Stats Exam 3 (Ch. 6,7, & 8)
Assume that adults have IQ scores that are normally distributed with a mean of μ=100 and a standard deviation σ=15. Find the probability that a randomly selected adult has an IQ between 84 and 116.
"between"= subtraction *put the bigger # on top, lower on the bottom* use z=(larger - mean)/S.D. - once find the answer, look for that number on the z chart. - do this for the smaller number too. z=(smaller - mean)/S.D. larger number z score - smaller number z score finally answer
Find the indicated critical value. z0.02 (always looking for the z-score to the RIGHT)
Looking for .9800 on the z-chart and thats the answer. 1 - 0.02 = .9800
What is the confidence interval estimate of the population mean μ? What does the confidence interval suggest about the effectiveness of the treatment?
-1.46 mg/dL < u < 7.66 mg/dL The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels.
Construct a 95% confidence interval of the mean drive-through service times at dinner for Restaurant X. Construct a 95% confidence interval of the mean drive-through service times at dinner for Restaurant Y. Compare the results
161.3 sec < u < 197.2 sec 139.1 sec < u < 167.3 sec The confidence interval estimates for the two restaurants overlap, so there does not appear to be a significant difference between the mean dinner times at the two restaurants.
What is the confidence interval for the population mean μ? Are the results between the two confidence intervals very different?
28.6 hg < u < 30.8 hg no, because the confidence interval limits are similar
Which of the following is NOT true when testing a claim about a proportion?
A conclusion based on a confidence interval estimate will be the same as a conclusion based on a hypothesis test.
What is the difference between a standard normal distribution and a nonstandard normal distribution?
Answer always- The standard normal distribution has a mean of 0 and a standard deviation of 1, while a nonstandard normal distribution has a different value for one or both of those parameters.
Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. (chart has a bell-shaped chart, with the number 0.1788 shaded to the LEFT) Find the indicated z score
Answer is going to be negative cause its to the LEFT. Find 0.1788 on the z chart. Than find the area to the right - 1 - (number found on the z chart)
Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. (chart has a bell-shaped chart, with the number 0.7673 shaded to the RIGHT) Find the indicated z score
Find the 0.7673 on the z chart and thats the indicated z score
Assume that adults have IQ scores that are normally distributed with a mean of 103.9 and a standard deviation 19.3. Find the first quartile Q1, which is the IQ score separating the bottom 25% from the top 75%. (Hint: Draw a graph.)
Find the first quartile (which is going to be 0.25 or .2500, look this up in the z chart) quartile=(z score)(S.D) + mean
Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find the probability of a bone density test score greater than −1.69. Sketch the region. Find the probability.
Going to be the graph that is shaded to the RIGHT and the number is NEGATIVE and to the LEFT To find the probability: (greater than = to the RIGHT, lesser than = to the LEFT) Find the z score of the number given (-1.69) and than do 1 - (the z score found).
Assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find P3, the 3rd percentile. This is the bone density score separating the bottom 3% from the top 97%. Which graph represents P3? Choose the correct graph below. The bone density score corresponding to P3 is...
Graph- The answer is going to be the graph that the shaded area is to the LEFT and P3 is on the LEFT side of the chart To find the bone density: look up .750 on the z chart and that's the answer
The statistics of n=22 and s=14.3 result in this 95% confidence interval estimate of σ: 11.0<σ<20.4. That confidence interval can also be expressed as (11.0, 20.4). Given that 15.7±4.7 results in values of 11.0 and 20.4, can the confidence interval be expressed as 15.7±4.7 as well?
No. The format implies that s=15.7, but s is given as 14.3. In general, a confidence interval for σ does not have s at the center.
The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 7 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes.
Step 1: Need to find the probability = total - favorite 7 - 1.25 Step 2: favorite / total 5.75 / 7 = 0.8214 round to the nearest 3 decimal place
What requirements are necessary for a normal probability distribution to be a standard normal probability distribution?
The mean and standard deviation have the values of μ=0 and o = 1.
