Qmeth Test 2 - 8/12/2020

Given: a i/Pi table, n = 803 I Pi 1 0.016 2 0.41 3 0.17 4 0.26 1) The expected count for outcome 1 is ________ 2) the expected count for outcome 2 is _________ 3) the expected count for outcome 3 is _________ 4) the expected count for outcome 4 is __________

(n)(Pi) 1) 128.48 = (0.16)(803) 2) 329.23 = (.41)(803) 3) 136.51 (.17)(803) 4) 208.78 (.26)(803)

If the consequences of making a Type I error are​ severe, would you choose the level of​ significance, α​, to equal​ 0.01, 0.05, or​ 0.10? Choose the correct answer below. A. 0.05 B. 0.01 C. 0.10

0.01 *choose a small a to make it difficult to reject H0

Determine the point estimate of the population​ proportion, the margin of error for the following confidence​ interval, and the number of individuals in the sample with the specified​ characteristic, x, for the sample size provided. Lower bound=0.253​, upper bound=0.827​, n= 1500 1) the point estimate of the population proportion is _______ 2) the margin of error is __________ 3) the number of individuals in the sample with the specified characteristic is ________

1) 0.54 P.E. = (lower + upper)/2 2) 0.287 (upper - lower)/2 3) 810 p-hat = x/n (rearrange to x = n(p-hat)), p-hat = P.E.

how to find the chi-square test statistic with two variables on statcrunch

1) inport data 2) stat 3) tables 4) contringency 5) data (raw data) or summary 6) select columns 7) select the row 8) row percent or column percent 9) the test statistic is referred to as value

Determine μx and σx from the given parameters of the population and sample size. μ=78​, σ=18​, n=81 1) μx = _______ 2) σx = _____

1) μx = 78 2) σx = 2

Twenty years​ ago, 49​% 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 252 of 800 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. 1) Because np01−p0=(a)________, (b)________ (≠, >, =, <​) 10, the sample size is (c)_________ (less than, greater than) ​5% of the population​ size, and the sample (d)_________________ (can be reasonably assumed to be random, is given to not be random, cannot be reasonably assumed to be random, is given to be random) the requirements for testing the hypothesis (e)________ (are not, are) satisfied. 2) H0 = H1 = 3) Find the test statistic z0 = _______ 4) P-value = ________ 5) Determine the conclusion for this hypothesis test. Choose the correct answer below. A. Since ​P-value<α​, reject the null hypothesis and conclude that there is sufficient evidence that parents feel differently today. B. Since ​P-value>α​, do not reject the null hypothesis and conclude that there is not sufficient evidence that parents feel differently today. C. Since ​P-value<α​, do not reject the null hypothesis and conclude that there is sufficient evidence that parents feel differently today. D. Since ​P-value>α​, reject the null hypothesis and conclude that there is not sufficient evidence that parents feel differently today.

1. (a) 199.9 (800)(.49)(.51), (b) >, (c) less than, (d) can be reasonably assumed to be random, (e) are 2. H0: p = .49 H1: p ≠ .49 3. on statcrunch, stat, proportion stat, one, summary, input 4. same as #3 5. A) Since ​P-value<α​, reject the null hypothesis and conclude that there is sufficient evidence that parents feel differently today.

The null and alternative hypotheses are given. Determine whether the hypothesis test is​ left-tailed, right-tailed, or​ two-tailed. What parameter is being​ tested? H0​: σ = 125 H1​: σ < 125 1. Is the hypothesis test​ left-tailed, right-tailed, or​ two-tailed? 2. what parameter is being tested? population proportion, population mean or population standard deviation

1. left-tailed test (right-tailed test = H0​: σ = 7 H1​: σ > 7 left-tailed test = H0​: σ = 7 H1​: σ < 7 two-tailed test = H0​: σ = 7 H1​: σ doesn't equal 7) 2. population standard deviation

Find the​ z-scores that separate the middle 23​% of the distribution from the area in the tails of the standard normal distribution. the z-scores are: _____,_______

1. stat 2. calculator 3. normal 4. click between rather than standard 5. set the answer as 0.23 so it gives you the two x values

Find the value of Za Z0.01 =

1. stat 2. calculator 3. normal 4. set as x>/- _______ (the area should be in the right tail) 4. set the answer to the subscript of z in order to find the x value

Determine the area under the standard normal curve that lies to the left of ​(a) Z=−1.48, ​(b) Z=−0.74​, ​(c) Z=1.43​, and​ (d) Z=0.76. a) the area to the left of Z = -1/48 is ________ b) the area to the left of Z = -0.74 is _______ c) the area to the left of Z = 1.43 is ________ d) the area to the left of Z = 0.76 is __________ explain how to do it on statcrunch

1. stat 2. calculator 3. normal 4. whether to put greater than or less then is indicated on the graph 5. set the x value with the indicated Z

Determine the area under the standard normal curve that lies to the left of ​(a) Z=0.09, (b) Z=−0.44​,(c) Z=−1.59​, and ​(d) Z=0.64.

1. stat 2. calculator 3. normal 4. whether to put greater than or less then is indicated on the graph 5. set the x value with the indicated Z

how to find test statistic (z0) and P-value on stat crunch when data is provided

1. stat 2. proportion stat 3. one sample 4. with data 5. select the column 6. success: write Liberal 7. input the hypothesis test for p data

Fill in the blank to complete the statement. The area under the normal curve to the right of μ equals​ _______.

1/2

Two​ researchers, Jaime and​ Mariya, are each constructing confidence intervals for the proportion of a population who is​ left-handed. They find the point estimate is 0.18. Each independently constructed a confidence interval based on the point​ estimate, but​ Jaime's interval has a lower bound of 0.097 and an upper bound of 0.212​, while​ Mariya's interval has a lower bound of 0.104 and an upper bound of 0.256. Which interval is​ wrong? Why? Choose the correct answer below. A. Jaime​'s interval is wrong because it is not centered on the point estimate. B. Mariya​'s interval is wrong because it does not include the point estimate. C. Mariya​'s interval is wrong because it is too wide. D. Jaime​'s interval is wrong because it is too narrow

A. Jaime​'s interval is wrong because it is not centered on the point estimate. *we know this by adding the lower and upper bounds and then dividing by 2 and seeing which person has the point estimate as the center

Complete the statement below. The points at x=​_______ and x=​_______ are the inflection points on the normal curve. A. The points are x=μ−σ and x=μ+σ. B. The points are x=μ−2σ and x=μ+2σ. C. The points are x=μ−3σ and x=μ+3σ.

A. The points are x=μ−σ and x=μ+σ.

To compute probabilities regarding the sample mean using the normal​ model, what size sample would be​ required? Choose the required sample size below. A. The sample size needs to be greater than 30. B. The normal model cannot be used if the shape of the distribution is unknown. C. The sample size needs to be less than 30. D. Any sample size could be used.

A. The sample size needs to be greater than 30.

The headline reporting the results of a poll​ stated, "Majority of Adults at Personal Best in the​ Morning." The results indicated that a survey of 1400 adults resulted in 55​% stating they were at their personal best in the morning. The​ poll's results were reported with a margin of error of 4​%. Explain why the​ poll's headline is accurate. Choose the correct answer below. A. More than 30 people were surveyed. B. All the values within the margin of error are greater than​ 50%. C. More than​ 50% of the respondents answered in the affirmative.

