=2.6.Find and interpret the​ p-value for this test.

Enter (STAT)+(TEST)+(1Z-TEST)+(STAT)+(Plug in U,Q,xbar,n, and >) The probability​ (assuming that H0 is​ true) of observing a value of the test statistic that is at least as contradictory to the null hypothesis is 0.911.

In a test of the hypothesis H0: μ=43 versus Ha: μ>43​, a sample of n=100 observations possessed mean x=42.3 and standard deviation s=5.2. Find and interpret the​ p-value for this test.

Enter (STAT)+(TEST)+(1Z-TEST)+(STAT)+(Plug in U,Q,xbar,n, and >) The probability​ (assuming that H0 is​ true) of observing a value of the test statistic that is at least as contradictory to the null hypothesis is .889

In a test of the hypothesis H0: μ=45 versus Ha: μ>45​, a sample of n=100 observations possessed mean x=44.3 and standard deviation s=4.7. Find and interpret the​ p-value for this test.

a. Find an estimate of p1​, the true proportion of​ super-experienced bidders who fall prey to the​ winner's curse. p1= 29/185 = .157 p2= 33/138= .239 ENTER +STAT+ B:Z-PROPZINT+(ENTER INFO) There is​ 90% confidence that the difference in the proportion of​ super- and​ less-experienced bidders who fall prey to the​ winner's curse is contained in the confidence interval. THIS CAN CHANGE Since this interval does not contain 0, there is evidence to indicate that there is a difference in the proportion of​ super- and​ less-experienced bidders who fall prey to the​ winner's curse.

In auction​ bidding, the​ "winner's curse" is the phenomenon of the winning​ (or highest) bid price being above the expected value of the item being auctioned. A study was conducted to see if​ less-experienced bidders were more likely to be impacted by the curse than​ super-experienced bidders. The study showed that of the 185 bids by​ super-experienced bidders, 29 winning bids were above the​ item's expected​ value, and of the bids by the 138 less-experienced bidders, 33 winning bids were above the​ item's expected value. Complete parts a through d.

H0​: μ=0.254 Ha​: μ≠0.254 you have to plug in the numbers find mean and standard deviation. use excel z<−2.576 or z> 2.576 its two tailed you have to divide .01 by 2 and use invnorm (0.05) to get 2.576 Reject the null hypothesis. The data show sufficient evidence at α=0.01 to conclude that the process is not operating satisfactorily. In this​ problem, α is the probability that the producer will think the process is not operating​ correctly, when in fact it is operating​ fine; the producer is at risk of fixing a nonexistent​ problem, which wastes money.​ Alternatively, β is the probability that the producer will think the process is operating​ correctly, when it is​ not; the producer will do​ nothing, and the consumer is at risk of receiving defective golf tees.

In quality-control applications of hypothesis​ testing, the null and alternative hypotheses are frequently specified as H0​: The production process is performing satisfactorily and Ha​: The process is performing in an unsatisfactory manner.​ Accordingly, α is sometimes referred to as the​ producer's risk, while β is called the​ consumer's risk. An injection molder produces plastic golf tees with a mean weight of 0.254 ounce. To investigate whether the injection molder is operating​ satisfactorily, 40 tees were randomly sampled from the last​ hour's production. Their weights​ (in ounces) are listed in the accompanying table. Complete parts a through g. a. Write H0 and Ha in terms of the true mean weight of the golf​tees, μ. f. Do the data provide sufficient evidence to conclude that the process is not operating​ satisfactory? g. In the context of the​ problem, explain why it makes sense to call α the​ producer's risk and β the​ consumer's risk.

ENTER STAT +2 SAMP T TEST +POOLED (ENTER X,Q, N) ENTER +STAT +(9) 2-SAMP T INT) +Pooled (ENTER INFO) With 95​% confidence, the difference in the means of the two populations is estimated to fall inside this interval. There is insufficient evidence to indicate that μ1−μ2 differs from 0 because the interval includes 0.

Independent random samples selected from two normal populations produced the sample means and standard deviations shown to the right. a. Assuming equal​ variances, conduct the test H0: μ1−μ2=0 against Ha: μ1−μ2≠0 using α=0.05. b. Find and interpret the 95​% confidence interval for μ1−μ2.

point estimator = x/n (120/1000) =.12 The sampling distribution of p is approximately normal. ENTER+(STAT)+(TEST) +(A) 1-propZint) +(enter x,n,cl%) One can be 90​%confident that the interval contains the true value of p. No. This proportion is below the lower limit of the confidence interval.

In​ 2010, a survey of 1000 homes in a region found that 120 had overestimated market values. Suppose you want to estimate​ p, the true proportion of homes in this region with market values that are overestimated. a. Find p​, the point estimate of p. b. Describe the sampling distribution of p. c. Find a 90​% confidence interval for p. d. Give a practical interpretation of the confidence​ interval, part c. e. Suppose a researcher claims that p=0.06. Is the claim​ believable? Explain.

pt is >greater than t0 to is given follow df on vertical axis of chart. and use the df and then follow for the split alpha. on horizontal axis.