In a clinical trial of a drug intended to help people stop smoking, 133 subjects were treated with the drug for 12 weeks, and 19 subjects experienced abdominal pain. If someone claims that more than 8% of the drug's users experience abdominal pain, that claim is supported with a hypothesis test conducted with a 0.05 significance level. Using 0.15 as an alternative value of p, the power of the test is 0.96. Interpret this value of the power of the test.
The power of 0.96 shows that there is a 9696% chance of rejecting the null hypothesis of p=0.080.08 when the true proportion is actually 0.150.15. That is, if the proportion of users who experience abdominal pain is actually 0.150.15, then there is a 9696% chance of supporting the claim that the proportion of users who experience abdominal pain is greater than 0.08.
Make a decision about the given claim. Use only the rare event rule, and make subjective estimates to determine whether events are likely. For example, if the claim is that a coin favors heads and sample results consist of 11 heads in 20 flips, conclude that there is not sufficient evidence to support the claim that the coin favors heads (because it is easy to get 11 heads in 20 flips by chance with a fair coin). Claim: The mean age of students in a large calculus class is greater than 20. A simple random sample of the students has a mean age of 20.4. Choose the correct answer below.
The sample is not unusual if the claim is true. The sample is not unusual if the claim is false. Therefore, there is not sufficient evidence to support the claim.
Claim: The standard deviation of pulse rates of adult males is less than 12 bpm. For a random sample of 142 adult males, the pulse rates have a standard deviation of 10.9 bpm. Find the value of the test statistic.
The value of the test statistic is 116.33
Claim: The mean pulse rate (in beats per minute) of adult males is equal to 69 bpm. For a random sample of 160 adult males, the mean pulse rate is 68.3 bpm and the standard deviation is 10.9 bpm. Find the value of the test statistic.
The value of the test statistic is -.81
Which of the following is NOT true of the confidence level of a confidence interval?
There is a 1−α chance, where α is the complement of the confidence level, that the true value of p will fall in the confidence interval produced from our sample.
The Ericsson method is one of several methods claimed to increase the likelihood of a baby girl. In a clinical trial, results could be analyzed with a formal hypothesis test with the alternative hypothesis of p>0.5, which corresponds to the claim that the method increases the likelihood of having a girl, so that the proportion of girls is greater than 0.5. If you have an interest in establishing the success of the method, which of the following P-values would you prefer: 0.999, 0.5, 0.95, 0.05, 0.01, 0.001? Why?
The P-value of 0.001 is preferred because it corresponds to the sample evidence that most strongly supports the alternative hypothesis that the method is effective.
Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. The results in the screen display are based on a 95% confidence level. Write a statement that correctly interprets the confidence interval.
We have 95% confidence that the limits of 13.05 Mbps and 22.15 Mbps contain the true value of the mean of the population of all data speeds at the airports.
A genetic experiment with peas resulted in one sample of offspring that consisted of 442 green peas and 169 yellow peas. a. Construct a 95% confidence interval to estimate of the percentage of yellow peas. b. It was expected that 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations?
a. .241 < p < .312 b. no, the confidence interval includes 0.25, so the true percentage could easily equal 25%
The values listed below are waiting times (in minutes) of customers at two different banks. At Bank A, customers enter a single waiting line that feeds three teller windows. At Bank B, customers may enter any one of three different lines that have formed at three teller windows. Answer the following questions. a. Construct a 90% confidence interval for the population standard deviation σ at Bank A. b. Construct a 90% confidence interval for the population standard deviation σ at Bank B. c. Interpret the results found in the previous parts. Do the confidence intervals suggest a difference in the variation among waiting times? Does the single-line system or the multiple-line system seem to be a better arrangement?
a. .36 min < o Bank A < .81 min b. 1.32 min < o Bank B < 2.98 min c. The variation appears to be significantly lower with a single line system. The single-line system appears to be better.
a. Given that Emily used a coin toss to select either her right hand or her left hand, what proportion of correct responses would be expected if the touch therapists made random guesses? b. Using Emily's sample results, what is the best point estimate of the therapists' success rate? c. Using Emily's sample results, construct a 95% confidence interval estimate of the proportion of correct responses made by touch therapists. d. What do the results suggest about the ability of touch therapists to select the correct hand by sensing energy fields?
a. .5 b. .456 c. .398 < p < .514 d. Since the confidence interval is not entirely above 0.5, there does not appear to be sufficient evidence that touch therapists can select the correct hand by sensing energy fields.
