ch 8 - 10 practice test

A simple random sample of size n=40 is drawn from a population. The sample mean is found to be x=121.9 and the sample standard deviation is found to be s=13.1. Construct a​ 99% confidence interval for the population mean.

The lower bound is 116.29. The upper bound is 127.51 Use statcrunch Stat> T Stat> One Sample> With summary> Enter sample mean> Enter sample std. dev > Enter sample size> Adjust confidence Interval > Compute.

Explain why the​ t-distribution has less spread as the number of degrees of freedom increases.

The​ t-distribution has less spread as the degrees of freedom increase​ because, as n​ increases, s becomes closer to σ by the law of large numbers.

Find the critical values χ2 1−α/2 and χ2 α/2 for a 98​% confidence level and a sample size of n=10.

χ2 1−α/2 = 2.088 χ2 α/2 = 21.666 Use stat>Cal> Chi-square> Enter n -1 (10-1=9) > select between > enter = 0.98 (confidence level) > compute.

Jan performed a study and obtained a​ p-value of 1.24. What conclusion should Jan​ make?

She made an error since it is not possible to get a​ p-value of 1.24.

Suppose the null hypothesis is rejected. State the conclusion based on the results of the test. Three years​ ago, the mean price of a​ single-family home was ​$243,703. A real estate broker believes that the mean price has decreased since then.

- There is sufficient evidence to conclude that the mean price of a​ single-family home has decreased. We know the above because when something is rejected, it means there is evidence to support. When it's not rejected it means there isn't enough evidence.

If we reject the null hypothesis when the statement in the null hypothesis is​ true, we have made a Type​ _______ error.

Answer: I

Determine if the following statement is true or false. When testing a hypothesis using the​ P-value Approach, if the​ P-value is​ large, reject the null hypothesis.

False

A simple random sample of size 20 is drawn from a population that is known to be normally distributed. The sample​ variance, s2​, is determined to be 12.1. Construct a​ 90% confidence interval for σ2.

Use stat > variance stats> one sample > w summary > enter sample variance provided along with sample size, adjust the confidence interval and compute. Lower = 7.63 Upper = 22.72 rounded 2 dec.

In a survey of 1002 adults, a polling agency​ asked, "When you​ retire, do you think you will have enough money to live comfortably or not. Of the 1002 surveyed, 530 stated that they were worried about having enough money to live comfortably in retirement. Construct a 95​% confidence interval for the proportion of adults who are worried about having enough money to live comfortably in retirement. Select the correct choice below and fill in the answer boxes to complete your choice.

Use stat crunch for Population proportion. Stat> Proportion stats> One sample> With summary> Enter number of success (530 in this case) > Enter number of observations (1002)in this case> Adjust the confidence interval if needed > Compute. Choose C: There is 95​% confidence that the true proportion of worried adults is between 0.498 and 0.560. rounded 3 dec.

The accompanying data represent the total travel tax​ (in dollars) for a​ 3-day business trip in 8 randomly selected cities. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts​ (a) through​ (c) below. ​(a) Determine a point estimate for the population mean travel tax. b) Construct and interpret a 95​% confidence interval for the mean tax paid for a​ three-day business trip. ​(c) What would you recommend to a researcher who wants to increase the precision of the​ interval, but does not have access to additional​ data?

a) 83.27 - this is the mean you find in stat > columns > select mean compute. b) One can be 95​% confident that the mean travel tax for all cities is between ​$73.06 and $93.48. We got 73.06 & 93.48 by using Stat> T Stats > One Sample > with DATA> select value & enter confidence interval then compute. c) The researcher could decrease the level of confidence.

Fill in the blanks to complete the statement. The​ _______ _______ is a statement of no change, no effect, or no difference.

null hypothesis .

Compute the critical value zα/2 that corresponds to a 98​% level of confidence.

- 2.33 Use stat normal calculator, select between, leave mean & std dev as is only enter the =0.98. compute.

