STATS EXAM 3

Percentages %

90% - 100% = 9% = 5% = 95% 95% - 100% = 5% 5% / 2 = 2.5% 95% + 2.5% = 97.5% 97% - 98% - 100% = 2% 99% - 100% = 1% 1% / 2 = 0.5% 99% + 0.5% = 99.5

Why does the sampling distribution of the mean follow a normal distribution for a large enough sample​ size, even though the population may not be normally​ distributed?

As the sample size gets large​ enough, the sampling distribution of the mean is approximately normally distributed.

Why is it not possible to have​ 100% confidence? Explain.

A​ 100% confidence interval is not possible unless either the entire population is sampled or an absurdly wide interval of estimates is provided.

If you use a 0.05 level of significance in a​ two-tail hypothesis​ test, what decision will you make if ZSTAT=+1.89​?

Determine the decision rule. Select the correct choice below and fill in the answer​ box(es) within your choice. Reject H0 if ZSTAT< −1.96 or ZSTAT > +1.96 State your conclusion. Choose the correct answer below. since ZSTAT does not fall into the rejection​ region, do not reject

Assuming that the population is normally​ distributed, construct a 95% confidence interval for the population​ mean, based on the following sample size of n=7. ​1, 2,​ 3, 4, 5, 6​, and 30 In the given​ data, replace the value 30 with 7 and recalculate the confidence interval. Using these​ results, describe the effect of an outlier​ (that is, an extreme​ value) on the confidence​ interval, in general.

Find a 95% confidence interval for the population​ mean, using the formula or technology. - import the data to excel - find the mean, standard deviation, sample size excel x: =average s: =stdev.s n: 7 use: https://www.socscistatistics.com/confidenceinterval/default2.aspx In the given​ data, replace the value 30 with 7. Find a 95%confidence interval for the population​ mean, using the formula or technology. - use the same data, but change 30 to 7 - use the same calculator. Using the results from the previous two​ steps, what is the effect of an outlier​ (that is, an extreme​ value) on the confidence​ interval, in​ general? A. The presence of an outlier in the original data increases the value of the sample mean and greatly inflates the sample standard​ deviation, widening the confidence interval.

How do you find the X value that corresponds to a given percentile of the normal​ distribution?

Find the Z value corresponding to the given​ percentile, and then use the equation X=μ+Zσ

An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.88 inch. The lower and upper specification limits under which the ball bearing can operate are 0.865 inch​ (lower) and 0.895 inch​ (upper). Past experience has indicated that the actual diameter of the ball bearings is approximately normally​ distributed, with a mean of 0.884 inch and a standard deviation of 0.006 inch. Suppose a random sample of 22 ball bearings are selected. Complete parts​ (a) through​ (e).

Mean: 0.884 σ: 0.006 n: 22 ST. DEV: σ/√n 0.006/√22 0.001279204 a. What is the probability that the sample mean is between the target and the population mean of 0.884​? 0.4991 EXCEL: =NORM.DIST X: 0.88 MEAN: 0.884 ST. DEV: 0.001279204 CUMULATIVE: TRUE 0.00088317 =0.5 - 0.00088317 =0.4991 b. What is the probability that the sample mean is between the lower specification limit and the​ target? 0.0008 EXCEL: =NORM.DIST X: 0.865 MEAN: 0.884 ST. DEV: 0.001279204 CUMULATIVE: TRUE 3.32702E-50 0.00088317 - 3.32702E-50 = 0.0008 c. What is the probability that the sample mean is greater than the upper specification​ limit? 0.000 EXCEL: =NORM.DIST X: 0.895 MEAN: 0.884 ST. DEV: 0.001279204 CUMULATIVE: TRUE 1 1 - 1 = 0.000 d. What is the probability that the sample mean is less than the lower specification​ limit? 0.0000 e. The probability is 88​% that the sample mean diameter will be greater than what​ value? 0.882 ​1 - 0.88 = 0.12 EXCEL =NORM.INV PROBABILITY: 0.12% MEAN: 0.884 ST. DEV: 0.001279204 0.882

The fill amount of bottles of a soft drink is normally​ distributed, with a mean of 2.0 liters and a standard deviation of 0.06 liter. Suppose you select a random sample of 25 bottles. a. What is the probability that the sample mean will be between 1.99 and 2.0 liters​? b. What is the probability that the sample mean will be below 1.98 liters​? c. What is the probability that the sample mean will be greater than 2.01 ​liters? d. The probability is 95​% that the sample mean amount of soft drink will be at least how​ much? e. The probability is 95​% that the sample mean amount of soft drink will be between which two values​ (symmetrically distributed around the​ mean)?