Use the sample data and confidence level given below to complete parts (a) through (d). A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n=902 and x=504 who said "yes." Use a 95% confidence level. a. Find the best point estimate of the population proportion p. b. Identify the value of the margin of error E. c. Construct the confidence interval. d. Write a statement that correctly interprets the confidence interval. Choose the correct answer below.
a. .559 b. .032 c. .527 < p < .591 d. one has 95% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.
a. Express the confidence interval in the format that uses the "less than" symbol. Given that the original listed data use one decimal place, round the confidence interval limits accordingly. b. Identify the best point estimate of μ and the margin of error c. In constructing the confidence interval estimate of μ, why is it not necessary to confirm that the sample data appear to be from a population with a normal distribution?
a. 13.05 Mbps < u < 22.15 Mbps b. the point estimate is 17.60, the margin of error is E = 4.55 Mbps c. Because the sample size of 50 is greater than 30, the distribution of sample means can be treated as a normal distribution.
a. Assume that nothing is known about the percentage of adults who have heard of the brand. b. Assume that a recent survey suggests that about 88% of adults have heard of the brand. c. Given that the required sample size is relatively small, could he simply survey the adults at the nearest college?
a. 188 b. 79 c. No, a sample of students at the nearest college is a convenience sample, not a simple random sample, so it is very possible that the results would not be representative of the population of adults.
A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in older subjects. After treatment with the drug, 14 subjects had a mean wake time of 96.6 min and a standard deviation of 44.4 min. Assume that the 14 sample values appear to be from a normally distributed population and construct a 90% confidence interval estimate of the standard deviation of the wake times for a population with the drug treatments. Does the result indicate whether the treatment is effective? a. Find the confidence interval estimate b. Does the result indicate whether the treatment is effective?
a. 33.85 min < o < 65.95 min b. No, the confidence interval does not indicate whether the treatment is effective.
The required sample size is ______ Would it be reasonable to sample this number of students?
a. 39 b. Yes. This number of IQ test scores is a fairly small number.
In a study of the accuracy of fast food drive-through orders, one restaurant had 31 orders that were not accurate among 365 orders observed. Use a 0.01 significance level to test the claim that the rate of inaccurate orders is equal to 10%. Does the accuracy rate appear to be acceptable? a. Identify the null and alternative hypotheses for this test. Choose the correct answer below. b. Identify the test statistic for this hypothesis test. c. Identify the P-value for this hypothesis test. d. Identify the conclusion for this hypothesis test. e. Does the accuracy rate appear to be acceptable?
a. H0: p = 0.1 H: p DOES NOT EQUAL 0.1 b. -.96 c. .337 d. Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the rate of inaccurate orders is equal to10%. e. Since there is not sufficient evidence to reject the claim that the rate of inaccurate orders is equal to 10%, it is plausible that the inaccuracy rate is 10%. This rate would be too high, the restaurant should work to lower the rate
In a study of 820 randomly selected medical malpractice lawsuits, it was found that 483 of them were dropped or dismissed. Use a 0.01 significance level to test the claim that most medical malpractice lawsuits are dropped or dismissed. a. Which of the following is the hypothesis test to be conducted? b. What is the test statistic? c. What is the P-value? d. What is the conclusion about the null hypothesis? e. What is the final conclusion?
a. H0: p = 0.5 H1: p > 0.5 b. z = 5.10 c. p-value = 0 d. Reject the null hypothesis because the P-value is less than or equal to the significance level, α. e. There is sufficient evidence to support the claim that most medical malpractice lawsuits are dropped or dismissed.
