Elementary Statistics: Picturing the World (7th Edition)
7th Edition
ISBN: 9780134683416
Author: Ron Larson, Betsy Farber
Publisher: PEARSON
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Chapter 8.3, Problem 17E
Testing the Difference Between Two Means In Exercises 9–20, (a) identify the claim and state H0 and Ha, (b) find the critical value(s) and identify the rejection region(s), (c) calculate
17. Product Ratings A company claims that its consumer product ratings (0–10) have changed from last year to this year. The table shows the company’s product ratings from the same eight consumers for last year and this year. At α = 0.05, is there enough evidence to support the company’s claim?
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P-values In Exercises 17–20, do the following:
a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.
b. Find the P- value. (See Figure 8-3 on page 364.)
c. Using a significance level of α = 0.05, should we reject H0 or should we fail to reject H0?
The test statistic of z = 1.00 is obtained when testing the claim that p > 0.3.
Test the claim that the mean GPA of Orange Coast students is larger than the mean GPA of
Coastline students at the 0.01 significance level.
The null and alternative hypothesis would be:
Ho:μο = μc Ho:μο με H1:po pc Hy:μο < με
Reject the null hypothesis
O Fail to reject the null hypothesis
(to 2 decimals)
(to 2 decimals)
Test the claim that the proportion of people who own cats is larger than 20% at the 0.005
significance level.
The null and alternative hypothesis would be:
H2:p = 0.2 Họ:p 2 0.2 H9:p 0.2 H9:p = 0.2 Hg: u 0.2 H1: u 0.2
The test is:
left-tailed two-tailed right-tailed
Based on a sample of 300 people, 24% owned cats
The test statistic is:
(to 2 decimals)
The p-value is:
(to 2 decimals)
Based on this we:
O Reject the null hypothesis
O Fail to reject the null hypothesis
Chapter 8 Solutions
Elementary Statistics: Picturing the World (7th Edition)
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- Test the claim that the mean GPA of night students is significantly different than the mean GPA of day students at the 0.05 significance level. The null and alternative hypothesis would be: Ho: PNPD Ho: UN ≤ HD Ha:N HD Ha:μN > HD The test is: right-tailed two-tailed left-tailed O Ho: PNPD Ho: PNPD Ho: PN Ha:PN PD Oarrow_forwardTest the claim that the proportion of people who own cats is larger than 50% at the 0.05 significance level. The null and alternative hypothesis would be: Ho:u = 0.5 Ho:p 0.5 Ho:p > 0.5 H:p 0.5 H:p> 0.5 H1:p > 0.5 H1:p #0.5 H1:p < 0.5 H:p < 0.5 The test is: two-tailed left-tailed right-tailed Based on a sample of 500 people, 52% owned cats The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: O Fail to reject the null hypothesis CO Reject the null hypothesis Check Answer op Carrow_forwardTest the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.01 significance level. The null and alternative hypothesis would be: = Ho:po Spc Ho:μο Σμα Ho:po = pc Hy:po > pc Hy:μο με Reject the null hypothesis Fail to reject the null hypothesis Ho: Po = μα Hy:μο # με (to 2 decimals) (to 2 decimals)arrow_forward
- Testing Claims About Variation. In Exercises 5–16, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Assume that a simple random sample is selected from a normally distributed population. Body Temperature Example 5 in Section 8-3 involved a test of the claim that humans have body temperatures with a mean equal to 98.6°F. The sample of 106 body temperatures has a standard deviation of 0.62°F. The conclusion in that example would change if the sample standard deviation s were 2.08°F or greater. Use a 0.01 significance level to test the claim that the sample of 106 body temperatures is from a population with a standard deviation less than 2.08°F. What does the result tell us about the validity of the hypothesis test in Example 5 in Section 8-3?arrow_forwardTesting Claims About Variation. In Exercises 5–16, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Assume that a simple random sample is selected from a normally distributed population. Mint Specs Listed below are weights (grams) from a simple random sample of “wheat” pennies (from Data Set 29 “Coin