Explain what a​ P-value is. What is the criterion for rejecting the null hypothesis using the​ P-value approach? What is the criterion for rejecting the null hypothesis using the​ P-value approach? Choose the correct answer below.

- A​ P-value is the probability of observing a sample statistic as extreme or more extreme than the one observed under the assumption that the statement in the null hypothesis is true. - If P-value <α​, reject the null hypothesis.

One​ year, the mean age of an inmate on death row was 40.4 years. A sociologist wondered whether the mean age of a​ death-row inmate has changed since then. She randomly selects 32 death-row inmates and finds that their mean age is 38.7​, with a standard deviation of 9.4. Construct a​ 95% confidence interval about the mean age. What does the interval​ imply? Choose the correct hypotheses. H0​:[_] [_] [_] H1​: [_] [_] [_] Fill in blanks: Construct a​ 95% confidence interval about the mean age. With​ 95% confidence, the mean age of a death row inmate is between ______ years and ______ years. What does the interval​ imply?

- H0​:μ =40.4 H1​: μ ≠ 40.4 - With​ 95% confidence, the mean age of a death row inmate is between 35.31 years and 42.09 years. To get the blanks use T stats, enter mean in this case was 38.7, std dev, sample size and select the confidence interval > compute. Round 2 dec. - A. Since the mean age from the earlier year is contained in the​ interval, there is not sufficient evidence to conclude that the mean age had changed.

In a survey conducted by the Gallup​ Organization, 1100 adult Americans were asked how many hours they worked in the previous week. Based on the​ results, a​ 95% confidence interval for the mean number of hours worked had a lower bound of 42.7 and an upper bound of 44.5. Provide two recommendations for decreasing the margin of error of the interval.

- Increase the sample size. - Decrease the confidence level.

Describe the sampling distribution of p. Assume the size of the population is 25,000. n=600​, p=0.2. Choose the phrase that best describes the shape of the sampling distribution of p hat below. Determine the mean of the sampling distribution of p hat. Determine the standard deviation of the sampling distribution of p hat.

- It's B: Approximately normal because n≤0.05N and np(1-p)>10. - mean for p hat is 0.2. - op = 0.016. To find the str dev of the sampling distribution of p hat do the below. Use desmos - sqrt all - 0.2(1-0.2) / 600.

A simple random sample of size n=53 is obtained from a population with μ=79 and σ=9. Does the population need to be normally distributed for the sampling distribution of x to be approximately normally​ distributed? Why? What is the sampling distribution of x​? What is the sampling distribution of x​? Select the correct choice below and fill in the answer boxes within your choice.

- It's C - No because the Central Limit Theorem states that regardless of the shape of the underlying​ population, the sampling distribution of x becomes approximately normal as the sample​ size, n, increases. - It's A - The sampling distribution of x is normal or approximately normal with μx=79 and σx=1.236 - you get this by doing 9/sqrt53.

Without doing any​ computation, decide which has a higher​ probability, assuming each sample is from a population that is normally distributed with μ=100 and σ=15. Explain your reasoning. ​(a)​ P(90≤x≤​110) for a random sample of size n=10 ​(b)​ P(90≤x≤​110) for a random sample of size n=20

- P(90≤x≤​110) for a random sample of size n=20 has a higher probability. As n​increases, the standard deviation decreases

Explain what​ "statistical significance" means.

- Statistical significance means that the result observed in a sample is unusual when the null hypothesis is assumed to be true.

Fill in the blanks to correctly complete the sentence below. Suppose a simple random sample of size n is drawn from a large population with mean μ and standard deviation σ. The sampling distribution of x has mean μx=​______ and standard deviation σx=​______.

- Suppose a simple random sample of size n is drawn from a large population with mean μ and standard deviation σ. The sampling distribution of x has mean μx = μ and standard deviation σx= σ / sqrt n.

Complete the sentence below. The​ _____ _____, denoted p​, is given by the formula p=​_____, where x is the number of individuals with a specified characteristic in a sample of n individuals.