B. All the values within the margin of error are greater than​ 50%.

Determine whether the following statement is true or false. Sample evidence can prove that a null hypothesis is true.

False *although sample data is used to test the null hypothesis, it cannot be stated with 100% certainty that the null hypothesis is true. it can only be determined whether the sample data supports or does not support the null hypothesis

True or False​: The population proportion and sample proportion always have the same value

False *the population proportion and sample proportion do not always have the same value

What effect does increasing the sample size have on the​ probability?

Increasing the sample size decreases the probability because σx decreases as n increases.

What effect does increasing the sample size have on the​ probability? Provide an explanation for this result.

Increasing the sample size decreases the probability because σx decreases as n increases.

Determine if the following statement is true or false. The normal curve is symmetric about its​ mean, μ.

The statement is true. The normal curve is a symmetric distribution with one​ peak, which means the​ mean, median, and mode are all equal.​ Therefore, the normal curve is symmetric about the​ mean, μ.

True or False? The mean of the sampling distribution of p-hat is p

True * the mean of the sample distribution of the sampling proportion equals the population proportion, p

Fill in the blank to complete the statement. The notation zα is the​ z-score that the area under the standard normal curve to the right of zα is​ _______.

a

A researcher studying public opinion of proposed Social Security changes obtains a simple random sample of 30 adult Americans and asks them whether or not they support the proposed changes. To say that the distribution of the sample proportion of adults who respond​ yes, is approximately​ normal, how many more adult Americans does the researcher need to sample in the following​ cases? ​1) 15​% of all adult Americans support the changes ​2) 20​% of all adult Americans support the changes

a required condition for normal approximation of the binomial is that np(1-p)>/- 10 1) the researcher must ask 49 more American adults (30)(.15)(.85) = 3.825, need to change the n so take 10/(.15)(.85) = n = 79 - 30 = 49 2) more 10/(.2)(.8) = 62.5 so 63-30 = 33 *always round up

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

if the correlation coefficient is _________ (greater than or less than) the critical value and has no outliers, its linear, the conditions for testing the hypothesis is satisfied

greater than (correlation coefficient > the critical value)

The​ _______ represents the expected proportion of intervals that will contain the parameter if a large number of different samples of size n is obtained. It is denoted​ _______.

level of confidence (1-a)(100%)

Compute the critical value zα/2 that corresponds to a 86​% level of confidence.

on stat crunch stat calculator normal between set the answer as .86 lease mean = 0 and standard deviation = 1 gives two answers, use the positive one

A​ ________ ________ is the value of a statistic that estimates the value of a parameter.

point estimate

Fill in the blank in the statement below. The procedure for constructing a confidence interval about a mean is​ _______, which means minor departures from normality do not affect the accuracy of the interval. potential answers are: robust non-normal confident symmetric

robust

The standard deviation of the sampling distribution of x-bar​, denoted σx​, is called the​ _____ _____ of the​ _____.

standard error of the mean

A normal score is the expected​ z-score of a data​ value, assuming the distribution of the random variable is normal. Is this statement true or​ false?

the statement is true

what is the test statistic for chi-square (chapter 12) and how do you find it

x^2, 0 (x with a with two subscripts, top and bottom) = the test statistic steps to finding it 1) import the data to statcrunch 2) create a expected column by going data, compute, expression (multiply the percentages given by the total observations for each column) 3) make sure their is an observe column 4) stat 5) goodness-of-fit 6) chi-square test 7) set the observed and expected column 8) chi-square = the test statistic

The graph of a normal curve is given. Use the graph to identify the value of μ and σ. - the graph shows a normal distribution with 100 at the center and 2 points between each line

μ = 100 σ = 2

The first significant digit in any number must be​ 1, 2,​ 3, 4,​ 5, 6,​ 7, 8, or 9. It was discovered that first digits do not occur with equal frequency. Probabilities of occurrence to the first digit in a number are shown in the accompanying table. The probability distribution is now known as​ Benford's Law. For​ example, the following distribution represents the first digits in 216 allegedly fraudulent checks written to a bogus company by an employee attempting to embezzle funds from his employer. 1) Because these data are meant to prove that someone is guilty of​ fraud, what would be an appropriate level of significance when performing a​ goodness-of-fit test? use a = _______ (0.01, 0.05, 0.1) 2) Using the level of significance chosen in part​ (a), test whether the first digits in the allegedly fraudulent checks obey​ Benford's Law. Do the first digits obey​ Benford's Law? What are the null and alternative​ hypotheses? A. H0​: The distribution of the first digits in the allegedly fraudulent checks does not obey​ Benford's Law. H1​: The distribution of the first digits in the allegedly fraudulent checks obeys​ Benford's Law. B. H0​: The distribution of the first digits in the allegedly fraudulent checks obeys​ Benford's Law. H1​: The distribution of the first digits in the allegedly fraudulent checks does not obey​ Benford's Law.

1) 0.01 2) B. H0​: The distribution of the first digits in the allegedly fraudulent checks obeys​ Benford's Law. H1​: The distribution of the first digits in the allegedly fraudulent checks does not obey​ Benford's Law. 3)

Describe the sampling distribution of p-hat. Assume the size of the population is 25,000. n=900​, p=0.3 1) Choose the phrase that best describes the shape of the sampling distribution of p below. A. Approximately normal because n≤0.05N and np(1−p)<10. B. Not normal because n≤0.05N and np(1−p)≥10. C. Approximately normal because n≤0.05N and np(1−p)≥10. D. Not normal because n≤0.05N and 2) Determine the mean of the sampling distribution of p-hat, μp-hat = ______ 3) Determine the standard deviation of the sampling distribution of p-hat. σp-hat = ​

1) C. Approximately normal because n≤0.05N and np(1−p)≥10. 2) μp-hat = p so 0.3 3) σp-hat = square root of (p(1-p))/n) ((.3)(.7))/900 = 0.015

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. Select all that apply. A. Increase the sample size. B. Decrease the confidence level. C. Increase the confidence level. D. Use fewer degrees of freedom. E. Decrease the sample size. F. Decrease the standard deviation of hours worked.

Increase the sample size. Decrease the confidence level.

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-bar has mean μx=​______ and standard deviation σx=​______.

μx = μ σ = σ/square root of n

A manufacturer of colored candies states that 13​% of the candies in a bag should be​ brown, 14​% yellow, 13​% red, 24​% blue, 20​% orange, and 16​% green. A student randomly selected a bag of colored candies. He counted the number of candies of each color and obtained the results shown in the table. Test whether the bag of colored candies follows the distribution stated above at α=0.05 level of significance. Using the level of significance α=0.05​, test whether the color distribution is the same. 1) Determine the null and alternative hypotheses. Choose the correct answer below. A. H0​: The distribution of colors is at most as uniform as stated by the manufacturer. H1​: The distribution of colors is more uniform than stated by the manufacturer. B. H0​: The distribution of colors is not the same as stated by the manufacturer. H1​: The distribution of colors is the same as stated by the manufacturer. C. H0​: The distribution of colors is at least as uniform as stated by the manufacturer. H1​: The distribution of colors is less uniform than stated by the manufacturer. D. H0​: The distribution of colors is the same as stated by the manufacturer. H1​: The distribution of colors is not the same as stated by the manufacturer.