Let t0 be a specific value of t. Use the table of critical values of t below to find t0-values such that following statements are true. a. Pt≥t0=​.025, where df=10 b. Pt≥t0=​.01, where df=17 c. Pt≤t0=​.005, where df=7 d. Pt≤t0=​.05, where df=

pt is <less that t0 Click the icon to view the table of critical values of t. 100-95 = 5 5/2 = 2.5 2.5/100 =0.025 follow df on vertical axis of chart. and use the df and then follow for the split alpha. on horizontal axis.

Let t0 be a specific value of t. Use the table of critical values of t below to find t0-values such that the following statements are true. a. P−t0<t<t0=​0.95, where df=5 b. Pt≤t0=​0.01, where df=19 c. Pt≤−t0 or t≥t0=​0.10, where df=20 d. Pt≤−t0 or t≥t0=​0.01, where df= 8

Reject H0 when H0 is​ true; insufficient evidence to reject H0 when H0 is​ true; reject H0when Ha is​ true; insufficient evidence to reject H0 when Ha is true

List the four possible results of the combinations of decisions and true states of nature for a test of hypothesis.

enter (Stat) + (test) +(8Tinterrval)+(STAT)+(plug in CL%, Xbar, SXdev) (2845,4491) b) With 95​% confidence, the true mean level of radon exposure in tombs in the region is between 2845 and 4491 Bq/m3.

Many ancient tombs were cut from limestone rock that contained uranium. Since most such tombs are not​ well-ventilated, they may contain radon gas. In one​ study, the radon levels in a sample of 12 tombs in a particular region were measured in becquerels per cubic meter Bq/m3. For this​ data, assume that xbar=3,668 Bq/m3 and s=1,296 Bq/m3. Use this information to​ estimate, with 95​% confidence, the true mean level of radon exposure in tombs in the region. Interpret the resulting interval. Assume that the sampled population is approximately normal. Interpret the interval. Select the correct choice below and fill in any answer boxes to complete your choice. ​(Round to the nearest integer as needed. Use ascending​ order.)

a. Compute the redemption rate for the sample of milkshake​ m-coupons. .032 b. Compute the redemption rate for the sample of donut​ m-coupons. . 011 c. Give a point​ estimate, p1−p2​, for the difference between the true redemption rates. Use p1 as the sample proportion of milkshake store​ m-coupons that have been redeemed and p2 as the sample proportion of donut store​ m-coupons that have been redeemed. .021 (the difference between the two points) d. Form a 90​% confidence interval for the difference between the true redemption rates. Give a practical interpretation of the result. ENTER +STAT+B: 2-PROPZINT+(ENTER INFO) FOR 90% INVERVAL With 90​% confidence, the confidence interval contains the true value of the difference between the population proportions of milkshake store​ m-coupons and donut store​ m-coupons that have been redeemed. The probability that an interval estimator encloses the population parameter is 0.90. ​Yes, because the lower limit of the confidence interval is greater than 0.01

Many companies now use text messaging on cell phones as a means of marketing their products. One way to do this is to send a redeemable discount coupon​ (called an​ m-coupon) via a text. The redemption rate of ​m-coupons—the proportion of coupons redeemed—was the subject of an article in a certain marketing journal. In a​ two-year study, over​ 8,000 mall shoppers participated by signing up to receive​ m-coupons. The researchers were interested in comparing the redemption rates of​ m-coupons for different products. In a sample of 2,445 m-coupons for products sold at a milkshake​ store, 78 were​ redeemed; in a sample of 6,607 m-coupons for products sold at a donut​ store, 73 were redeemed. Complete parts a through g.

The experiment consists of n​ identical, independent​ trials, where there are only two possible​ outcomes, S​ (for Success) and F​ (for Failure). The probability of S remains the same from trial to trial. The variable x is the number of​ S's in n trials. p= just the 64% and convert to .64 u = mutiple .64 by n o mutiple (1-.63)* (.64) * n

Many primary care doctors feel overworked and burdened by potential lawsuits. In​ fact, a group of researchers reported that 64​% of all general practice physicians do not recommend medicine as a career. Let x represent the number of sampled general practice physicians who do not recommend medicine as a career. Complete parts a through d.

The regression coefficient β0 does not have a practical interpretation. The regression coefficient β1 is the estimated increase​ (or decrease) in sweetness index for each​ 1-unit increase in pectin.

One manufacturer has developed a quantitative index of the​ "sweetness" of orange juice.​ (The higher the​ index, the sweeter the​ juice). Is there a relationship between the sweetness index and a chemical measure such as the amount of​ water-soluble pectin​ (parts per​ million) in the orange​ juice? Data collected on these two variables for 24 production runs at a juice manufacturing plant are shown in the accompanying table. Suppose a manufacturer wants to use simple linear regression to predict the sweetness​ (y) from the amount of pectin​ (x).

a) enter (Stat) +(1edit)+(plug in numbers) enter (Stat) + (test) +(8Tinterrval)+(data)+(plug in CL%) (.242, .281) (three decimal places) b) With 95​% confidence, the true mean bubble rising velocity is between .242 and . 281. (plug in numbers) c) No. The discovered mean velocity lies well outside the confidence interval calculated in part a.​ Thus, it is very unlikely that these data were generated using this sparging rate. ( its outside the interval)