ST. DEV = 0.06/√25 = 0.012 a. The probability is 0.297 EXCEL: =NORM.DIST X: 1.99 MEAN: 2.0 ST. DEV: 0.012 CUMULATIVE: TRUE 0.20232838 2.0 / 2 = 1 1 / 2 = 0.5 0.5 - 0.20232838 = 0.297 b. The probability is 0.047 EXCEL: =NORM.DIST X: 1.98 MEAN: 2.0 ST. DEV: 0.012 CUMULATIVE: TRUE 0.047 c. The probability is 0.202 EXCEL: =NORM.DIST X: 2.01 MEAN: 2.0 ST. DEV: 0.012 CUMULATIVE: TRUE 0.79767162 1 - 0.79767162 = 0.202 d. There is a 95​% probability that the sample mean amount of soft drink will be at least 1.980 ​liter(s). 95% - 100% = 5% EXCEL =NORM.INV PROBABILITY: 5% MEAN: 2.0 ST. DEV: 0.012 1.980 e. There is a 95​% the probability that the sample mean amount of soft drink will be between 1.976 liter(s) and 2.023 liter(s). 95% - 100% = 5% 5% / 2 = 2.5% 95% + 2.5% = 97.5% EXCEL =NORM.INV PROBABILITY: 2.5% MEAN: 2.0 ST. DEV: 0.012 1.976 EXCEL =NORM.INV PROBABILITY: 97.5% MEAN: 2.0 ST. DEV: 0.012 2.023

Suppose the defendant in a particular judicial system is presumed guilty until proven innocent. What are the null and alternative​ hypotheses? What are the meanings of the risks of committing either a Type I or Type II​ error?

State the null and alternative hypotheses. H0​: The defendant is guilty H1​: The defendant is innocent What are the meanings of the risks of committing either a Type I or Type II​ error? - A Type I error would be not convicting a guilty person. A Type II error would be convicting an innocent person.

A consumer organization wants to estimate the actual tread wear index of a brand name of tires that claims​ "graded 150​" on the sidewall of the tire. A random sample of n=18 indicates a sample mean tread wear index of 135.5 and a sample standard deviation of 21.8. Complete parts​ (a) through​ (c).

a. Assuming that the population of tread wear indexes is normally​ distributed, construct a 95% confidence interval estimate of the population mean tread wear index for tires produced by this manufacturer under this brand name. use: https://www.socscistatistics.com/confidenceinterval/default2.aspx b. Do you think that the consumer organization should accuse the manufacturer of producing tires that do not meet the perfomance information on the sidewall of the​ tire? Yes, because a grade of 150 is not in the interval. or No, because a grade of 150 is in the interval. c. Explain why an observed tread wear index of 148 for a particular tire is not​ unusual, even though it is outside the confidence interval developed in​ (a). It is not unusual because it is only 0.57 standard deviations above the sample mean.

A bottled water distributor wants to estimate the amount of water contained in 1​-gallon bottles purchased from a nationally known water bottling company. The water bottling​ company's specifications state that the standard deviation of the amount of water is equal to 0.05 gallon. A random sample of 50 bottles is​ selected, and the sample mean amount of water per 1​-gallon bottle is 0.975 gallon. Complete parts​ (a) through​ (d).

a. Construct a 99​% confidence interval estimate for the population mean amount of water included in a​ 1-gallon bottle. use: https://www.socscistatistics.com/confidenceinterval/default3.aspx b. On the basis of these​ results, do you think that the distributor has a right to complain to the water bottling​ company? Why? Yes, because a​1-gallon bottle containing exactly​ 1-gallon of water lies outside the 99​% confidence interval. c. Must you assume that the population amount of water per bottle is normally distributed​ here? Explain. No, because the Central Limit Theorem almost always ensures that X is normally distributed when n is large. In this​ case, the value of n is large. d. Construct a 90​% confidence interval estimate. How does this change your answer to part​(b)? use: https://www.socscistatistics.com/confidenceinterval/default3.aspx How does this change your answer to part​ (b)? A​ 1-gallon bottle containing exactly​ 1-gallon of water lies outside the 90​% confidence interval. The distributor still has a right to complain to the bottling company.