1) D. H0​: The distribution of colors is the same as stated by the manufacturer. H1​: The distribution of colors is not the same as stated by the manufacturer.

In a​ survey, 600 adults in a certain country were asked how many hours they worked in the previous week. Based on the​ results, a​ 95% confidence interval for mean number of hours worked was lower​ bound: 41.6 and upper​ bound: 45.1. Which of the following represents a reasonable interpretation of the​ result? For those that are not​ reasonable, explain the flaw. 1) There is a​ 95% chance the mean number of hours worked by adults in this country in the previous week was between 41.6 hours and 45.1 hours. 2) We are​ 95% confident that the mean number of hours worked by adults in this country in the previous week was between 41.6 hours and 45.1 hours. 3) 95% of adults in this country worked between 41.6 hours and 45.1 hours last week. 4) We are​ 95% confident that the mean number of hours worked by adults in a particular area of this country in the previous week was between 41.6 hours and 45.1 hours. potential answers = A. Flawed. This interpretation implies that the mean is only for last week. B. Flawed. This interpretation makes an implication about individuals rather than the mean. C. Correct. This interpretation is reasonable. D. Flawed. This interpretation implies that the population mean varies rather than the interval. E. Flawed. The interpretation should be about the mean number of hours worked by adults in the whole​ country, not about adults in the particular area.

1) Flawed. This interpretation implies that the population mean varies rather than the interval 2) Correct. This interpretation is reasonable 3) Flawed. This interpretation makes an implication about individuals rather than the mean 4) Flawed. The interpretation should be about the mean number of hours worked by adults in the whole​ country, not about adults in the particular area.

According to a food​ website, the mean consumption of popcorn annually by Americans is 64 quarts. The marketing division of the food website unleashes an aggressive campaign designed to get Americans to consume even more popcorn. 1) Determine the null and alternative hypotheses that would be used to test the effectiveness of the marketing campaign. H0: H1: 2) A sample of 809 Americans provides enough evidence to conclude that marketing campaign was effective. Provide a statement that should be put out by the marketing department. A. There is not sufficient evidence to conclude that the mean consumption of popcorn has risen. B. There is not sufficient evidence to conclude that the mean consumption of popcorn has stayed the same. C. There is sufficient evidence to conclude that the mean consumption of popcorn has risen. D. There is sufficient evidence to conclude that the mean consumption of popcorn has stayed the same. 3. Suppose, in​ fact, the mean annual consumption of popcorn after the marketing campaign is 64 quarts. Has a Type I or Type II error been made by the marketing​ department? If we tested this hypothesis at the α=0.1 level of​ significance, what is the probability of committing this​ error? Select the correct choice below and fill in the answer box within your choice. ​(Type an integer or a decimal. Do not​ round.) A. The marketing department committed a Type I error because the marketing department rejected the null hypothesis when it was true. The probability of making a Type I error is ________. B. The marketing department committed a Type I error because the marketing department did not reject the alternative hypothesis when the null hypothesis was true. The probability of making a Type I error is _________. C. The marketing department committed a Type II error because the marketing department rejected the null hypothesis when it was true. The probability of making a Type II error is _________. D. The marketing department committed a Type II error because the marketing department did not reject the alternative hypothesis when the null hypothesis was true. The probability of making a Type II error is __________.

1) H0: mean = 64 H1: mean > 64 2) (c) There is sufficient evidence to conclude that the mean consumption of popcorn has risen. 3) The marketing department committed a Type I error because the marketing department rejected the null hypothesis when it was true. The probability of making a Type I error is 0.1 (given by the a value given).

In a​ study, researchers wanted to measure the effect of alcohol on the hippocampal​ region, the portion of the brain responsible for​ long-term memory​ storage, in adolescents. The researchers randomly selected 15 adolescents with alcohol use disorders to determine whether the hippocampal volumes in the alcoholic adolescents were less than the normal volume of 9.02cm3. An analysis of the sample data revealed that the hippocampal volume is approximately normal with no outliers and x=8.19cm3 and s=0.7cm3. Conduct the appropriate test at the α=0.01 level of significance. 1) H0 = H1 = 2) t-statistic (t0) = _______ 3) P-value = _________ 4) __________ (reject, do not reject) the null hypothesis. there _______ (is, is not) sufficient evidence to claim that the mean hippocampal volume is ________ (less than or greater than) ______ cm^3

1) H0: mean = 9.02 H1: mean < 9.02 2 and 3) on statcrunch stat t-stats one sample with summary sample mean = 8.19 sample std. dev = 0.7 sample size = 15 hypothesis test 4) reject, is, less than, 9.02

For students who first enrolled in two year public institutions in a recent​ semester, the proportion who earned a​ bachelor's degree within six years was 0.394. The president of a certain college believes that the proportion of students who enroll in her institution have a higher completion rate. 1) Determine the null and alternative hypotheses. H0: H1: 2) Explain what it would mean to make a Type I error. A. The president rejects the hypothesis that the proportion of students who earn a​ bachelor's degree within six years is 0.394​, ​when, in​ fact, the proportion is greater than 0.394. B. The president fails to reject the hypothesis that the proportion of students who earn a​ bachelor's degree within six years is 0.394​, ​when, in​ fact, the proportion is greater than 0.394. C. The president rejects the hypothesis that the proportion of students who earn a​ bachelor's degree within six years is 0.394​, ​when, in​ fact, the proportion is 0.394. D. The president fails to reject the hypothesis that the proportion of students who earn a​ bachelor's degree within six years is 0.394​, ​when, in​ fact, the proportion is 0.394. ​3) Explain what it would mean to make a Type II error. A. The president fails to reject the hypothesis that the proportion of students who earn a​ bachelor's degree within six years is 0.394​, ​when, in​ fact, the proportion is 0.394. B. The president rejects the hypothesis that the proportion of students who earn a​ bachelor's degree within six years is 0.394​, ​when, in​ fact, the proportion is 0.394. C. The president rejects the hypothesis that the proportion of students who earn a​ bachelor's degree within six years is 0.394​, ​when, in​ fact, the proportion is greater than 0.394. D. The president fails to reject the hypothesis that the proportion of students who earn a​ bachelor's degree within six years is 0.394​, ​when, in​ fact, the proportion is greater than 0.394.