A random sample of 854 births included 434 boys. Use a 0.10 significance level to test the claim that 51.2% of newborn babies are boys. Do the results support the belief that 51.2% of newborn babies are boys? a. Identify the null and alternative hypotheses for this test. Choose the correct answer below. b. Identify the test statistic for this hypothesis test. c. Identify the P-value for this hypothesis test. d. Identify the conclusion for this hypothesis test. e. Do the results support the belief that 51.2% of newborn babies are boys?
a. H0: p = 0.512 H1: p DOES NOT EQUAL 0.512 b. The test statistic for this hypothesis test is -.22 c. The P-value for this hypothesis test is .826 d. Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that 51.2% of newborn babies are boys. e. The results do not support the belief that 51.2% of newborn babies are boys; the results merely show that there is not strong evidence against the rate of 51.2%.
In a recent court case it was found that during a period of 11 years 889 people were selected for grand jury duty and 37% of them were from the same ethnicity. Among the people eligible for grand jury duty, 78.3% were of this ethnicity. Use a 0.05 significance level to test the claim that the selection process is biased against allowing this ethnicity to sit on the grand jury. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution. a. Which of the following is the hypothesis test to be conducted? d. What is the conclusion on the null hypothesis? e. Does the jury selection system appear to be fair?
a. H0: p = 0.783 H1: p < 0.783 b. z = -29.87 c. p-value = .0001 d. reject the null hypothesis because the P-value is less than or equal to the significance level, α. e. There is sufficient evidence to support the claim that the selection process is biased against allowing this ethnicity to sit on the grand jury. The jury selection process appears to be unfair.
Trials in an experiment with a polygraph include 97 results that include 22 cases of wrong results and 75 cases of correct results. Use a 0.05 significance level to test the claim that such polygraph results are correct less than 80% of the time. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution. a. Let p be the population proportion of correct polygraph results. Identify the null and alternative hypotheses. Choose the correct answer below. b.The test statistic is z= c. The P-value is d. Identify the conclusion about the null hypothesis and the final conclusion that addresses the original claim.
a. H0: p=0.80 H1: p< 0.80 b. -.66 c. .2546 d. Fail to reject, IS NOT
A certain drug is used to treat asthma. In a clinical trial of the drug, 24 of 254 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 10% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts (a) through (e) below. a. Is the test two-tailed, left-tailed, or right-tailed? b. What is the test statistic? c. What is the P-value? d. What is the null hypothesis, and what do you conclude about it? Decide whether to reject the null hypothesis. Choose the correct answer below. e. What is the final conclusion?
a. Left-tailed test b. z = -0.29 c. P-value = 0.3848 d. H0: p = 0.1 Fail to reject the null hypothesis because theP-value is greater than the significancelevel, α. e. There is not sufficient evidence to support the claim that less than 10% of treated subjects experienced headaches.
Assume a significance level of α=0.1 and use the given information to complete parts (a) and (b) below. Original claim: The mean pulse rate (in beats per minute) of a certain group of adult males is 71 bpm. The hypothesis test results in a P-value of 0.0053. a. State a conclusion about the null hypothesis. (Reject H0 or fail to reject H0.) Choose the correct answer below. b. Without using technical terms, state a final conclusion that addresses the original claim. Which of the following is the correct conclusion?
a. Reject H0 because theP-value is less than or equal to α. b. There is sufficient evidence to warrant rejection of the claim that the mean pulse rate (in beats per minute) of the group of adult males is 71 bpm.
Identify the type I error and the type II error that corresponds to the given hypothesis. The proportion of people who write with their left hand is equal to 0.27. a. Which of the following is a type I error? b. Which of the following is a type II error?
a. Reject the claim that the proportion of people who write with their left hand is 0.27 when the proportion is actually 0.27. b. Fail to reject the claim that the proportion of people who write with their left hand is 0.27 when the proportion is actually different from 0.27.
he test statistic of z=−2.26 is obtained when testing the claim that p<0.82. a. Using a significance level of α=0.10, find the critical value(s). b. Choose the correct conclusion below
a. The critical value(s) is/are z= -1.28 b. Reject H0. There is sufficient evidence to support the claim that p<0.82.