One method of purification of molten salt involves oxidation. An important aspect of the purification process is the rising velocity of oxygen bubbles in the molten salt. An experiment was conducted in which oxygen was inserted​ (at a designated sparging​ rate) into molten salt and photographic images of the bubbles taken. A random sample of 25 images yielded the data on bubble velocity​ (measured in meters per​ second) shown in the accompanying table. Complete parts a and b below. a. Use technology to find a 95​% confidence interval for the mean bubble rising velocity of the population. Interpret the result. The confidence interval Interpret the result. Select the correct choice below and fill in any answer boxes to complete your choice. ​(Round to three decimal places as needed. Use ascending​ order.) b. The researchers discovered that the mean bubble rising velocity is μ=0.338 when the sparging rate of oxygen is 3.32×10^−6. Do you believe that the data in the table were generated at this sparging​ rate? Explain. Choose the correct answer below.

The null hypothesis would be that the drug is not safe and the alternative hypothesis would be that the drug is safe. The Type I error for this problem would be to conclude that the drug is safe when the drug is actually not safe. The Type II error for this problem would be to conclude that the drug is not safe when the drug is actually safe. In this application α is the probability of concluding that a drug is safe when actually it is not safe and β is the probability of concluding that a drug is not safe when the drug actually is safe. The value of α should be minimized to minimize the risk of falsely concluding the drug is safe when it is not safe and could hurt people.

Pharmaceutical companies spend billions of dollars per year on research and development of new drugs. The pharmaceutical company must subject each new drug to lengthy and involved testing before receiving the necessary permission from the Food and Drug Administration​ (FDA) to market the drug. The​ FDA's policy is that the pharmaceutical company must provide substantial evidence that a new drug is safe prior to receiving FDA​ approval, so that the FDA can confidently certify the safety of the drug to potential consumers. Use this information to complete parts a through c below. a. If the new drug testing were to be placed in a test of hypothesis​ framework, would the null hypothesis be that the drug is safe or​ unsafe? The alternative​ hypothesis? b. Given the choice of null and alternative hypotheses in part a​, describe Type I and Type II errors in terms of this application. Define α and β in terms of this application. c. If the FDA wants to be very confident that the drug is safe before permitting it to be​ marketed, is it more important that α or β be​ small? Choose the correct answer below.

H0​: μ=76 Ha​: μ>76 Enter (STAT)+(EDIT) +(Enter 3 hardness levels) Enter (STAT)+(TEST)+(2T-TEST)+(DATA)+(plug in uo and <>=) Reject H0. There is sufficient evidence at the α=0.10 level of significance to conclude that the true mean hardness value is greater than 76 J/cm2.

Polyester resins reinforced with fiberglass are used to fabricate wall panels of restaurants. It is theorized that adding cement kiln dust​ (CKD) to the polyester composite will increase wall panel hardness. In a​ study, hardness​ (joules [J] per squared​ centimeters) was determined for three polyester composite mixtures that used a​ 40% CKD weight ratio. The hardness values were reported as 82​,83​,and 79 J/cm2.Research has shown that the mean hardness value of polyester composite mixtures that use a​ 20% CKD weight ratio is μ=76 J/cm2. In your​ opinion, does using a​ 40% CKD weight ratio increase the mean hardness value of polyester composite​ mixtures? Support your answers statistically.

H0​: β1 = 0 Ha​: β1 > 0

Refer to the accompanying simple linear regression relating y=average state standardized test math score in 2014 with x=average state standardized test math score in 2010 for each of 51 states. Complete parts a through c below. Click the icon to view the simple linear regression printout.

Find z score for .99 ENTEr+2nd+VARS+INVNORM+.99 = zscore (XBAR 1 - XBAR 2 ) - Z SCORE SQRT ( S1/n1 + S2/n2) (XBAR 1 - XBAR 2 ) + Z SCORE SQRT ( S1/n1 + S2/n2) Since the confidence interval is entirely greater than​ zero, there is evidence that​ buy-side analysis has the greater mean forecast error The two samples are random and independent.

Researchers compared earnings forecasts of​ buy-side and​ sell-side financial analysts. Data were collected on 4,675 forecasts made by​ buy-side analysts and 57,315 forecasts made by​ sell-side analysts, and the relative absolute forecast error was determined for each. The mean and standard deviation of forecasts errors for both types of analysts are given in the accompanying table. Complete parts a through c below. Click the icon to view the table of summary statistic

ENTER +STAT +(9) 2-SAMP T INT) +Pooled (ENTER INFO) b. Based on the​ interval, part a​, which group has the shorter mean response​ time? Does this result support the​ researchers' last name effect​ theory? Explain. A. The R-Z group has the shorter response time because the interval is entirely greater than zero. This supports the​ researchers' theory.