- The sample proportion ​denoted p hat​, is given by the formula p hat= x/n​, where x is the number of individuals with a specified characteristic in a sample of n individuals.

A​ government's congress has 944​members, of which 101 are women. An alien lands near the congress building and treats the members of congress as as a random sample of the human race. He reports to his superiors that a​ 95% confidence interval for the proportion of the human race that is female has a lower bound of 0.087 and an upper bound of 0.127. What is wrong with the​ alien's approach to estimating the proportion of the human race that is​ female?

- The sample is not a simple random sample.

In a previous​ year, 56​% of females aged 15 and older lived alone. A sociologist tests whether this percentage is different today by conducting a random sample of 650 females aged 15 and older and finds that 374 are living alone. Is there sufficient evidence at the α=0.05 level of significance to conclude the proportion has​ changed? Fill in blanks. Because np(1−p)=[__]​ [_] 10, the sample size is __________ 5% of the population​ size, and the sample ___________________________, the requirements for testing the hypothesis __________ satisfied. Fill in blanks What are the null and alternative​ hypotheses? H0​: [_] [_] [_] versus H1​: [_] [_] [_] Find the test statistic for this hypothesis test. z0 = Determine the​ P-value for this hypothesis test. State the conclusion for this hypothesis test.

Fill in blanks. Because np(1−p)=160.2 >​10, the sample size is less than 5% of the population​ size, and the sample is given to be random, the requirements for testing the hypothesis are satisfied. **by using desmos to get the 160.2- np (1- p) - H0 P = 0.56 H1 P ≠ 0.56 - z0 = 0.79 Round 2 dec. **use stat > proportion stat > one sample > w summary > enter number of success & observation, adjust the H0 & H1 to match above and leave confidence level as is > compute. - p-value = 0.429 rounded 3 dec. P is given w step above. -Do not reject H0. There is not sufficient evidence at the α=0.05 level of significance to conclude that the proportion of females who are living alone has changed.

Twenty years​ ago, 51​% of parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. A recent survey found that 217 of 700 parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. Do parents feel differently today than they did twenty years​ ago? Use the α=0.1 level of significance. Fill in blanks. Because np(1−p)=[__]​ [_] 10, the sample size is __________ 5% of the population​ size, and the sample ___________________________, the requirements for testing the hypothesis __________ satisfied. Fill in blanks What are the null and alternative​ hypotheses? H0​: [_] [_] [_] versus H1​: [_] [_] [_] Find the test statistic. z0= Find the​ P-value. Determine the conclusion for this hypothesis test. Choose the correct answer below.

Fill in blanks. Because np(1−p)=174.91 >​10, the sample size is less than 5% of the population​ size, and the sample can be reasonably assumed to be random, the requirements for testing the hypothesis are satisfied. **you get 174.9 by using desmos - np (1- p) P = 51% > 0.51 N = 700 700 (0.51)(1-0.51). - H0 P = 0.51 H1 P ≠ 0.51 - z0 = -10.59 **use stat > proportion stat > one sample > w summary > enter number of success & observation, adjust the H0 & H1 to match above and leave confidence level as is > compute. Round 2 dec. - P-value = 0.000 Rounded 3 dec. Step above also gave the p-value. - Since ​P-value<α​, reject the null hypothesis and conclude that there is sufficient evidence that parents feel differently today.

In a​ survey, 37​% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too​ high, so he randomly selected 200 pet owners and discovered that 69 of them spoke to their pet on the telephone. Does the veterinarian have a right to be​ skeptical? Use the α=0.1 level of significance. Fill in blanks. Because np(1−p)=[__]​ [_] 10, the sample size is __________ 5% of the population​ size, and the sample ___________________________, the requirements for testing the hypothesis __________ satisfied. Fill in blanks What are the null and alternative​ hypotheses? H0​: [_] [_] [_] versus H1​: [_] [_] [_] Determine the test​ statistic, z0. Determine the critical​ value(s). Select the correct choice below and fill in the answer box to complete your choice. a. zα= b. ±zα/2 = ± Does the veterinarian have a right to be​ skeptical?