Assume that the sample is a simple random sample obtained from a normally distributed population of IQ scores of statistics professors. Use the table below to find the minimum sample size needed to be 99% confident that the sample standard deviation s is within 40% of σ. Is this sample size practical?
a. The minimum sample size needed is 22. b. Yes, because the sample size is small enough to be practical for most applications
Assume that the sample is a simple random sample obtained from a normally distributed population of flight delays at an airport. Use the table below to find the minimum sample size needed to be 99% confident that the sample standard deviation is within 30% of the population standard deviation. A histogram of a sample of those arrival delays suggests that the distribution is skewed, not normal. How does the distribution affect the sample size?
a. The minimum sample size needed is 38. b. The computed minimum sample size is not likely correct.
In a certain survey, 520 people chose to respond to this question: "Should passwords be replaced with biometric security (fingerprints, etc)?" Among the respondents, 55% said "yes." We want to test the claim that more than half of the population believes that passwords should be replaced with biometric security. Complete parts (a) through (d) below. a. Are any of the three requirements violated? Can a test about a population proportion using the normal approximation method be used? b. It was stated that we can easily remember how to interpret P-values with this: "If the P is low, the null must go." What does this mean? c. Another memory trick commonly used is this: "If the P is high, the null will fly." Given that a hypothesis test never results in a conclusion of proving or supporting a null hypothesis, how is this memory trick misleading? d. Common significance levels are 0.01 and 0.05. Why would it be unwise to use a significance level with a number like 0.0483?
a. The sample observations are not a random sample, so a test about a population proportion using the normal approximating method cannot be used. b. This statement means that if the P-value is very low, the null hypothesis should be rejected. c. This statement seems to suggest that with a high P-value, the null hypothesis has been proven or is supported, but this conclusion cannot be made. d. Choosing this specific of a significance level could give the impression that the significance level was chosen specifically to reach a desired conclusion.
The test statistic of z=1.11 is obtained when testing the claim that p≠0.327 a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed. b. Find the p-value c. Using a significance level of α=0.05, should we reject H0 or should we fail to reject H0?
a. This is a two-tailed test b. P-value = .267 c. Fail to reject H0. There is not sufficient evidence to support the claim that p≠0.327.
Assume that human body temperatures are normally distributed with a mean of 98.20°F and a standard deviation of 0.63°F. a. A hospital uses 100.6°F as the lowest temperature considered to be a fever. What percentage of normal and healthy persons would be considered to have a fever? Does this percentage suggest that a cutoff of 100.6°F is appropriate? b. Physicians want to select a minimum temperature for requiring further medical tests. What should that temperature be, if we want only 5.0% of healthy people to exceed it? (Such a result is a false positive, meaning that the test result is positive, but the subject is not really sick.)
a. Yes, because there is a small probability that a normal and healthy person would be considered to have a fever. b. look in 6.2 notes, problem 8
a. What is the number of degrees of freedom that should be used for finding the critical value tα/2? b. Find the critical value tα/2 corresponding to a 95% confidence level. c. Give a brief general description of the number of degrees of freedom.
a. df = 49 b. ta/2 = 2.01 c. The number of degrees of freedom for a collection of sample data is the number of sample values that can vary after certain restrictions have been imposed on all data values.
A survey found that women's heights are normally distributed with mean 63.2 in. and standard deviation 3.2 in. The survey also found that men's heights are normally distributed with mean 67.7 in. and standard deviation 3.3 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 55 in. and a maximum of 64 in. Complete parts (a) and (b) below. a. Find the percentage of men meeting the height requirement. What does the result suggest about the genders of the people who are employed as characters at the amusement park? b. If the height requirements are changed to exclude only the tallest 50% of men and the shortest 5% of men, what are the new height requirements?
a. look in 6.2 notes problem 7 for A and B
a. Assume that nothing is known about the percentage to be estimated. b. Assume prior studies have shown that about 45% of full-time students earn bachelor's degrees in four years or less. c. Does the added knowledge in part (b) have much of an effect on the sample size?
a. n = 1691 b. 1674 c. No, using the additional survey information from part (b) only slightly reduces the sample size.
a. Find the sample size using the range rule of thumb to estimate o. b. Assume that σ=10.9 bpm, based on the value s=10.9 bpm from the sample of 146 male pulse rate c. Compare the results from parts (a) and (b). Which result is likely to be better?
a. n = 173 b. n = 80 c. The results from part (a) is larger than the results from part (b). The result from part (b) is likely to be better because it uses a better estimate of o.
