Question

A survey asked orchestra members to estimate the number of hours they practice on a typical day. From the results, it was determined that 25 pianists reported practicing an average of 4.1 hours a day with a standard deviation of 0.77, while 31 percussionists reported practicing an average of 3.7 hours a day with a standard deviation of 1.03. Using alpha = 0.05, test the claim that pianists practice longer than percussionists.

d. t-test or z-test? Explain how you know. What would be the critical value(s)?

e. Calculate the test statistic. What decision is justified by your work?

f. Summarize your results.

Please help, thank you!!

Answer 1

Since the population standard deviations are unknown for both samples, we will use a t-test.

Entire test:

T-test for two Means – Unknown Population Standard Deviations - Equal Variance

The following information about the samples has been provided: a. Sample Means : Xˉ1​=4.1 and Xˉ2​=3.7 b. Sample Standard deviation: s1=0.77 and s2=1.03 c. Sample size: n1=25 and n2=31 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ1 =μ2 Ha: μ1 >μ2 This corresponds to a Right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) The degrees of freedom Assuming that the population variances are equal, the degrees of freedom are given by n1+n2-2=25+31-2=54. (3a) Critical Value Based on the information provided, the significance level is α=0.05, and the degree of freedom is 54. Therefore the critical value for this Right-tailed test is tc​=1.6736. This can be found by either using excel or the t distribution table. (3b) Rejection Region The rejection region for this Right-tailed test is t>1.6736 (4)Test Statistics The t-statistic is computed as follows: (5) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0.0565 (6) The Decision about the null hypothesis (a) Using traditional method Since it is observed that t=1.6113 < tc​=1.6736, it is then concluded that the null hypothesis is Not rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.0565, and since p=0.0565>0.05, it is concluded that the null hypothesis is Not rejected. (7) Conclusion It is concluded that the null hypothesis Ho is Not rejected. Therefore, there is Not enough evidence to claim that the population mean μ1​ is greater than μ2, at the 0.05 significance level.

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