1) H0: p = 0.394 H1: p > 0.394 2) C. The president rejects the hypothesis that the proportion of students who earn a​ bachelor's degree within six years is 0.394​, ​when, in​ fact, the proportion is 0.394. (rejects the null hypothesis when its true) 3) D. The president fails to reject the hypothesis that the proportion of students who earn a​ bachelor's degree within six years is 0.394​, ​when, in​ fact, the proportion is greater than 0.394. (fails to reject the null hypothesis when the alternative is true)

Several years​ ago, the mean height of women 20 years of age or older was 63.7 inches. Suppose that a random sample of 45 women who are 20 years of age or older today results in a mean height of 65.1 inches. ​1) State the appropriate null and alternative hypotheses to assess whether women are taller today. H0 = H1 = ​2) Suppose the​ P-value for this test is 0.05. Explain what this value represents. A. There is a 0.05 probability of obtaining a sample mean height of exactly 65.1 inches from a population whose mean height is 63.7 inches. B. There is a 0.05 probability of obtaining a sample mean height of 63.7 inches or taller from a population whose mean height is 65.1 inches. C. There is a 0.05 probability of obtaining a sample mean height of 65.1 inches or shorter from a population whose mean height is 63.7 inches. D. There is a 0.05 probability of obtaining a sample mean height of 65.1 inches or taller from a population whose mean height is 63.7 inches. ​3) Write a conclusion for this hypothesis test assuming an α=0.10 level of significance. A. Do not reject the null hypothesis. There is sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today. B. Reject the null hypothesis. There is not sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today. C. Reject the null hypothesis. There is sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today. D. Do not reject the null hypothesis. There is not sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today.

1) H0​: μ=63.7 in. versus H1​: μ>63.7 in. 2) D. There is a 0.05 probability of obtaining a sample mean height of 65.1 inches or taller from a population whose mean height is 63.7 inches. 3) C. Reject the null hypothesis. There is sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today.

Suppose a researcher is testing the hypothesis H0​: p=0.6 versus H1​: p<0.6 and she finds the​ P-value to be 0.22. 1) Explain what this means. Would she reject the null​ hypothesis? Why? A. If the​ P-value for a particular test statistic is 0.22​, she expects results no more extreme than the test statistic in exactly 22 of 100 samples if the null hypothesis is true. B. If the​ P-value for a particular test statistic is 0.22​, she expects results no more extreme than the test statistic in about 22 of 100 samples if the null hypothesis is true. C. If the​ P-value for a particular test statistic is 0.22​, she expects results at least as extreme as the test statistic in exactly 22 of 100 samples if the null hypothesis is true. D. If the​ P-value for a particular test statistic is 0.22​, she expects results at least as extreme as the test statistic in about 22 of 100 samples if the null hypothesis is true. 2) Choose the correct conclusion below. A. Since this event is​ unusual, she will reject the null hypothesis. B. Since this event is not​ unusual, she will not reject the null hypothesis. C. Since this event is​ unusual, she will not reject the null hypothesis. D. Since this event is not​ unusual, she will reject the null hypothesis.

1) If the​ P-value for a particular test statistic is 0.22​, she expects results at least as extreme as the test statistic in about 22 of 100 samples if the null hypothesis is true. 2) B. Since this event is not​ unusual, she will not reject the null hypothesis.

A simple random sample of size n=49 is obtained from a population with μ=38 and σ=3. Does the population need to be normally distributed for the sampling distribution of x-bar to be approximately normally​ distributed? Why? What is the sampling distribution of x-bar​? 1) Does the population need to be normally distributed for the sampling distribution of x-bar to be approximately normally​ distributed? Why? 2) What is the sampling distribution of x​? the sample distribution of x-bar is _______________ (normal or approximately norm, unicorn, skewed left, follows students t-distribution) with μx=_______ and σx= ________

1) No because the Central Limit Theorem states that regardless of the shape of the underlying​ population, the sampling distribution of x-bar becomes approximately normal as the sample​ size, n, increases. 2) The sampling distribution of x-bar is normal or approximately normal with μx=38 and σx=0.429 (3/square root of 49)

According to a​ study, the proportion of people who are satisfied with the way things are going in their lives is 0.80. 1) 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. 2) ​Explain why the sample​ proportion, p-hat​, is a random variable. What is the source of the​ variability? 3) Since the sample size is (a)__________ (no less, no more) than​ 5% of the population size and np(1−​p)= (b)______ ≥​10, the distribution of p-hat is (c)____________ (skewed left, uniform, approximately normal, skewed right) with μp= (d)_______ and σp= (e)________ 4) the probability that proportion who satisfied with the way things are going int heir life exceeds 0.84 is _____ 5) the probability that 74 or fewer in the sample of 100 are satisfied is (a)_______, which (b)________ (is or is not) unusual because this probability (c)_______ (is or is not) less than (d)_______%

1) The response is qualitative because the responses can be classified based on the characteristic of being satisfied or not. 2) 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. 3) (a) no more, (b) 16 (.8)(.2)(100) (c) approximately normal (d) 0.8 (e) 0.04 (square of (.8)(.2)/100) 4) do this in statcrunch under normal 5) (a) 0.0668 (from statcrunch), (b) is not, (c) is not, (d) 5

Test the hypothesis using the​ P-value approach. Be sure to verify the requirements of the test. H0: p=0.89 versus H1: p≠0.89 n=500, x=440, α=0.1 1) Is np01−p0≥10​? Select the correct choice below and fill in the answer box to complete your choice. ​(Type an integer or a decimal. Do not​ round.) A. Yes, because np(1−p)= ________ B. No, because np(1−p)= ________ 2) p-hat = ________ 3) find the test statistic; z0 = _____ 4) P-value = _________ 5) _________ (reject or do not reject) the null hypothesis because the P-value is ___________ (greater than or less than) a

1) Yes, because np(1−p)= 48.95 (500)(.11)(.89) 2) p-hat = 0.88 (440/500) 3) z0 = -0.71 4) P-value = 0.475 *for both 3 and 4 go to statcrunch (stat, proportion stat, one, with summary, input) 4) do not reject, greater

Explain what a​ P-value is. What is the criterion for rejecting the null hypothesis using the​ P-value approach? 1) Explain what a​ P-value is. Choose the correct answer below. A. A​ P-value is the value used to designate the area α in either the​ left- or​ right-tail of the normal curve. B. A​ P-value is the number of standard deviations that the observed proportion is from the proportion stated in the null hypothesis. C. 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. 2) What is the criterion for rejecting the null hypothesis using the​ P-value approach? Choose the correct answer below. A. If P-value<−zα for a​ left-tailed test, or if ​P-value>zα for a​ right-tailed test, or if P-value<−zα/2 or P-value>zα/2 for a​ two-tailed test, then reject the null hypothesis. B. If ​P-value>α​, reject the null hypothesis. C. If ​P-value<α​, reject the null hypothesis.

1) c. 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. 2) c. If ​P-value<α​, reject the null hypothesis.