Claim: A majority of adults would erase all of their personal information online if they could. A software firm survey of 676 randomly selected adults showed that 61% of them would erase all of their personal information online if they could. Complete parts (a) and (b) below. a. Express the original claim in symbolic form. Let the parameter represent the adults that would erase their personal information. b. Identify the null and alternative hypotheses.
a. p > 0.5 b. H0: p = 0.5 H1: p > 0.5
What is different about the normality requirement for a confidence interval estimate of σ and the normality requirement for a confidence interval estimate of μ?
a. stricter b. greater c. less
Claim: The mean pulse rate (in beats per minute) of adult males is equal to 69.2 bpm. For a random sample of 153 adult males, the mean pulse rate is 69.7 bpm and the standard deviation is 11.1 bpm. Complete parts (a) and (b) below. a. Express the original claim in symbolic form.
a. u = 69.2 bpm b. H0: u = 69.2 bpm H1: u = 69.2 bpm
Pulse rates of women are normally distributed with a mean of 77.5 beats per minute and a standard deviation of 11.6 beats per minute. Answer the following questions. What are the values of the mean and standard deviation after converting all pulse rates of women to z scores using z=(x−μ)σ?
answer always u=0 o=1 Answer always- The z scores are numbers without units of measurement
What conditions would produce a negative z-score?
answer always: a z-score corresponding to an area located entirely in the left side of the curve
A magazine provided results from a poll of 1000 adults who were asked to identify their favorite pie. Among the 1000 respondents, 11% chose chocolate pie, and the margin of error was given as ±3 percentage points. Describe what is meant by the statement that "the margin of error was given as ±3 percentage points."
answer always: the statement indicates that the interval 11% + 3% is likely to contain the true population percentage of people that prefer chocolate pie
About ______% of the area is between z=−2 and z=2 (or within 2 standard deviations of the mean).
answer will be 95.44 (when finding the area between z=-2 and z=2) 1. find z score for -2 and 2 (will be the same) 2. add the z scores together 3. 1 - (z score) 4. convert the decimal to percentage
Use the given information to find the number of degrees of freedom, the critical values χ2L and χ2R, and the confidence interval estimate of σ. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Nicotine in menthol cigarettes 98% confidence; n=27, s=0.21 mg.
df = 26 xL2 = 12.198 xR2 = 45.642 The confidence interval estimate of o is .16 mg < o < .31 mg
A normal distribution is informally described as a probability distribution that is "bell-shaped" when graphed. Draw a rough sketch of a curve having the bell shape that is characteristic of a normal distribution.
looks like a regular graph. (low, medium, low); two tails. "bell-shaped."
Express the confidence interval 0.111<p<0.999 in the form p±E.
p^ +- E = .555 plus or minus .444
A magazine provided results from a poll of 500 adults who were asked to identify their favorite pie. Among the 500 respondents, 13% chose chocolate pie, and the margin of error was given as ±3 percentage points. What values do p, q, n, E, and p represent? If the confidence level is 90%, what is the value of α?
p^ = the sample proportion q^ = found from evaluating 1 - p^ n = the sample size E = the margin of error p = the population proportion a = .1 if the confidence level is 90%
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
pick ta/2 NOT za/2. ta/2 = 2.67
Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.
use x=z*o+u z is the number in the graph o is given u is given find z in the z chart and than put it in the z spot
For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of scores that are a. significantly high (or at least 2 standard deviations above the mean). b. significantly low (or at least 2 standard deviations below the mean). c. not significant (or less than 2 standard deviations away from the mean).
using answers from the problem before a. 2.28% b. 2.28% c. 95.44%
Find the critical value zα/2 that corresponds to the confidence level 99%.
z = 2.58
Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. The area of the shaded region is
z=(x-u)/o all of the numbers should be given in the equation x=95 u=100 o=15 look for the number found from this equation in the z chart