Fill in blanks. Because np(1−p)=46.6 >​10, the sample size is less than 5% of the population​ size, and the sample is given to be random, the requirements for testing the hypothesis are satisfied **by using desmos to get the 46.6- np (1- p) - H0​: p = 0.37 versus H1​: p < 0.37 -zo = -A. za = -1.28 rounded 2 dec. You get this by using normal cal in statcrunch, only enter the = 0.1 (your level of significance given) > compute. Use what's given in P X < part. - The veterinarian does not have a right to be skeptical. There is not sufficient evidence to conclude that the true proportion of pet owners who talk to their pets on the telephone is less than 37​%.

A baseball​ pitcher's most popular pitch is a​ four-seam fastball. The data below represent the pitch speed​ (in miles per​ hour) for a random sample of 15 of his​ four-seam fastball pitches. ​(a) Is​ "pitch speed" a quantitative or qualitative​ variable? Why is it important to know this when determining the type of confidence interval you may​ construct? ​(b) Draw a normal probability plot to verify that​ "pitch speed" could come from a population that is normally distributed. Which normal probability plot below represents the​ data? ​(c) Draw a boxplot to verify the data set has no outliers. ​(d) Are the requirements for constructing a confidence interval for the mean pitch speed of the​ pitcher's four-seam fastball​ satisfied? ​(e) Construct and interpret a​ 95% confidence interval for the mean pitch speed of the​ pitcher's four-seam fastball. Select the correct choice below​ and, if​ necessary, fill in the answer boxes to complete your choice. f) Do you believe that a​ 95% confidence interval for the mean pitch speed of​ four-seam fastballs for all pitchers in a league would be narrower or​ wider? Why?

The variable​ "pitch speed" is a quantitative variable. This is important to know because confidence intervals for a mean are constructed on quantitative data while confidence intervals for a proportion are constructed on qualitative data with two possible outcomes. b) Stat crunch. Go to Graph > QQ Plot > select variable, correlation & normal quantiles on y-axis> compute. Then select the correct graph. c) graph>box plot> select variable and hozontal compute. select the correct graph on the test question. d) A​ t-interval can be constructed because there are no outliers and the data can reasonably be assumed to come from a normal distribution. e) The​ 95% confidence interval is 88.12 ,91.04. Use Stat first generate your mean & std dev. Them go to T Stats > one sample > w summary > enter mean, std dev, sample size and confidence interval. > compute. f) The interval for all pitchers in a league would be wider because the variability between pitchers is most likely greater than the variability between pitches for a pitcher.

The mean waiting time at the​ drive-through of a​ fast-food restaurant from the time an order is placed to the time the order is received is 88.2 seconds. A manager devises a new​ drive-through system that he believes will decrease wait time. As a​ test, he initiates the new system at his restaurant and measures the wait time for 10 randomly selected orders. The wait times are provided in the table to the right. Complete parts​ (a) and​ (b) below. ​(a) Because the sample size is​ small, the manager must verify that the wait time is normally distributed and the sample does not contain any outliers. The normal probability plot is shown below and the sample correlation coefficient is known to be r=0.988. Are the conditions for testing the hypothesis​ satisfied? ​(b) Is the new system​ effective? Conduct a hypothesis test using the​ P-value approach and a level of significance of α=0.01. First determine the appropriate hypotheses. H0​: [_] [_] 88.2 H1​: [_] [_] 88.2 Find the test statistic. Find the​ P-value Use the α=0.01 level of significance. What can be concluded from the hypothesis​ test?

Yes, the conditions are satisfied. The normal probability plot is linear ​enough, since the correlation coefficient is greater than the critical value. In​addition, a boxplot does not show any outliers. The graph attached shows it's linear, we know the correlation is 0.988 and by looking at the table of the critical value we know the correlation is greater. You compare it to the critical value that is noted for your sample size. b) H0​: μ = 88.2 H1​: μ < 88.2 - t0 = -2.10 rounded 2 dec. First open the table provided(image attached) in stat crunch, select columns to get the mean & std dev. Then use Stat T stats, enter mean, std dev, sample size, H0 / H1 and leave confidence interval as is. > compute. - p value= 0.033 rounded 3 dec. Found with the above. - The​ P-value is greater than the level of significance so there is not sufficient evidence to conclude the new system is effective.