Construct a 90​% confidence interval of the population proportion using the given information. x=75 n=250 1) lower bound is ________ 2) upper bound is ________

1) lower bound = 0.252 2) upper bound = 0.348 on statcrunch - stat - proportion stats - one sample - with summary - input # of successes and # of observations - click condolence interval and input .9

A random sample of 1001 adults in a certain large country was asked​ "Do you pretty much think televisions are a necessity or a luxury you could do​ without?" Of the 1001 adults surveyed, 534 indicated that televisions are a luxury they could do without. 1) Obtain a point estimate for the population proportion of adults in the country who believe that televisions are a luxury they could do without. p-hat = ________ 2) verify the requirements for constructing a confidence interval about p are satisfied. The sample (a)__________ (is stated to be, is stated not to be, can be assumed to be, can not be assumed to be) a simple random​ sample, the value of (b)______________ is (c)______, which is (d)____________ (less than, greater than or equal to) 10, and the (e)__________ (sample size, population size, population proportion, sample proportion) (f)_____________ (cannot be assumed to be, can be assumed to be, is stated to not be, is stated to be) less than or equal to​ 5% of the (g)_____________ (population size, population proportion, sample proportion, sample size) 3) Construct and interpret a 95​% confidence interval for the population proportion of adults in the country who believe that televisions are a luxury they could do without. Select the correct choice below and fill in any answer boxes within your choice. ____________(there is a or we are) _____% _________ (probability or confident) the proportion of adults in the country who believe that televisions are a luxury they could do without is between ______ and ________ 4) it is ____________ (likely, possible but not likely, not possible) that a supermajority (more than 60%) of adults in the country believe that television is a luxury they could be without because the 95% confidence internal ___________ (contains or does not contain) _________ 5) Use the results of part​ (3) to construct a 95​% confidence interval for the population proportion of adults in the country who believe that televisions are a necessity. Lower bound: ________ Upper bound: ________

1) p-hat = 0.533 (534/1001) 2) (a) is stated to be, (b) (n)(p-hat)(1-p-hat), (c) 249.160, (d) greater than or equal to, (e) sample size, (f) can be assumed to be, (g) population size 3) We are 95% confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between 0.503 and 0.564 (do this on statcrunch by going to stat, proportion stat, one sample, with summary, ...) 4) it is possible but not likely that a supermajority (more than 60%) of adults in the country believe that television is a luxury they could be without because the 95% confidence internal does not contain 0.6 (refer to the question before because the limits don't include 0.6) 5) Lower bound: 0.436 Upper bound: 0.497 (basically take 1001 - 534 and use that number in statcrunch)

Suppose a simple random sample of size n=39 is obtained from a population with μ=66 and σ=14. ​1) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities regarding the sample​ mean? Assuming the normal model can be​ used, describe the sampling distribution x-bar (choices are: a: since the sample size is large enough, the population distribution does not need to be normal, b: there are no requirements on the shape of the distribution of the population, c: the population must be normally distributed, d: the population must be normally distributed and the sample must be large) ​2) Assuming the normal model can be​ used, determine ​P(x<69.9​). 3) Assuming the normal model can be​ used, determine ​P(x≥67.3​).

1) part 1, a: since the sample size is large enough, the population distribution does not need to be normal (this answer differs with the similar question option) part 2, Approximately normal​, with μx=66 and σx=14/square root of 39 2 and 3) just plug into statcrunch

In a survey of 2045 adults in a certain country conducted during a period of economic​ uncertainty, 69​% thought that wages paid to workers in industry were too low. The margin of error was 8 percentage points with 90​% 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. 1) We are 90​% confident 69​% 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? 2) We are 82​% to 98​% confident 69​% 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? 3) We are 90​% confident that the interval from 0.61 to 0.77 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? 4) In 90​% 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 0.61 and 0.77. Is the interpretation​ reasonable? * the potential options A. The interpretation is reasonable. B. The interpretation is flawed. The interpretation indicates that the level of confidence is varying. C. The interpretation is flawed. The interpretation suggests that this interval sets the standard for all the other​ intervals, which is not true. D. The interpretation is flawed. The interpretation provides no interval about the population proportion.

1) the interpretation is flawed. The interpretation provides no interval about the population proportion 2) the interpretation is flawed. The interpretation indicates that the level of confidence is varying 3) The interpretation is reasonable 4) The interpretation is flawed. The interpretation suggests that this interval sets the standard for all the other​ intervals, which is not true.

Test the hypothesis using the​ P-value approach. Be sure to verify the requirements of the test. H0​: p=0.6 versus H1​: p>0.6 n=100​; x=65​, α=0.05 1) Is np01−p0≥​10? (yes or no) 2) use technology to find the P-value. P-value = _______ 3) ___________(reject, do not reject) the null hypothesis, because the P-value is _______ (greater than or less than) than a

1) yes (100)(.6)(.4) = 24 > 10 2) on statcrunch stat proportion stat one sample with summary input x and n hypothesis test for p 3) do not reject, greater than (P-value</- a, reject the null hypothesis If P-value > a, fail to reject the null hypothesis)

Determine the point estimate of the population mean and margin of error for the confidence interval. Lower bound is 20​, upper bound is 30. 1. The point estimate of the population mean is ______ 2.

1. 25 (30+20)/2 = 25 2. 5 (30-20)/2 = 5

Some have argued that throwing darts at the stock pages to decide which companies to invest in could be a successful​ stock-picking strategy. Suppose a researcher decides to test this theory and randomly chooses 150 companies to invest in. After 1​ year, 81 of the companies were considered​ winners; that​ is, they outperformed other companies in the same investment class. To assess whether the​ dart-picking strategy resulted in a majority of​ winners, the researcher tested H0​: p=0.5 versus H1​: p>0.5 and obtained a​ P-value of 0.1636. Explain what this​ P-value means and write a conclusion for the researcher.​ (Assume α is 0.1 or​ less.) 1. Choose the correct explanation below. A. About 16 in 100 samples will give a sample proportion as high or higher than the one obtained if the population proportion really is greater than 0.5. B. About 16 in 100 samples will give a sample proportion as high or higher than the one obtained if the population proportion really is 0.5. C. About 81 in 150 samples will give a sample proportion as high or higher than the one obtained if the population proportion really is 0.5. D. About 81 in 150 samples will give a sample proportion as high or higher than the one obtained if the population proportion really is greater than 0.5. 2. Choose the correct conclusion below. A. Because the​ P-value is​ large, do not reject the null hypothesis. There is not sufficient evidence to conclude that the​ dart-picking strategy resulted in a majority of winners. B. Because the​ P-value is​ small, reject the null hypothesis. There is sufficient evidence to conclude that the​ dart-picking strategy resulted in a majority of winners. C. Because the​ P-value is​ large, reject the null hypothesis. There is sufficient evidence to conclude that the​ dart-picking strategy resulted in a majority of winners. D. Because the​ P-value is​ small, do not reject the null hypothesis. There is not sufficient evidence to conclude that the​ dart-picking strategy resulted in a majority of winners.

1. B) About 16 in 100 samples will give a sample proportion as high or higher than the one obtained if the population proportion really is 0.5. 2. A. Because the​ P-value is​ large, do not reject the null hypothesis. There is not sufficient evidence to conclude that the​ dart-picking strategy resulted in a majority of winners.

The null and alternative hypotheses are given. Determine whether the hypothesis test is​ left-tailed, right-tailed, or​ two-tailed. What parameter is being​ tested? H0​: σ = 7 H1​: σ < 7 1. What type of test is being conducted in this​ problem? right-tailed test, left-tailed test or two tailed test 2. what parameter is being tested? population proportion, population mean or population standard deviation

1. left-tailed test (right-tailed test = H0​: σ = 7 H1​: σ > 7 left-tailed test = H0​: σ = 7 H1​: σ < 7 two-tailed test = H0​: σ = 7 H1​: σ doesn't equal 7) 2. population standard deviation (based on the symbol used)

Assume the random variable X is normally distributed with mean μ=50 and standard deviation σ=7. Compute the probability. Be sure to draw a normal curve with the area corresponding to the probability shaded. P(X>41) 1. which of the following normal curves corresponds to P(X>41)? 2. P(X>41)=?