Researchers investigated the speed with which consumers decide to purchase a product. The researchers theorized that consumers with last names that begin with letters later in the alphabet will tend to acquire items faster than those whose last names begin with letters earlier in the alphabet—called the last name effect. MBA students were offered tickets to a basketball game. The first letter of the last name of respondents and their response times were noted. The researchers compared the response times for two​ groups: (1) those with last names beginning with a​ letter, A-​I, and​ (2) those with last names beginning a​ letter, R-Z. Summary statistics for the two groups are provided in the accompanying table. Complete parts a and b below Click the icon to view the summary statistics.

All March malaria patients and all August malaria patient ( 30/151)- (35/406) = Since 0 is not contained in the confidence​ interval, conclude that there is a difference in the malaria admission rates.

Researchers investigated whether the malaria admission rate in a certain country is higher in some months than in others. In a sample of 151 hospital patients admitted in March​, 30 were treated for malaria. In an independent sample of 406 patients admitted in August ​(5 months​ later), 35 were treated for malaria. Complete parts a through d below. a. Describe the two populations of interest in this study. Choose the correct answer below.

H0: p1−p2=0 Ha: p1−p2>0 ENTER+STAT+6:2-PROPZTEST+ENTER INFO Since the​ p-value is greater than the given value of α​, there is insufficient evidence to reject H0.

Response rates to Web surveys are typically​ low, partially due to users starting but not finishing the survey. The factors that influence response rates were investigated in a certain journal. In a designed​ study, Web users were directed to participate in one of several surveys with different formats. For​ example, one format utilized a welcome screen with a white background and another format utilized a welcome screen with a red background. The​ "break-off rates," the

H0​: p = 0.075 Ha​: p < 0.075

Several years​ ago, a government agency reported the default rate​ (the proportion of borrowers who default on their​ loans) on a certain type of loan at 0.075. Set up the null and alternative hypotheses to determine if the default rate this year is less than 0.075.

ho: u = 0 ha: u > 0 Enter (STAT)+(TEST)+(2T-TEST)+(STAT)+(Plug in U, Q, xbar ,n, and >) t = 4.54 p = 0.0003 Ho is rejected at the α=0.01 level of significance. There is sufficient evidence to conclude that the average annualized return for all stock screeners is positive. The population must be approximately normal. The sample must be randomly selected.

Stock screeners are automated tools used by investment companies to help clients select stocks to invest in. The data on the annualized percentage return on investment for 13 randomly selected stock screeners are given below. You want to determine whether μ​, the average annualized return for all stock​ screeners, is positive. A technology printout of the analysis is shown to the right. Complete parts a through e below.

H0​: p=0.50​; Ha​: p>0.50 Enter (2nd)+(VARS)+(INVNORM) =alpha = rejection space. sign is also > Enter (STAT)+(TEST)+(51-PropZTEST) +(TYPE x,n,>) you have to count all the improvements for the X reject the null hypothesis because the test statistic is in the rejection region. Assuming p=0.50​, the​ p-value is the probability that p is greater than the observed value for a random sample of 39 observations.

Suppose 39 women used a skin cream for 22 weeks. At the end of the​ period, a dermatologist judged whether each woman exhibited skin improvement. The results are shown to the right​ (where I=improved skin and N=no ​improvement). Complete parts a and b.

​Yes, because p satisfies Ha: p>0.50 and is not very close to 0.50. ENTER (2nd) +(VARS)+(3INVNORM) plug in a and enter ENTER+(STAT)+(TEST) +(A) 1-propZint) +(enter x,n,cl%)] do not reject the null hypothesis because the test statistic is not in the rejection region. Assuming p=0.40​, the​ p-value is the probability that p is less than 0.36 for a random sample of 100 observations.

Suppose a random sample of 100 observations from a binomial population gives a value of p=0.57 and you wish to test the null hypothesis that the population parameter p is equal to 0.50 against the alternative hypothesis that p is greater than 0.50.Complete parts a through c.

First multipy 0.32 x 200 to get x ENTER+(STAT)+(TEST) +(5) 1-propZtest) +(enter x,n,cl%)] Reject the null hypothesis because the test statistic is in the rejection region.​Therefore, there is sufficient evidence at the 0.10 level of significance to indicate that the population proportion is less than 0.40. ENTER+(STAT)+(TEST) +(5) 1-propZtest) +(enter x,n,cl%)] Assuming p=0.40​, the​ p-value is the probability that p is less than0.32for a random sample of 200observations.

Suppose a random sample of 200 observations from a binomial population has produced p=0.32 and we wish to test H0: p=0.40 against the alternative Ha: p<0.40. Complete parts a through d. a. Calculate the value of the​ z-statistic for this test. b. Note that the numerator of the​ z-statistic is the same as if n=200​, p=0.17​, and we wish to test H0: p=0.25 against the alternative Ha: p<0.25. Considering​ this, why is the absolute value of z for the original test smaller than that calculated for the revised​ test? Select the correct choice below and fill in the answer boxes to complete your choice.

point estimator = x/n (76/100) =.76 ENTER+(STAT)+(TEST) +(A) 1-propZint) +(enter x,n,cl%) We are 99​% confident that the true proportion of drivers using cell phones is inside this interval.