Weights” in Appendix B). U.S. Mint specifications now require a standard deviation of 0.0230 g for weights of pennies. Use a 0.01 significance level to test the claim that wheat pennies are manufactured so that their weights have a standard deviation equal to 0.0230 g. Does the Mint specification appear to be met?arrow_forwardTest the claim that the proportion of people who own cats is larger than 70% at the 0.005 significance level. The null and alternative hypothesis would be: Ho: u 0.7 Ho: µ = 0.7 H: µ > 0.7 H:p 0.7 H:p > 0.7 H1 :p <0.7 H : µ < 0.7 H1: µ # 0.7 The test is: right-tailed left-tailed two-tailed Based on a sample of 300 people, 74% owned cats The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: O Fail to reject the null hypothesis O Reject the null hypothesisarrow_forward
- Test the claim that the mean placement test score for night students (N) is larger than the mean placement test score for day students (D) at the 0.05 significance level.The alternative hypothesis would be: Ha:μN>μDHa:μN>μD Ha:μN≠;μDHa:μN≠;μD Ha:pN≠;pDHa:pN≠;pD Ha:pN>pDHa:pN>pD Ha:μN<μDHa:μN<μD Ha:pN<pDHa:pN<pD I The test is: two-tailed right-tailed left-tailed Use a 0.05 alpha level. The sample consisted of 65 night students, with a sample mean GPA of 70 and a standard deviation of 11, and 45 day students, with a sample mean GPA of 66 and a standard deviation of 11.The test statistic is: The critical value is:arrow_forwardTest the claim that the mean GPA of night students is smaller than 2.6 at the 0.005 significance level. The null and alternative hypothesis would be: Ho:µ > 2.6 Ho:p = 0.65 Ho:u 0.65 H1:µ 2.6 H1:p > 0.65 H1:µ # 2.6 H1:p < 0.65 The test is: right-tailed left-tailed two-tailed Based on a sample of 80 people, the sample mean GPA was 2.59 with a standard deviation of 0.08 The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: O Fail to reject the null hypothesis O Reject the null hypothesis Quortion Holn: O Post to forumarrow_forwardTest the claim that the mean GPA of night students is smaller than 2.2 at the 0.005 significance level. The null and alternative hypothesis would be: Ho: u 0.55 Ho:p 2.2 o H1:µ > 2.2 H1:p 0.55 H1:p # 0.55 H1: µ 7 2.2 H1: p < 2.2 The test is: two-tailed right-tailed left-tailed o Based on a sample of 65 people, the sample mean GPA was 2.18 with a standard deviation of 0.06 The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Rased on thic we P Type here to search HEWLETT - PAC ARD 12 esc 14 * 144 @ #3 $ % 1 3. 4 & 8. tab Q W T. Y. lock D F G C V.arrow_forward
- Test the claim that the mean GPA of night students is significantly different than 2 at the 0.01 significance level. The null and alternative hypothesis would be: Ho:µ = 2 Ho:µ 0.5 H,:µ > 2 H:p 2 H :p + 0.5 H¸ :p 0.5 The test is: left-tailed right-tailed two-tailed Based on a sample of 75 people, the sample mean GPA was 2.03 with a standard deviation of 0.08 The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: OFail to reject the null hypothesis O Reject the null hypothesisarrow_forwardTesting Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim. Lead in Medicine Listed below are the lead concentrations (in μ g/g) measured in different Ayurveda medicines. Ayurveda is a traditional medical system commonly used in India. The lead concentrations listed here are from medicines manufactured in the United States (based on data from “Lead, Mercury, and Arsenic in US and Indian Manufactured Ayurvedic Medicines Sold via the Internet,” by Saper et al., Journal of the American Medical Association, Vol. 300, No. 8). Use a 0.05 significance level to test the claim that the mean lead concentration for all such medicines is…arrow_forwardTest the claim that the proportion of people who own cats is smaller than 60% at the 0.025 significance level. The null and alternative hypothesis would be: Ho: p=0.6 Ho:μ ≥ 0.6 Ho: p 0.6 The test is: left-tailed right-tailed two-tailed Based on a sample of 600 people, 53% owned cats The test statistic is: The p-value is: Based on this we: Ho:p>0.6 Ho: 0.6 Ho: μ 0.6 Reject the null hypothesis O Fail to reject the null hypothesis (to 2 decimals) (to 2 decimals) =arrow_forward
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