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
10. SAT Scores An instructor for a SAT preparation course claims that the course will improve the test scores of students. The table shows the critical reading scores for 10 students the first two times they took the SAT. Before taking the SAT for the second time, the students took the instructor’s course to try to improve their critical reading SAT scores. At α = 0.01, is there enough evidence to support the instructor’s claim?
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Elementary Statistics: Picturing the World (7th Edition)
- (c) Calculate the test statistic. d) Decide whether to reject or fail to reject the null hypothesis. Then interpret the decision in the context of the original claim.arrow_forward(c) Calculate the test statistic. d) Decide whether to reject or fail to reject the null hypothesis. Then interpret the decision in the context of the original claim.arrow_forwardTest 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)arrow_forward
- 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:PN 2 PD Ho:PN = PD Ho: PN PD H1:UN > HD H1:uN Next Question M hparrow_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 = Pc Ho:Po > Pc Ho:µ0 µc H1: Ho Pc H1:µo + HC The test is: two-tailed left-tailed right-tailed The sample consisted of 55 Orange Coast students, with a sample mean GPA of 3.46 and a standard deviation of 0.08, and 55 Coastline students, with a sample mean GPA of 3.49 and a standard deviation of 0.07. 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 Reject the null hypothesisarrow_forwardTest 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 hypothesisarrow_forward
- ASSESSMENT: A. Classify the random variables as discrete or continuous. 1. number of defective computers produced by a manufacturer 2. weight of newborns each year in a hospital 3. amount of paint utilized in a building project 4. average amount of electricity consumed per household per month 5. number of deaths per year attributed to lung cancerarrow_forwardIllustrate the null hypothesis for a correlated t test?arrow_forwardAnalysis of whether two categorical variables, X = course subject and Y = grade earned by a student was conducted. Researchers selected n = 800 students for their study using m = 4 distinct courses and k = 6 grades, from A to F and W. They evaluated a x2 test statistic for testing independence between X and Y equal to TS = 29.64 At the significance level a = 0.01, do researchers have sufficient evidence that X and Y are dependent? 1. Show critical value (or values) needed for this procedure 2. Formulate the rejection rule 3. State your decision: whether the independence hypothesis should be rejected at the significance level a = 0.01 Solutionarrow_forward
- provide an example of hypothesis testing using the t score testarrow_forwardTest the claim that the proportion of people who own cats is smaller than 80% at the 0.10 significance level. The null and alternative hypothesis would be: 0.8 Но: д — 0.8 Но:д 0.8 Но:р > 0.8 Нo:р H1: µ 0.8 H1:p > 0.8 The test is: left-tailed right-tailed two-tailed Based on a sample of 800 people, 73% owned cats The p-value is: (to 2 decimals) Based on this we: O Reject the null hypothesis Fail to reject the null hypothesisarrow_forwardQuestion below pleasearrow_forward
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