1. since its > the tail should be in the right side, can put this into stat crunch to check 2. in statcrunch stat calculator normal set the mean and standard deviation set > 41

Determine the area under the standard normal curve that lies between ​(a) Z=−0.14 and Z=0.14​, ​(b) Z=−0.61 and Z=0​, and ​(c) Z=0.81 and Z=1.49. ​(a) The area that lies between Z=−0.14 and Z=0.14 is nothing.

1. stat 2. calculator 3. normal 4. go to between not standard 5. set the x value with the indicated Z

Find the​ Z-score such that the area under the standard normal curve to the right is 0.45. The approximate​ Z-score that corresponds to a right tail area of 0.45 is ___________

1. stat 2. calculator 3. normal 4. set the greater than or less than so that the area is in the right tail 5. set 0.45 as the answer so we can solve for the x value

The null and alternative hypotheses are given. Determine whether the hypothesis test is​ left-tailed, right-tailed, or​ two-tailed. 1. What parameter is being​ tested? H0​: σ=4.2 H1​: σ≠4.2 (right-tailed, left-tailed, and two-tailed) 2. What parameter is being tested? (standard deviation (σ), mean, porportion (p))

1. two-tailed (right-tailed test = H0​: σ = 7 H1​: σ > 7 left-tailed test = H0​: σ = 7 H1​: σ < 7 two-tailed test = H0​: σ = 7 H1​: σ doesn't equal 7) 2. σ, standard deviation

State the conclusion based on the results of the test. According to the Federal Housing Finance​ Board, the mean price of a​ single-family home two years ago was ​$299,400. A real estate broker believes that because of the recent credit​ crunch, the mean price has increased since then. The null hypothesis is rejected. Choose the correct answer below. A. There is sufficient evidence to conclude that the mean price of a​ single-family home has increased from its level two years ago of ​$299,400. B. There is not sufficient evidence to conclude that the mean price of a​ single-family home has decreased from its level two years ago of ​$299,400. C. There is sufficient evidence to conclude that the mean price of a​ single-family home has decreased from its level two years ago of ​$299,400. D. There is not sufficient evidence to conclude that the mean price of a​ single-family home has increased from its level two years ago of ​$299,400.

A. There is sufficient evidence to conclude that the mean price of a​ single-family home has increased from its level two years ago of ​$299,400.

Determine if the following statement is true or false.​ Why? The expected frequencies in a​ chi-square test for independence are found using the formula below. Expected frequency= ((row total)(column total))(table total) A. True. It is a simplification of multiplying the proportion of a row variable by the proportion of the column variable to find the proportion for a​ cell, then multiplying by the table total. B. True. The area to the right of this value in the corresponding​ chi-square distribution gives the​ P-value for the test. C. False. This formula gives the expected proportion. The​ (table total) factor in the denominator on the right should be squared. D. False. The expected frequencies for a​ chi-square test are given by Oi−EiEi​, where i represents cell number.

A. True. It is a simplification of multiplying the proportion of a row variable by the proportion of the column variable to find the proportion for a​ cell, then multiplying by the table total.

What does ​"95​% ​confidence" mean in a 95​% confidence​ interval? A. The probability that the value of the parameter lies between the lower and upper bounds of the interval is 95​%. The probability that it does not is 5​%. B. If 100 different confidence intervals are​ constructed, each based on a different sample of size n from the same​ population, then we expect 95 of the intervals to include the parameter and 5 to not include the parameter. C. The value of the parameter lies within 95​% of a standard deviation of the estimate. D. The confidence interval includes 95​% of all possible values for the parameter.

B. If 100 different confidence intervals are​ constructed, each based on a different sample of size n from the same​ population, then we expect 95 of the intervals to include the parameter and 5 to not include the parameter.

What happens to the probability of making a Type II​ error, β​, as the level of​ significance, α​, ​decreases? Why? Choose the correct answer below A. The probability decreases. Type I and Type II errors are proportional. B. The probability increases. Type I and Type II errors are inversely related. C. The probability decreases. The more careful researcher​ is, the lower the chances of making any error. D. The probability increases. The sum of α and β always equals 1.

B. The probability increases. Type I and Type II errors are inversely related.

State the conclusion based on the results of the test. According to the​ report, the mean monthly cell phone bill was ​$48.35 three years ago. A researcher suspects that the mean monthly cell phone bill is higher today. The null hypothesis is rejected. Choose the correct answer below. A. There is sufficient evidence to conclude that the mean monthly cell phone bill is less than its level three years ago of ​$48.35. B. There is not sufficient evidence to conclude that the mean monthly cell phone bill is higher than its level three years ago of ​$48.35. C. There is sufficient evidence to conclude that the mean monthly cell phone bill is higher than its level three years ago of ​$48.35.

C) There is sufficient evidence to conclude that the mean monthly cell phone bill is higher than its level three years ago of ​$48.35.

A group conducted a poll of 2093 likely voters just prior to an election. The results of the survey indicated that candidate A would receive 45​% of the popular vote and candidate B would receive 40​% of the popular vote. The margin of error was reported to be 5​%. The group reported that the race was too close to call. Use the concept of a confidence interval to explain what this means. What does it mean to say the race was too close to​ call? A. Since the poll results do not show that one of the candidates has more than​ 50% of the popular​ vote, the poll cannot predict the winner. B. Since the difference between the percentages of the popular vote for the candidates is less than the 15​% of voters that are​ undecided, the poll cannot predict the winner. C. The margin of error suggests candidate A may receive between 40​% and 50​% of the popular vote and candidate B may receive between 35​% and 45​% of the popular vote. Because the poll estimates overlap when accounting for margin of​ error, the poll cannot predict the winner. D. Since the estimated proportions depend on the level of​ confidence, the candidate predicted to receive the greater percentage of the popular vote changes for different levels of​ confidence, so the poll cannot predict the winner.

C. The margin of error suggests candidate A may receive between 40​% and 50​% of the popular vote and candidate B may receive between 35​% and 45​% of the popular vote. Because the poll estimates overlap when accounting for margin of​ error, the poll cannot predict the winner.

Suppose there are n independent trials of an experiment with k>3 mutually exclusive​ outcomes, where pi represents the probability of observing the ith outcome. What would be the formula of an expected count in this​ situation? Choose the correct answer below. A. The expected counts for each possible outcome are given by Ei=n. B. The expected counts for each possible outcome are given by Ei=npi. C. The expected counts for each possible outcome are given by Ei=npi. D. The expected counts for each possible outcome are given by Ei=pi.

C. The expected counts for each possible outcome are given by Ei=npi.

State the requirements to perform a​ goodness-of-fit test. Choose the correct answer below. Select all that apply. A. at least​ 90% of expected frequencies ≥5 B. at least​ 90% of expected frequencies ≥10 C. at least​ 80% of expected frequencies ≥5 D. at least​ 80% of expected frequencies ≥10 E. all expected frequencies≥1 F. all expected frequencies≥5

C. at least​ 80% of expected frequencies ≥5 E. all expected frequencies≥1

Explain what​ "statistical significance" means. A. Statistical significance means that the null hypothesis claims the population proportion is equal to something other than 0.5. B. Statistical significance means that the scenario being analyzed will have a meaningful​ real-world impact. C. Statistical significance means that the tools used to measure the data introduce error that needs to be accounted for when considering whether or not to reject the null hypothesis. D. Statistical significance means that the result observed in a sample is unusual when the null hypothesis is assumed to be true. E. Statistical significance means that the sample standard deviation is unusually​ small, resulting in an unusually large test statistic.