The table below contains a certain social media​ company's penetration values​ (the percentage of a​ country's population that use this social​ media) for 14 countries. Complete parts a through c below. 34 58 11 11 28 25 25 29 29 46 80 43 16 33

a. Construct a 99​% confidence interval estimate for the population mean social media penetration. excel: mean: =average standard deviation: =stdev.s sample size: use: https://www.socscistatistics.com/confidenceinterval/default2.aspx b. What assumption do you need to make about the population distribution to construct the interval in​ (a)? The social media penetration is normally distributed. c. Given the data​ presented, do you think the assumption needed in​ (a) is​ valid? Explain. ​No, the data indicate the population distribution is skewed to the right.

A bottled water distributor wants to determine whether the mean amount of water contained in​ 1-gallon bottles purchased from a nationally known water bottling company is actually 1 gallon. You know from the water bottling company specifications that the standard deviation of the amount of water is 0.02 gallon. You select a random sample of 50 bottles, and the mean amount of water per​ 1-gallon bottle is 0.996 gallon. Complete parts​ (a) through​ (d) below.

a. Is there evidence that the mean amount is different from 1.0 ​gallon? (Use α=0.01​.) Let μ be the population mean. Determine the null​ hypothesis, H0​, and the alternative​ hypothesis, H1. H0: μ = 1 H1: μ ≠ 1 What is the test​ statistic? ZSTAT = X − μ / σ/√n 0.996 - 1 / 0.02/√50 = -1.41 What​ is/are the critical​ value(s)? (Use α=0.01​.) 0.01/2 = 0.005 - use the t distributuion chart - it's going to be above the % Z = -2.576, 2.58 What is the final​ conclusion? Fail to reject H0. There is not sufficient evidence that the mean amount is different from 1.0 gallon. b. Compute the​ p-value and interpret its meaning. What is the​ p-value? 0.161 = 0.157 or 0.159 - use an online p value calculator. use: https://www.socscistatistics.com/pvalues/normaldistribution.aspx Interpret the meaning of the​ p-value. Choose the correct answer below. Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the mean amount is equal to 1 gallon. c. Construct a 99​% confidence interval estimate of the population mean amount of water per​ 1-gallon bottle. use: https://www.socscistatistics.com/confidenceinterval/default3.aspx d. Compare the results of​ (a) and​ (c). What conclusions do you​ reach? The results of​ (a) and​ (c) are the same

An orange juice producer buys oranges from a large orange grove that has one variety of orange. The amount of juice squeezed from these oranges is approximately normally​ distributed, with a mean of 5.0 ounces and a standard deviation of 0.24 ounce. Suppose that you select a sample of 36 oranges. a. What is the probability that the sample mean amount of juice will be at least 4.68 ​ounces? b. The probability is 78​% that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population​ mean? c. The probability is 79​% that the sample mean amount of juice will be greater than what​ value?

a. The probability is 1.000 ST. DEV: σ/√n 0.24 / √36 = 0.04 EXCEL: =NORM.DIST X: 4.68 MEAN: 5 ST. DEV: 0.04 CUMULATIVE: TRUE 6.22096E-16 1 - 6.22096E-16 = 1.000 b. There is a 78​% probability that the sample mean amount of juice will be contained between 4.95 ounce(s) and 5.04 ounce(s). 78% - 100% = 22% 22% / 2 = 11% 11% - 100% = 89% EXCEL =NORM.INV PROBABILITY: 11% MEAN: 5 ST. DEV: 0.04 4.95 EXCEL =NORM.INV PROBABILITY: 89% MEAN: 5 ST. DEV: 0.04 5.04 c. There is a 79​% probability that the sample mean amount of juice will be greater than 4.96 ounce(s). 79% - 100% = 21% EXCEL =NORM.INV PROBABILITY: 21% MEAN: 5 ST. DEV: 0.04 4.96

An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.77 inch. The lower and upper specification limits under which the ball bearings can operate are 0.76 inch and 0.78 ​inch, respectively. Past experience has indicated that the actual diameter of the ball bearings is approximately normally​ distributed, with a mean of 0.775 inch and a standard deviation of 0.007 inch. Complete parts​ (a) through​ (e) below.

