Question

1. A professor hypothesizes that students who earn a C or better in her class spend more time outside of class studying than students who receive a D or F. She collects the following data from two samples of students. What does she conclude?

Number of hours studying per week for C or better students:
8, 4, 4, 2, 1, 5, 3, 2, 3

Number of hours studying per week for D or F students:
6, 2, 1, 0, 3, 2, 1

1. Null hypothesis:
2. Alternative hypothesis:
3. Statistical test (be specific!):
4. Significance level: alpha = .05
5. degrees of freedom:
6. Critical region (t-value):
7. Calculated t (show your work):
8. Decision:

Answer 1

(a) The step-by-step instructi ons to obtain two-sample - test using Minitab18 Type the vari ables X and Y into the columns C1 and C2 of the Minitab18 worsheet. Choose the Stat ->Basic stati stics 2- sample t Choose Each sample in its column *Select the column C1 and C2 into the variables box. Choose the confi dence level 95. Choose the Hypothesized difference:0.0 Hypothesized difference. Choose the Alternative hypothesis: Difference> Click ok The two-sample - test of the populati on mean using Mintab18:

Two-Sample T-Test and Cl: X, Y Method Hr:mean of X Pmean of Y Difference: Hp - Equal variances are assumed for this analysis. Descriptive Statistics SE Mean Sample N Mean StDev X 3.56 2.07 0.69 Y 7 2.14 1.95 0.74 Estimation for Difference Pooled 95% Lower Bound for Difference Difference StDev 2.02 1.41 -0.38 Test Null hypothesis Alternative hypothesis Ho: HH0 - H: H: >0 - T-Value P-Value DF 1.39 14 0,093

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(a) Let and ^ be the populati on means and let x and x be the means of the respective samples The summary statisti cs of the Two-sample -test output of the Minitab18: Sample: 9, , = 3.56, s = 2.07 Sample2: 27, x, = 2.14, s2 =1.95. Step 1: The nul hypothesis is right-tailed test The null hypothesis is given below Ho (The two population We need to use equal) means are (b) The alternative hypothesis is H (The populati is greater than ,) on mean (c) Step2: Select the distribution to use: Here, the two samples independent, a and a are unknown but equal, and the sample are normally di stributed Hence, all conditions sizes are small but both populations are are sati sfied. Consequently we will use the t distribution (d) Step3: Determine the rejection andnon-rejection regi ons. sign in the alternative hypothesis indicates that the test is right-tailed The 0.05 The significance level is (e) The formula for calculating the degrees of freedom is df n2-2 =9+7-2 = 16-2 - 14 df 14 Therefore, the degrees of freedom is df = 14 (f The critical value From the t distribution table, the critical value of t for df 14 and 0.05 area in the right-tail of the t distributi on is 1761310136

(g) Step 4: Cal culate the value of the t-test stati stic. The value of the test statistic t for x- x, is computed as follows. The pooled standard deviati on (-1)(n-) S= -2 (9-1)4.277778+(7-1)3.809524 9+7-2 (8) (0.13425)+39(0.098735) 16-2 34.22222 22.85714 14 57.07937 = 14 = /4.077098 = 2.019182 s= 2.019182 Step5: The standard error of the difference between the means 1 1 = 2.019182. L 61 2.019182/0.111111+0.142857 = 2.019182 0.253968 = 2.019182 (0.503953) = 1.017572 두=11.017572 Step6: The formula for calculating the t-statistic is (3.555556-2.142857)-(0) 1.017572 1.412698413 1.017572 = 1.388302775 t=1.3883 (h Step7: The conclusion and the decisionis based on the critical value method: 1.3883 06 14 = 1.761310136. Since, the test stati stic is We do not reject the null hypothesis Ho There is not sufficient evidence to conclude that the population mean of the number of hours studying per week for C or better students is greater than that of the population mean of the number of hours studying per week for D or F stu dents