The reading speed of second grade students in a large city is approximately​ normal, with a mean of 89 words per minute​ (wpm) and a standard deviation of 10 wpm. Complete parts​ (a) through​ (f). ​(a) What is the probability a randomly selected student in the city will read more than 93 words per​ minute? Interpret this probability. Select the correct choice below and fill in the answer box within your choice. (b) What is the probability that a random sample of 12 second grade students from the city results in a mean reading rate of more than 93 words per​ minute? Interpret this probability. Select the correct choice below and fill in the answer box within your choice (c) What is the probability that a random sample of 24 second grade students from the city results in a mean reading rate of more than 93 words per​ minute? Interpret this probability. Select the correct choice below and fill in the answer box within your choice. ​(d) What effect does increasing the sample size have on the​ probability? Provide an explanation for this result. ​(e) A teacher instituted a new reading program at school. After 10 weeks in the​ program, it was found that the mean reading speed of a random sample of 22 second grade students was 91.2 wpm. What might you conclude based on this​ result? Select the correct choice below and fill in the answer boxes within your choice. ​(f) There is a​ 5% chance that the mean reading speed of a random sample of 22 second grade students will exceed what​ value?

a) 0.3446 - Use stat normal calculator. Interpretation: If 100 different students were chosen from this​ population, we would expect 34 to read more than 93 words per minute. b) 0.0829 Use stat normal calculator but in the std dev this time I put 10/sqrt 12 to get the correct answer. Interpretation - If 100 different samples of n=12 students were chosen from this​ population, we would expect 8 ​sample(s) to have a sample mean reading rate of more than 93 words per minute. c) 0.0250 same thing stat normal cal and same as pt B, update the std dev with a sqrt of what C is asking for (this case 24). Interpretation: If 100 different samples of n=24 students were chosen from this​ population, we would expect 3 (this was rounded) sample(s) to have a sample mean reading rate of more than 93 words per minute. d) Increasing the sample size decreases the probability because σx decreases as n increases e & f - got it wrong. Need to review.

According to a​ report, the mean of monthly cell phone bills was ​$49.99 three years ago. A researcher suspects that the mean of monthly cell phone bills is higher today. ​(a)Determine the null and alternative hypotheses. ​(b)Explain what it would mean to make a Type I error. ​(c)Explain what it would mean to make a Type II error. (a) State the hypotheses. Fill in blanks. H0​:[_] [_] $[_] H1​: [_] [_] $[_] ​(b) Explain what it would mean to make a Type I error. Choose the correct answer below. ​(c) Explain what it would mean to make a Type II error. Choose the correct answer below.

a) H0​: μ = $49.99 H1​: μ > $49.99 For above we know it's μ because in the sentence it's mentioned $49.99 is the mean. H0 is always "=". H1 ">" is greater than in this case, we know this because the sentence says "higher today" meaning greater. And our number is $49.99 since it was provided. For B & C remember the following to determine the answer. Type 1 error - rejecting H0 in favor of H1, when in fact H0 is true. Type II error- Not rejecting H0 when in H1 is true. b) The sample evidence led the researcher to believe the mean of the monthly cell phone bill is higher than ​$49.99​, when in fact the mean of the bill is $49.99 c) The sample evidence did not lead the researcher to believe the mean of the monthly cell phone bill is higher than ​$49.99​, when in fact the mean of the bill is higher than $49.99.

A simple random sample of size n=64 is obtained from a population with μ=82 and σ=16. ​(a) Describe the sampling distribution of x. ​(b) What is P x>84.5​? ​(c) What is P x≤77.1​? ​(d) What is P 80.7<x<85​?

a) The distribution is approximately normal. Mean = 82. σx = 2 - to find use desmos 16/sqrt 64. b) 0.1056 - Find using normal calculator in stat.crunch. c) 0.0071 - Find using normal calculator in stat.crunch. d) 0.6753 - Find using normal calculator in stat.crunch.