Suppose data collected by observers at randomly selected intersections across the country revealed that in a sample of 100 ​drivers, 76 were using their cell phone. a. Give a point estimate of​ p, the true driver cell phone use rate​ (that is, the true proportion of drivers who are using their cell phone while​ driving). b. Compute a 99​% confidence interval for p. c. Give a practical interpretation of the​ interval, part b.

random sample is selected from the target population. This is the correct answer. The population from which the sample is selected has a distribution that is approximately normal. Multiply p by 2 The​ p-value for the​ two-tailed test would be 0.056. Since the​ p-value is greater than or equal to α=​0.05, the null hypothesis is not rejected. There is insufficient evidence to support the alternative hypothesis that μ≠100. ​(Round to three decimal places as​ needed.)

Suppose you conduct a​ t-test for the null hypothesis H0​: μ=100 versus the alternative hypothesis Ha​: μ>100 based on a sample of 19 observations. The test results are t=2.04 and ​p-value=0.028. a. What assumptions are necessary for the validity of this​ procedure? b. Interpret the results of the test. c. Suppose the alternative hypothesis had been the​ two-tailed Ha​: μ≠100. If the​ t-statistic were​ unchanged, then what would the​ p-value be for this​ test? Interpret the​ p-value for the​ two-tailed test.

y=β0+β1x+ε ENTER +STAT+EDIT+ENTER INFORMATION ENTER +STAT+CALC+ 4:lingREG(ax+B) but flip the awnsers Since a drive with distance 0 yards is outside the range of the sample​ data, the​ y-intercept has no practical interpretation. For each additional yard in​ distance, the accuracy is estimated to change by the value of the slope. e. A golfer is practicing a new swing to increase her average driving distance. If the golfer is concerned that her driving accuracy will be​ lower, which of the two​ estimates, y-intercept or​ slope, will help determine if the​ golfer's concern is​ valid? The slope will help determine if the​ golfer's concern is valid because the sign of the slope determines whether the accuracy increases or decreases with distance.

The average driving distance​ (yards) and driving accuracy​ (percent of drives that land in the​ fairway) for 8 golfers are recorded in the table to the right. Complete parts a through e below.

23/1000= p 39/1000 = p

The chances of a tax return being audited are about 23 in 1,000 if an income is less than​ $100,000 and 39 in 1,000 if an income is​ $100,000 or more. Complete parts a through e.

H0​: μ1−μ2=0 Ha​: μ1−μ2≠0 ENTEr+2nd+VARS+INVNORM+.99 = zscore (XBAR 1 - XBAR 2 ) - Z SCORE SQRT ( S1/n1 + S2/n2) (XBAR 1 - XBAR 2 ) + Z SCORE SQRT ( S1/n1 + S2/n2) ENTER STAT +2 SAMP Z TEST + (DATA) +POOLED (ENTER X,Q, N) Get data from pop up Do not reject H0. There is insufficient evidence to conclude that the mean amount of surplus that hay producers are willing to sell differs for the two regions.

The conversion of biomass to energy is critical for producing transportation fuels. To determine how willing producers are to supply biomass products such as surplus​ hay, researchers conducted a survey of producers in regions A and B. Independent samples of 571 region A producers and 628 region B producers participated in the survey. Each producer was asked to give the maximum proportion of hay produced that he or she would be willing to sell to the biomass market. Summary statistics for the two groups of producers are listed in the accompanying table. Does the mean amount of surplus that hay producers are willing to sell to the biomass market differ for the two​ regions, A and​ B? Use α=0.05 to make the comparison

H0: p=0.71 vs. Ha: p≠0.71 z score -2.83 Reject the null hypothesis. There is sufficient evidence to conclude that the true detection rate for pictures of PTWs is not 0.71.

The factors that affect the visibility of powered​ two-wheelers (PTWs)—motorcycles—on the road were investigated in a study. A visual search study was conducted in which viewers were presented with pictures of driving scenarios and asked to identify the presence or absence of a PTW. Of interest to the researchers is the detection​ rate, that​ is, the proportion of pictures showing a PTW in which the viewer actually detected the presence of the PTW. Suppose​ that, in​ theory, the true detection rate for pictures of PTWs is 0.71. The study revealed that in a sample of 2,476 pictures that included a​ PTW, only 1,694 were detected by the viewers. Use this result to test the theory at α=0.10. Identify the null and alternative hypotheses for this test. Let p be the true proportion of pictures showing a PTW in which the viewer actually detected the presence of the PTW. Choose the correct answer below

H0​: μ0=0 Ha​: μ0≠0 Enter (Stat) +(Test) +(1Z-Test)+(Stat)+(put in u0, xbar, stdev not equal uo) Calculate the value of the test statistic z. @0.01 Do not reject H0. There is insufficient evidence to conclude that the true mean​ point-spread error for all games differs from 0.