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

Explain the difference between statistical significance and practical significance. Choose the correct answer below. A. Statistical significance means that the hypothesis test being performed is useful for building theoretical foundations for other statistical work. Practical significance means that the particular application of the hypothesis test is of great importance to the real world. B. Statistical significance refers to the type of hypothesis test needed to analyze a​ population, with some tests being more important than Z tests. Practical significance refers to how difficult a desired hypothesis test is to perform in an​ application, with some tests being easier to perform than others. C. Statistical significance refers to how an unusual event is unlikely to actually appear in a real world​ application, such as every entry in a sample of size 50 having the same value. Practical significance refers to how an unusual event is likely to actually appear in a real world​ application, such as a rejection of a null hypothesis using data that looks feasible. D. Statistical significance means that the sample statistic is not likely to come from the population whose parameter is stated in the null hypothesis. Practical significance refers to whether the difference between the sample statistic and the parameter stated in the null hypothesis is large enough to be considered important in an application.

Statistical significance means that the sample statistic is not likely to come from the population whose parameter is stated in the null hypothesis. Practical significance refers to whether the difference between the sample statistic and the parameter stated in the null hypothesis is large enough to be considered important in an application. *Note that a statistically significant result may be of no practical significance.

The data from a simple random sample with 25 observations was used to construct the plots given below. The normal probability plot that was constructed has a correlation coefficient of 0.941. Judge whether a​ t-interval could be constructed using the data in the sample. The normal probability plot ____________ (does not suggest, suggests) the data could come from a normal population because 0.941 ______ (<,>) _______ and the box-plot _________ (shows, does not show) outliers, so a​ t-interval _________ (could not, could) be constructed. *provided are two links, one shows the critical values of the correlation coefficient and the other shows a box plot that has a mean of 10 and ranges for like 0 to 35 (skewed left with a outlier to the right)

The normal probability plot does not suggest the data could come from a normal population because 0.941 < 0.959 (from the critical values table corresponding to 25 observations) and the box-plot shows outliers, so a​ t-interval could not be constructed.

Suppose the null hypothesis is not rejected. State the conclusion based on the results of the test. Three years​ ago, the mean price of a​ single-family home was ​$243,715. A real estate broker believes that the mean price has increased since then. Which of the following is the correct​ conclusion? A. There is sufficient evidence to conclude that the mean price of a​ single-family home has increased. B. There is not sufficient evidence to conclude that the mean price of a​ single-family home has not changed. C. There is not sufficient evidence to conclude that the mean price of a​ single-family home has increased. D. There is sufficient evidence to conclude that the mean price of a​ single-family home has not changed.

There is not sufficient evidence to conclude that the mean price of a single-family home has increased

State the conclusion based on the results of the test. The mean of the pressure required to open a certain valve is known to be μ=8.1 psi. Due to changes in the manufacturing​ process, the​ quality-control manager feels that the average pressure has changed. The null hypothesis was rejected. Choose the correct answer below. A. There is sufficient evidence that the mean of the pressure required to open a certain valve has not changed. B. There is not sufficient evidence that the mean of the pressure required to open a certain valve has not changed. C. There is sufficient evidence that the mean of the pressure required to open a certain valve has changed. D. The mean of the pressure has changed. E. There is not sufficient evidence that the mean of the pressure required to open a certain valve has changed. F. The mean of the pressure has not changed.

There is sufficient evidence that the mean of the pressure required to open a certain valve has changed

Suppose the null hypothesis is rejected. State the conclusion based on the results of the test. Six years​ ago, 12.6​% of registered births were to teenage mothers. A sociologist believes that the percentage has increased since then. Which of the following is the correct​ conclusion? A. There is sufficient evidence to conclude that the percentage of teenage mothers has remained the same. B. There is sufficient evidence to conclude that the percentage of teenage mothers has increased. C. There is not sufficient evidence to conclude that the percentage of teenage mothers has remained the same. D. There is not sufficient evidence to conclude that the percentage of teenage mothers has increased.

There is sufficient evidence to conclude that the percentage of teenage mothers has increased

Fill in the blank to complete the statement. If we do not reject the null hypothesis when the statement in the alternative hypothesis is​ true, we have made a Type​ _______ error.

Type II *A Type I error occurs if the null hypothesis is rejected​ when, in​ fact, the null hypothesis is true. A Type II error occurs if the null hypothesis is not rejected​ when, in​ fact, the alternative hypothesis is true.

A random sample of 16 undergraduate students receiving student loans was​ obtained, and the amounts of their loans for the school year were recorded. Use a normal probability plot to assess whether the sample data could have come from a population that is normally distributed. Using the correlation coefficient of the normal probability​ plot, is it reasonable to conclude that the population is normally​ distributed?

Yes. The correlation between the expected​z-scores and the observed​data, 0.98​, exceeds the critical​value, 0.941. Therefore, it is reasonable to conclude that the data come from a normal population. (it is reasonable to conclude that the population is normally distributed if the observed data exceeds the critical value to find the observed correlation coefficient on statcrunch 1. graph 2. QQ plot 3. Highlight observed value 4. Check correlation statistic and normal quantiles on y-axis 5. Check to see if the correlation given matches that in the critical values of the correlation coefficient table to find the expected we look at the critical values table for correlation coefficient.

Suppose the lengths of human pregnancies are normally distributed with μ=266 days and σ=16 days. Complete parts ​(a) and​ (b) below. a) The figure to the right represents the normal curve with μ=266 days and σ=16 days. The area to the left of X=250 is 0.1587. Provide two interpretations of this area. 1) the proportion of human pregnancies that last ______ (more/less) than _______ days is ________ 2) the probability that a randomly selected human pregnancy lasts __________ (more/less) than ____ days is _________ b) the figure to the right represents the normal curve with μ=266 days and σ=16 days. The area between x=235 and x=285 is 0.8561. 1) The proportion human pregnancies that last __________ (between or less than/more than) _____ or _____ days is _________ 2) the probability that a randomly selected human pregnancy lasts ________ (between or less than/more than) _______ and _______ days is ______

a) 1) the proportion of human pregnancies that last less (this is because the shaded region on the graph is in the left tail) than 250 days is 0.1587 (given in the problem) 2) the probability that a randomly selected human pregnancy lasts more than 250 days is 0.1587 (this is the same question as above) b) 1) the proportion human pregnancies that between 235 and 285 days is 0.8561 2) the probability that a randomly selected human pregnancy lasts between 235 and 285 days is 0.8561

The graph to the right is the uniform probability density function for a friend who is x minutes late. (its like a box shape with the y-intersection at 1/30 and the x-intersection at 30 - two straight lines go out form those point and intersect - y-axis is density and x-axis is time in minutes) ​(a) The probability that the friend is between 20 and 25 minutes late is __________ ​(b) It is 10 A.M. There is a 30​% probability the friend will arrive within ___________ minutes

a) 0.167 (this is 1/6 of the squares area) b) 9 (30 mins)(.3)