In a survey of 2075 adults in a certain country conducted during a period of economic​ uncertainty, 66​% thought that wages paid to workers in industry were too low. The margin of error was 2 percentage points with 95​% confidence. For parts​ (a) through​ (d) below, which represent a reasonable interpretation of the survey​ results? For those that are not​ reasonable, explain the flaw. ​(a) We are 95​% confident 66​% of adults in the country during the period of economic uncertainty felt wages paid to workers in industry were too low. Is the interpretation​ reasonable? (b) We are 93​% to 97​% confident 66​% of adults in the country during the period of economic uncertainty felt wages paid to workers in industry were too low. Is the interpretation​ reasonable? ​(c) We are 95​% confident that the interval from 0.64 to 0.68 contains the true proportion of adults in the country during the period of economic uncertainty who believed wages paid to workers in industry were too low. Is the interpretation​ reasonable? ​(d) In 95​% of samples of adults in the country during the period of economic​ uncertainty, the proportion who believed wages paid to workers in industry were too low is between 0.64 and 0.68. Is the interpretation​ reasonable?

a) The interpretation is flawed. The interpretation provides no interval about the population proportion b) The interpretation is flawed. The interpretation indicates that the level of confidence is varying. c) The interpretation is reasonable. d) The interpretation is flawed. The interpretation suggests that this interval sets the standard for all the other​ intervals, which is not true.

According to a study conducted by a statistical​ organization, the proportion of people who are satisfied with the way things are going in their lives is 0.72. Suppose that a random sample of 100 people is obtained. Complete parts​ (a) through​ (e) below. (a) Suppose the random sample of 100 people is​ asked, "Are you satisfied with the way things are going in your​ life?" Is the response to this question qualitative or​ quantitative? Explain. ​(b) Explain why the sample​ proportion, p hat​, is a random variable. What is the source of the​ variability? ​(c) Describe the sampling distribution of p​ hat, the proportion of people who are satisfied with the way things are going in their life. Be sure to verify the model requirements. ​(d) In the sample obtained in part​ (a), what is the probability that the proportion who are satisfied with the way things are going in their life exceeds 0.74​? ​(e) Using the distribution from part​ (c), would it be unusual for a survey of 100 people to reveal that 66 or fewer people in the sample are satisfied with their​ lives?

a) The response is qualitative because the responses can be classified based on the characteristic of being satisfied or not. b)The sample proportion p hat is a random variable because the value of p hat varies from sample to sample. The variability is due to the fact that different people feel differently regarding their satisfaction. c) Since the sample size is no more than​ 5% of the population size and np(1−​p)=20.16≥​10, the distribution of p hat is approximately normal with μp=0.72 and σp=0.045 To find 20.16 use desmos enter - n x mean x 1-mean (100(0.72)(-0.72) = 20.16 To find 0.045 use desmos enter - sqrt 0.72 (1-0.72) / 100. d) 0.3283 (rounded 4 decimals, except is wasn't.) To get the above use stat normal cal. Enter mean (0.72), std dev (0.045) & > 0.74 (find this in the question for d.). e) The probability that 66 or fewer people in the sample are satisfied is 0.0912​ (found using stat normal), which is not unusual because this probability is not less than 5%.