The final scores of games of a certain sport were compared against the final point spreads established by odds makers. The difference between the game outcome and point spread​ (called a​ point-spread error) was calculated for 260 games. The sample mean and sample standard deviation of the​ point-spread errors are x=−1.1 and s=11.8. Use this information to test the hypothesis that the true mean​ point-spread error for all games differs from 0. Conduct the test at α=0.01 and interpret the result.

a) 95/100 = .95 b) b) The researchers can be 95​% confident that the true mean HRV for all hypertensive police officers in the city is between 6.5 and 26.6. A similar statement can be made about the police officers without hypertension. c) The phrasing "95​% confident" means that 95​% of confidence intervals constructed from similarly collected samples will contain the true population mean. d) smaller, smaller , critical value

The heart rate variability​ (HRV) of police officers was the subject of research published in a biology journal. HRV is defined as the variation in the time intervals between heartbeats. A measure of HRV was obtained for each in a sample of 380 police officers from the same city.​ (The lower the measure of​ HRV, the more susceptible the officer is to cardiovascular​ disease.) For the 71 officers diagnosed with​ hypertension, a 95​% confidence interval for the mean HRV was (6.5​,126.6​). For the 309 officers that are not hypertensive, a 95​% confidence interval for the mean HRV was (148.9​,193.2​). Use this information to complete parts a through d below. a. What confidence coefficient was used to generate the confidence​ intervals? b. Give a practical interpretation of both of the 95​% confidence intervals. Use the phrase ​"95​% confident" in your answer. c. When you say you are "95​% confident," what do you​ mean? d. If you want to reduce the width of each confidence​ interval, should you use a smaller or larger confidence​ coefficient? Explain. A ___confidence coefficient should be used. This will result in a ___ standard _____value and thus a narrower confidence interval

H0​: μ=μ0 Ha​: μ>u0 Enter (Stat) +(EDIT)+(input information in L1) Enter (Stat) +(Test) +(2T-Test)+(DATA)+(put in u0 >uo) Enter (Stat) +(EDIT)+(input information in L1) Enter (Stat) +(Test) +(2T-Test)+(DATA)+(put in u0 >uo) Agree. There is sufficient evidence to support the claim.

The maximum allowable mean level of a toxin in​ smelters, herbicide production​ facilities, and other places where the toxin is used is 0.003 milligrams per cubic meter of air. Suppose smelters at two plants are being investigated to determine whether they are meeting standards. Two analyses of the air are made at each​ plant, and the results are shown below. A claim is made that the standard is being violated at Plant 2 but not at Plant 1. Do you​ agree? Assume α=0.05. Find the test statistic for Plant 1

Use calculator for z score (2nd)+(VARS)+(3InvNorm)+(.percentage) (2nd)+(VARS)+(3InvNorm)+(0.05) =1.64 Use Q/Sqrt(n) * Z Score + xbar (upper Bound) 3.2/Sqrt(121) * 1.64 + 34.6(upper Bound) = 35.08 Use 3.2/Sqrt(121) * 1.64 - 34.6 (lower bound) = 34.12 Use absolute value no negative b) same formula just change n c) Quadrupling the sample size while holding the confidence coefficient fixed decreases the width of the confidence interval by a factor of 2

The mean and standard deviation of a random sample of n measurements are equal to 34.6 and 3.2​, respectively. a. Find a 90​% confidence interval for μ if n=121. b. Find a 90​% confidence interval for μ if n=484. c. Find the widths of the confidence intervals found in parts a and b. What is the effect on the width of a confidence interval of quadrupling the sample size while holding the confidence coefficient​ fixed?

It is inappropriate because the number of successes in the sample is less than 15. add +2 to n and to x 3+2 = 5 343+2 ENTER+(STAT)+(TEST) +(A) 1-propZint) +(enter x,n,cl%)] One can be 90​% confident that the interval contains the true population proportion of all mountain casualties that require a femoral shaft splint.

The most common injury that occurs among mountain climbers is trauma to the lower extremity​ (leg). Consequently, rescuers must be proficient in immobilizing and splinting of fractures. The researchers examined the likelihood of needing certain types of splints. A Mountain Rescue study reported that there were 3 shaft splints needed among 343 live casualties. The researchers will use this study to estimate the proportion of all mountain casualties that require a femoral shaft splint. Complete parts a and b below. a. Is the sample large enough to apply the​ large-sample estimation method of this​ section? Show why or why not.

Enter +STAT+EDIT (INPUT FOR L1-L2= L3 ENTER + STAT + TEST+(8: TINTERVAL) +(DATA)+(L3) Since 0 is not contained in the confidence​ interval, conclude that there is a difference in means and that μ1<μ2​, which means the interval supports the​ researcher's conclusion.

The potential of using solar panels constructed above national highways to generate energy was explored in a particular engineering journal.​ Two-layer solar panels​ (with 1 meter separating the​ panels) were constructed above sections of both​ east-west and​ north-south highways in a certain country. The amount of energy​ (kilowatt-hours) supplied to the​ country's grid by the solar panels above the two types of highways was determined each month. The data for several randomly selected months are provided in the accompanying table. The researchers concluded that the​ two-layer solar panel energy generation is more viable for the​ north-south oriented highways as compared to​ east-west oriented roadways. Compare the mean solar energy amounts generated for the two types of highways using a 90​% confidence interval. Does the interval support the​ researchers' conclusion? Find a 90​% confidence interval for μd.Let μd=μ1−μ2​, where μ1 is the mean solar energy generated for the​ east-west highways and μ2is the mean solar energy generated for the​ north-south highways.

enter (Stat) +(1edit)+(plug in numbers) enter (Stat) + (test) +(8Tinterrval)+(data)+(plug in CL%) (3.62,8.38) b) enter (Stat) + (test) +(8Tinterrval)+(stat)+(plug in CL%) +(change n) (5.01,7) c) As the sample size​ increases, the width decreases.