a. What is the probability that a ball bearing is between the target and the actual​ mean? EXCEL: =NORM.DIST X: 0.77 MEAN: 0.775 ST. DEV: 0.007 CUMULATIVE: TRUE 0.237525262 EXCEL: =NORM.DIST X: 0.775 MEAN: 0.775 ST. DEV: 0.007 CUMULATIVE: TRUE 0.5 SUBTRACT 0.5 - 0.237525262 = 0.2624 b. What is the probability that a ball bearing is between the lower specification limit and the​ target? EXCEL: =NORM.DIST X: 0.76 MEAN: 0.775 ST. DEV: 0.007 CUMULATIVE: TRUE 0.016062286 SUBTRACT 0.237525262 - 0.016062286 = 0.2214 c. What is the probability that a ball bearing is above the upper specification​ limit? EXCEL: =NORM.DIST X: 0.78 MEAN: 0.775 ST. DEV: 0.007 CUMULATIVE: TRUE 0.7624744738 SUBTRACT 1 - 0.7624744738 = 0.2375 d. What is the probability that a ball bearing is below the lower specification​ limit? - SAME NORM.DIST FROM QB. 0.0160 e. Of all the ball​ bearings, 94​% of the diameters are greater than what​ value? 94% - 100% = 6% EXCEL: =NORM.INV PROBABILITY: 6% MEAN: 0.775 ST. DEV: 0.007 0.764

Webrooming, researching products online before buying them in​ store, has become the new norm for some consumers and contrasts with​ showrooming, researching products in a physical store before purchasing online. A recent study reported that most shoppers have a specific spending limit in place while shopping online. Findings indicate that men spend an average of $245 online before they decide to visit a store. Assume that the spending limit for men is normally distributed and that the standard deviation is $18. Complete parts​ (a) through​ (d) below.

a. What is the probability that a male spent less than ​$211 online before deciding to visit a​ store? EXCEL X: 211 MEAN: 245 ST. DEV: 18 CUMULATIVE: TRUE 0.0294 b. What is the probability that a male spent between $265 and ​$291 online before deciding to visit a​ store? EXCEL =NORM.DIST X: 265 MEAN: 245 ST. DEV: 18 CUMULATIVE: TRUE 0.86673974 EXCEL =NORM.DIST X: 291 MEAN: 245 ST. DEV: 18 CUMULATIVE: TRUE 0.99469908 SUBTRACT 0.86673974 - 0.99469908 =0.1279 c. Ninety-seven percent of the amounts spent online by a male before deciding to visit a store are less than what​ value? EXCEL =NORM.INV PROBABILITY: 97% MEAN: 245 ST. DEV: 18 278.85 d. Ninety-five percent of the amounts spent online by a male before deciding to visit a store are between what two values symmetrically distributed around the​ mean? 95% - 100% = 5% 5% / 2 = 2.5% EXCEL =NORM.INV PROBABILITY: 2.5% MEAN: 245 ST. DEV: 18 209.72 95% + 2.5% = 97.5% EXCEL =NORM.INV PROBABILITY: 97.5% MEAN: 245 ST. DEV: 18 280.27

According to reports by interns to an employment​ website, the mean monthly pay of interns at Company X is $6,559. Suppose that the intern monthly pay is normally​ distributed, with a standard deviation of $600. Complete parts​ (a) through​ (e) below.

a. What is the probability that the monthly pay of an intern at Company X is less than ​$6,350​? EXCEL: =NORM.DIST X: 6350 MEAN: 6559 ST. DEV: 600 CUMULATIVE: TRUE 0.3637 b. What is the probability that the monthly pay of an intern at Company X is between ​$6,250 and $6,700​? EXCEL: =NORM.DIST X: 6250 MEAN: 6559 ST. DEV: 600 CUMULATIVE: TRUE 0.30327651 EXCEL: =NORM.DIST X: 6700 MEAN: 6559 ST. DEV: 600 CUMULATIVE: TRUE 0.59289563 0.59289563 - 0.30327651 = 0.2896 c. What is the probability that the monthly pay of an intern at Company X is greater than ​$7,100​? 0.1836 d. Ninety-eight percent of the monthly intern pays are higher than what​ value? 98% - 100% = 2% EXCEL =NORM.INV PROBABILITY: 2% MEAN: 6559 ST. DEV: 600 5326 e. Ninety-five percent of the monthly intern pays are between what two​ values, symmetrically distributed around the​ mean? 95% - 100% = 5% 5% / 2 = 2.5% 95% + 2.5% = 97.5% EXCEL =NORM.INV PROBABILITY: 2.5% MEAN: 6559 ST. DEV: 600 5383 EXCEL =NORM.INV PROBABILITY: 97.5% MEAN: 6559 ST. DEV: 600 7734