The shape of the distribution of the time required to get an oil change at a 10​-minute ​oil-change facility is unknown.​ However, records indicate that the mean time is 11.8 minutes​, and the standard deviation is 4.2 minutes. Complete parts​ (a) through​ (c) below. ​(a) To compute probabilities regarding the sample mean using the normal​ model, what size sample would be​ required? Choose the required sample size below. ​(b) What is the probability that a random sample of n=45 oil changes results in a sample mean time less than 10 minutes? (c) Suppose the manager agrees to pay each employee a​ $50 bonus if they meet a certain goal. On a typical​ Saturday, the​ oil-change facility will perform 45 oil changes between 10 A.M. and 12 P.M. Treating this as a random​ sample, at what mean​ oil-change time would there be a​ 10% chance of being at or​ below? This will be the goal established by the manager.

a) The sample size needs to be greater than 30. b) The probability is approximately 0.0020. Rounded 4 decimals. Use stat normal cal. Enter the std dev as o / sqrt n (4.2/sqrt(45). Then update the P x < 10, then compute. c) There would be a​ 10% chance of being at or below 11 minutes. Rounded one decimal. To find the 11, use stat normal cal. Enter mean and std dev (like in pt B). Leave PX BLANK. Enter = 0.10 (10%). Compute. Round one decimal.

The acceptable level for insect filth in a certain food item is 4 insect fragments​ (larvae, eggs, body​ parts, and so​ on) per 10 grams. A simple random sample of 40 ​ten-gram portions of the food item is obtained and results in a sample mean of x=4.2 insect fragments per​ ten-gram portion. Complete parts​ (a) through​ (c) below (a) Why is the sampling distribution of x approximately​ normal? ​(b) What is the mean and standard deviation of the sampling distribution of x assuming μ=4 and σ=4​? ​(c) What is the probability a simple random sample of 40 ​ten-gram portions of the food item results in a mean of at least 4.2 insect​ fragments? Is this result​ unusual? What might we​ conclude?

a) The sampling distribution is approximately normal because the sample size is large enough. b) μx = 4 (this is mean (μ), doesn't change) σx = 0.316 - to find use desmos sqrt4/sqrt40. *both rounded 3 decimals. c) ​P(x≥4.2​)= 0.2634. rounded 4 decimals. Use stat normal cal, enter mean & std dev from pt B, then P(x>4.2) - then compute. - This result is not unusual because its probability is large. - Since this result is not unusual, it is not reasonable to conclude that the population mean is higher than 4.

Complete parts ​(a​) through ​(d) for the sampling distribution of the sample mean shown in the accompanying graph. LOADING... Click the icon to view the graph. (a) What is the value of μx​? ​(b) What is the value of σx​? ​(c) If the sample size is n=16​, what is likely true about the shape of the​ population? ​(d) If the sample size is n=16​, what is the standard deviation of the population from which the sample was​ drawn?

a) mean = 700, you determine this by looking at the graph provided. b) std dev= 20. you get this by also looking at the graph, so the mean 700 + 20 = 720. c) It's B = the shape of the population is approximately normal. d) Std dev of the population is = 80 **Use desmos, use the 20 from pt B then multiply it by sqrt 16. 20 x sqrt16 = 80.

A researcher wishes to estimate the proportion of adults who have​ high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.03 with 90​% confidence if ​(a) she uses a previous estimate of 0.54​? ​(b) she does not use any prior​ estimates?

a) n = 747 Use stat crunch. Stat > Proportion Stat> Power / Sample size > Enter Confidence level = 0.90 > Width = 0.06 (0.03+0.03) > Target proportion = 0.54 > Compute b) 752 - no idea how. It's supposed to be the above, but doesn't work.

Suppose Jack and Diane are each attempting to use a simulation to describe the sampling distribution from a population that is skewed right with mean 60 and standard deviation 15. Jack obtains 1000 random samples of size n=4 from the​ population, finds the mean of the​ means, and determines the standard deviation of the means. Diane does the same​ simulation, but obtains 1000 random samples of size n=35 from the population. Complete parts​ (a) through​ (c) below. (a) Describe the shape you expect for Jack​'s distribution of sample means. Describe the shape you expect for Diane​'s distribution of sample means. Choose the correct answer below. (b) What do you expect the mean of Jack​'s distribution to​ be? What do you expect the mean of Diane​'s distribution to​ be?

a) ​Jack's distribution is expected to be skewed right​, but not as much as the original distribution. Diane​'s distribution is expected to be approximately normal. b) Jack​'s distribution is expected to have a mean of 60. Diane​'s distribution is expected to have a mean of 60. c) Jack​'s distribution is expected to have a standard deviation of 7.5. Diane​'s distribution is expected to have a standard deviation of 2.54. You get these by std dev 15 / sqrt over the sample size (n) to get Jacks & Dianes std devs.