The random sample shown below was selected from a normal distribution. b. Assume that sample mean x and sample standard deviation s remain exactly the same as those you just calculated but that are based on a sample of n=25 observations. Repeat part a. What is the effect of increasing the sample size on the width of the confidence​ intervals? What is the effect of the sample size on the width of the confidence​ interval?

a. List the values x may assume. 1, 4​, 6​, 7​, 9 (input all numbers on top row) b. What value of x is most​ probable? nothing (pick the one with the highest number probability) add all probabilities up;

The random variable x has the following discrete probability distribution. Complete parts a through f.

a) 100-(percentage) = Alpha Alpha / 2 = divided alpha 100-95 = 5 5/2 = 2.5 2.5/100= .025 Use calculator for z score (2nd)+(VARS)+(3InvNorm)+(.percentage) (2nd)+(VARS)+(3InvNorm)+(.025) = 1.96 Use Q/Sqrt(n) * Z Score + xbar (upper Bound) 68/Sqrt(267) * 1.96 + 22 (upper Bound) = 30.157 (three decimal) Use 68/Sqrt(267) * 1.96 - 22 (lower bound) = 13.843 Use absolute value (no negative) a2 ) One can be​ 95% confident the true mean change in the Mathematics score lies within the above interval b) Use Q/Sqrt(n) * Z Score + xbar (upper Bound) 46/Sqrt(267) * 1.96 + 9 (upper Bound) = 14.518 (three decimal) Use 46/Sqrt(267) * 1.96 - 9 (lower bound) = 3.482 Use absolute value (no negative) b2) One can be​ 95% confident the true mean change in the Verbal score lies within the above interval. c) Verbal is more likely to have this true mean change because it falls within the​ 95% confidence interval for Verbal. Which ever is in the 95 confidence level 6 is in (3.482 , 14.518) verbal interval

The results of a study of 267 students who paid a private tutor to help them improve their scores on standardized test are shown below. The changes in both the Mathematics and Verbal scores for these students are reproduced in the table. Complete parts a through c below. a. Construct and interpret a​ 95% confidence interval for the population mean change in the Mathematics score for students who pay a private tutor. a2 )Interpret this interval. Choose the correct answer below b. Construct and interpret a​ 95% confidence interval for the population mean change in the Verbal score for students who pay a private tutor. b2) Interpret this interval. Choose the correct answer below c. Suppose the true population mean change in score on one of the standardized tests for all students who paid a private tutor is 6. Which of the two​ tests, Mathematics or​ Verbal, is most likely to have this mean​ change? Explain.

Enter +STAT+EDIT (INPUT FOR L1-L2= L3 ENTER + STAT + TEST+(2;T-TEST) +(DATA)+(L3) Reject H0 at α = 0.05. There is sufficient evidence to show that cameras at traffic stops produce a significant decrease in the mean number of crashes caused by red light running per intersection per year.

To combat red light running crashes many states have installed red light cameras at dangerous intersections to photograph the license plates of vehicles that run the red light. How effective are​ photo-red enforcement programs in reducing​ red-light-running crash incidents at​ intersections? A​ state's department of transportation conducted a comprehensive study of its​ photo-red enforcement program. In one portion of the​ study, the department provided crash data both before and after installation of red light cameras at several intersections. The data​ (measured as the number of crashes caused by​ red-light-running per intersection per​ year) for 13 intersections in a specific​ county, are given in the accompanying table. Analyze the data for the department. What do you​ conclude?

H0: p1−p2=0 Ha: p1−p2≠0 ENTER+STAT+6:2-PROPZTEST+ENTER INFO Do not reject H0. There is insufficient evidence to conclude that the percentage of TV commercials that use religious symbolism has changed since the 1998 study.

University professors conducted a study of television commercials. The key research question was​ "Do television advertisers use religious symbolism to sell goods and​ services?" In a sample of 791 TV commercials collected in​ 1998, only 14 commercials used religious symbolism. Of the 1,503 TV commercials examined in the more recent​ study, 45 commercials used religious symbolism. Conduct an analysis to determine if the percentage of TV commercials that use religious symbolism has changed since the 1998 study assuming a significance level of 0.01. If you detect a​ change, estimate the magnitude of the difference and attach a measure of reliability to the estimate.

follow table and deduct from 1 a. x±1.96 σ/n 95​% ​(Round to two decimal places as​ needed.) b. x±1.645 σ/n 90% ​(Round to two decimal places as​ needed.) c. x±2.575 σ/n 99​% ​(Round to two decimal places as​ needed.) d. x±1.282 σ/n 80​% ​(Round to two decimal places as​ needed.) e. x±0.99 σ/n 67.78​% ​(Round to two decimal places as​ needed.)

What is the confidence level of each of the following confidence intervals for μ​? Complete parts a through e.