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean μ=297 days and standard deviation σ=29 days. Complete parts​ (a) through​ (f) below. a) the probability that a randomly selected pregnancy lasts less than 285 days is approximately ________ interpret this probability: if 100 pregnant individuals were selected independently from this population, we would expect ______ pregnancies to last ________ (more than, exactly, less than) 286 days b) Suppose a random sample of 18 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies. the sampling distribution of x-bar is ___________ with μx = _______ and σx = ________ c) the probability that the mean of a random sample of 18 pregnancies is less than 286 days is approximately ________ If 100 independent random samples of size n=18 pregnancies were obtained from this​ population, we would expect ________ sample(s) to have a sample mean of 286 days (exactly, or less, or more) d) The probability that the mean of a random sample of 29 pregnancies is less than 286 days is approximately __________ If 100 independent random samples of size n=29 pregnancies were obtained from this​ population, we would expect ________ sample(s) to have a sample mean of 286 days (exactly, or less, or more) e) What might you conclude if a random sample of 29 pregnancies resulted in a mean gestation period of 286 days or​ less? This result would be __________ (expected, unusual), so the sample likely came from a population whose mean gestation period is ________ (equal to, greater than, less than) 297 days. f) The probability that a random sample of size 17 will have a mean gestation period within 8 days of the mean is___________

a) just pug into statcrunch (stat, calculator, normal, etc) if 100 pregnant individuals were selected independently from this population, we would expect 35 pregnancies to last less than 286 days b) the sampling distribution of x-bar is normal with μx = 297 and σx = 6.8354 (29/square root of 18) c) the probability that the mean of a random sample of 18 pregnancies is less than 286 days is approximately 0.0538 (on statcrunch) If 100 independent random samples of size n=18 pregnancies were obtained from this​ population, we would expect 5 sample(s) to have a sample mean of 286 days or less d) use statcrunch to find by changing the standard deviation based on the sample If 100 independent random samples of size n=29 pregnancies were obtained from this​ population, we would expect 2 sample(s) to have a sample mean of 286 days or less e) This result would be unusual so the sample likely came from a population whose mean gestation period is less than 297 days. (less than comes from the question and unusual is because the probability is less than 0.05) f) 0.7446 (use between on statcrunch)

Based on the results of parts ​(a)-​(c), write a few sentences that explain the difference between​ "accepting" the statement in the null hypothesis versus​ "not rejecting" the statement in the null hypothesis. One should (a)__________ (accept, not reject) rather than (b)_________ (not reject, accept) the null hypothesis. If one (c)_________ (does not reject, accepts) the null​ hypothesis, this indicates that the (d)_________ (sample mean, population mean) is a specific​ value, such as 102​, 103​, or 104​, and so the same data have been used to conclude that the (e)____________ (population mean, sample mean) is three different values. If one (f)____________ (does not reject, accepts) the null​ hypothesis, this indicates that the (g)____________ (sample mean, population mean) could be 102​, 103​, or 104 or even some other​ value; we are simply not ruling them out as the value of the (h)____________ (population mean, sample mean). ​Therefore, (i)____________ (not rejecting, accepting) the null hypothesis can lead to contradictory​ conclusions, whereas (j)___________ (not rejecting, accepting) does not.

a) not reject b) accept c) accepts d) population mean e) population mean f) does not reject g) population mean h) population mean i) accepting j) not rejecting

Use the accompanying data table to ​ (a) draw a normal probability​ plot (b) the correlation is _________ (c) the critical value is _________

a) on statcrunch 1. Graph 2. QQ plot 3. Highlight observed value 4. Check correlation statistic and normal quantiles on y-axis 5. Check to see if the correlation given matches that in the critical values of the correlation coefficient table b) given at the top of the graph c) given by the table and in the row with how many values there are

Suppose a simple random sample of size n=64 is obtained from a population with μ=86 and σ=8. ​(a) Describe the sampling distribution of x-bar. (choices: uniform, skewed right, skewed left, approximately normal, the shape of the distribution is unknown ​(b) What is P x-bar>87.35​? ​(c) What is P x-bar≤83.7​? ​(d) What is P 84.65<x-bar<88.05​?

a) the distribution is approximately normal b - d) on stat crunch stat calculator normal input mean a 86 and sample standard deviation as 1 *remember to find σx but σ/square root of n

The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal​ distribution, with a mean of 19 minutes and a standard deviation of 3 minutes. ​(a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take​ longer, the customer will receive the service for​ half-price. What percent of customers receive the service for​ half-price? ​(b) If the automotive center does not want to give the discount to more than 5​% of its​ customers, how long should it make the guaranteed time​ limit?

a) this is normal but find the area in the right tail, or the percent of people how it takes more than 20 minutes for b) set the answer to 0.05 for 5% (5% in the right tail) and get the x value that corresponds

Determine whether the following statement is true or false. To construct a confidence interval about the​ mean, the population from which the sample is drawn must be approximately normal.

false

Determine whether the following graph can represent a normal density function. the curve is normal shaped but the ends go below the x-axis

no

Determine whether the following graph can represent a normal density function. the curve is normal shaped in the middle but the ends slope up after hitting the x-axis

no

Fill in the blank. A​ _______ is a graph that plots observed data versus normal scores.

normal probability plot

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

null hypothesis

The number of chocolate chips in an​ 18-ounce bag of chocolate chip cookies is approximately normally distributed with mean 1252 and standard deviation 129 chips. a) the probability that a randomly selected bag contains between 1100 and 1400 chocolate chips is _________ b) The probability that a randomly selected bag contains fewer than 1000 chocolate chips is _________ c) The proportion of bags that contains more than 1200 chocolate chips is __________ d) A bag that contains 1475 chocolate chips is in the _____th percentile.

on statcrunch 1. stat 2. calculator 3. normal 4. set the mean and standard deviation 5. decide if your using standard or between 6. decide if its talking about the right or left tail *the first three questions are normal and the last one you just have to convert the decimal into a percentile (remember to make sure to have the area in the left tail for this one)*

Assume the random variable X is normally distributed with mean μ=50 and standard deviation σ=7. The 86th percentile =

on statcrunch 1. stat 2. calculator 3. normal 4. set the mean and standard deviation 5. set the <> so that the tail is to the left (that's how percentile are) 5. the 86th percentile means 0.86 in the answer place to give the x-value

Assume the random variable X is normally​ distributed, with mean μ=52 and standard deviation σ=9. The 9th percentile =

on statcrunch 1. stat 2. calculator 3. normal 4. set the mean and standard deviation 5. set the <> so that the tail is to the left (that's how percentile are) 5. the 9th percentile means 0.09 in the answer place to give the x-value

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.

sample proportion p-hat = x/n

A study was conducted that resulted in the following relative frequency histogram. Determine whether or not the histogram indicates that a normal distribution could be used as a model for the variable. - the graph is skewed right

the histogram is not bell-shaped, so a normal distribution could not be used as a model for the variable

Is the statement below true or​ false? The distribution of the sample​ mean, x​-bar, will be normally distributed if the sample is obtained from a population that is normally​ distributed, regardless of the sample size.

true