Suppose a simple random sample of size n=200 is obtained from a population whose size is N=25,000 and whose population proportion with a specified characteristic is p=0.8. Complete parts ​(a) through​ (c) below. (a) Describe the sampling distribution of p. Choose the phrase that best describes the shape of the sampling distribution below. Determine the mean of the sampling distribution of p hat. Determine the standard deviation of the sampling distribution of p hat. b) What is the probability of obtaining x=170 or more individuals with the​ characteristic? That​ is, what is ​P(p≥0.85​)? ​(c) What is the probability of obtaining x=148 or fewer individuals with the​ characteristic? That​ is, what is ​P(p≤0.74​)?

a)Approximately normal because n≤0.05N and np(1-p)>10 Mean = 0.8 Std. dev. = 0.028284 (rounded 6 decimals) You find this by using desmos. sqrt: 0.8 (1-0.8) / 200 (n). b) 0.0385. Use stat normal cal, enter mean & std dev from pt A, enter > 0.85 in p and compute. Don't worry about the x value they give you in the question. c) 0.0169 Use stat normal cal, enter mean & std dev from pt A, enter < 0.74 in p and compute. Don't worry about the x value they give you in the question both C & B are rounded 4 decimals.

According to a survey in a​ country, 23​% of adults do not own a credit card. Suppose a simple random sample of 400 adults is obtained. Complete parts​ (a) through​ (d) below. ​(a) Describe the sampling distribution of p hat​, the sample proportion of adults who do not own a credit card. Choose the phrase that best describes the shape of the sampling distribution of p hat below. Determine the mean of the sampling distribution of p hat. Determine the standard deviation of the sampling distribution of p hat. ​(b) What is the probability that in a random sample of 400 ​adults, more than 27​% do not own a credit​ card? (c) What is the probability that in a random sample of 400 adults, between 22​% and 27​% do not own a credit​ card? (d) Would it be unusual for a random sample of 400 adults to result in 88 or fewer who do not own a credit​ card? Why? Select the correct choice below and fill in the answer box to complete your choice.

a)Approximately normal because n≤0.05N and np(1-p)>10 Mean = 0.23 rounded 2 dec. std dev = 0.021 rounded 3 dec. Found using desmos - sqrt 0.23(1-0.23) / 400 b) 0.0284 rounded 4 dec.- use stat normal cal. If 100 different random samples of 400 adults were​ obtained, one would expect 3 (rounded from b's first answer) to result in more than 27​% not owning a credit card c) 0.6546 rounded 4 dec. use stat crunch normal cal. If 100 different random samples of 400 adults were​ obtained, one would expect 65 (found using by rounding above) to result in between 22​% and 27​% not owning a credit card. d) The result is not unusual because the probability that p hat is less than or equal to the sample proportion is 0.3170 (rounded4 dec)​, which is greater than​ 5%. Found above using stat normal, in P (x < 88/400) = > compute.

Suppose a simple random sample of size n=1000 is obtained from a population whose size is N=2,000,000 and whose population proportion with a specified characteristic is p=0.79. Complete parts​ (a) through​ (c) below. (a) Describe the sampling distribution of p hat. ​(b) What is the probability of obtaining x=820 or more individuals with the characteristic?

a)Approximately​ normal, μp=0.79 and σp≈0.0129 To get op use Desmos sqrt 0.79 (1-0.79) / 1000. **remember to round to the 4th decimal for your answer.** b) 0.0100 To get answer use stat crunch normal calculator. In the P(x> enter 820/1000) = 0.01002045. c) 0.0605 To get answer use stat crunch normal calculator. In the P(x < enter 770/1000) = 0.06052426.