​H0: μ=5.24 vs. Ha​: μ< 5.24 F. t < -2.764 Enter (STAT)+(TEST)+(2T-TEST)+(STAT)+(Plug in U, Q, xbar ,n, and <) t = -4.93 Reject H0. There is sufficient evidence to indicate μ<5.24 Mpa. We must assume that the sample was random and selected from a population with a distribution that is approximately normal.

When bonding​ teeth, orthodontists must maintain a dry field. A new bonding adhesive has been developed to eliminate the necessity of a dry field.​ However, there is concern that the new bonding adhesive is not as strong as the current​ standard, a composite adhesive. Tests on a sample of 11 extracted teeth bonded with the new adhesive resulted in a mean breaking strength​ (after 24​ hours) of x=4.63 Mpa and a standard deviation of s=0.41 Mpa. Orthodontists want to know if the true mean breaking strength is less than 5.24 Mpa, the mean breaking strength of the composite adhesive. b. Find the rejection region for the test using α=0.01. Choose the correct answer below.

The null hypothesis is the status quo hypothesis and the alternative hypothesis is the research hypothesis.

Which​ hypothesis, the null or the​ alternative, is the​ status-quo hypothesis? Which is the research​ hypothesis?

Yes. As long as a sample is sufficiently large that the Central Limit Theorem​ applies, the confidence interval will be valid regardless of the shape of the population distribution.

Will a​ large-sample confidence interval be valid if the population from which the sample is taken is not normally​ distributed? Explain.

No, because the Central Limit Theorem states that the sampling distribution of x is approximately normally distributed only if the sample size is large enough.

Will the sampling distribution of x always be approximately normally​ distributed? Explain. Choose the correct answer below.

There is a small sample size. The population from which the sample is selected has a distribution that is approximately normal. The distribution curves are both​ mound-shaped and symmetric.​ However, the​ t-test statistic curve is flatter than the​ z-statistic curve because the​ t-test is much more sensitive to the sample size. t>2.718 where df​= 11 = 0.010 =0.010 0.050 us student t- table follow degree find df then find the z score within bracket the = one gets doubled.

a. Consider testing H0​: μ=80. Under what conditions should you use the​ t-distribution to conduct the​ test? b. In what ways are the distributions of the​ z-statistic and​ t-test statistic​ alike? How do they​ differ? a. Consider testing H0​: μ=80. Under what conditions should you use the​ t-distribution to conduct the​ test? Select all of the conditions that apply. b. In what ways are the distributions of the​ z-statistic and​ t-test statistic​ alike? How do they​ differ?

enter (Stat) + (test) +(8Tinterrval)+(STAT)+(plug in CL%, Xbar, SXdev) (2.413, 3.987) b) The water department can be 95​% confident that the mean lead level in drinking water for all residents in the town is within this interval. c) The phrase "95​% confidence​ interval" means that if many intervals were constructed using the same methods and the same sample size, then 95% of those intervals would contain the true population mean.

​Periodically, a town water department tests the the drinking water of homeowners for contaminants such as lead. The lead levels in water specimens collected for a sample of 10 residents of the town had a mean of 3.2 ​mg/L and a standard deviation of 1.1 mg/L. Complete parts a through c. a. Construct a 95​% confidence interval for the mean lead level in water specimens from the town. b. Interpret the interval in terms of this application. Choose the correct answer below. c. What is meant by the phrase "95​% confidence​ interval"?

a) enter (Stat) + (test) +(8Tinterrval)+(STAT)+(plug in xbar ,sxdev, n, CL%) ( 1.851,4.729) b) The water department can be 90​% confident that the mean lead level in drinking water for all residents in the town is within this interval. c) same sample size, 90%, population mean

​Periodically, a town water department tests the the drinking water of homeowners for contaminants such as lead. The lead levels in water specimens collected for a sample of 10 residents of the town had a mean of 3.3 ​mg/L and a standard deviation of 2.5 mg/L. Complete parts a through a. Construct a 90​% confidence interval for the mean lead level in water specimens from the town. b. Interpret the interval in terms of this application. Choose the correct answer below. c. What is meant by the phrase ​"90​% confidence​ interval"? The phrase ​"90​% confidence​ interval" means that if many intervals were constructed using the same methods and the same sample size, then 90​% of those intervals would contain the true population mean.

a) find 90​% confidence interval for upper and lower on picture b) With 90​% confidence, the true mean failure time of used colored display panels is between the endpoints of the confidence interval. c) 90/100 = 0.9

​Wear-out failure time of electronic components is often assumed to have a normal distribution. A lot of 50 used display panels was purchased by an outlet store. Each panel displays 12 to 18 color characters. Prior to the​ acquisition, the panels had been used for about​ one-third of their expected lifetimes. The failure times​ (in years) for a sample of 50 used panels and an SPSS printout of the analysis are given in the accompanying tables. Complete parts a through c. a. Locate a 90​% confidence interval for the true mean failure time of used colored display panels on the printout. b. Give a practical interpretation of the​interval, part a c. In repeated sampling of the population of used colored display​panels, where a 90​% confidence interval for the mean failure time is computed for each​sample, what proportion of all the confidence intervals generated will capture the true mean failure​time?