Question

Sample 2 11 n X Assume that both populations are normally distributed a) Test whether , at the = 0.01 level of significance for the given sample data b) Construct a 50% confidence interval about 4-12 Sample 1 19 5078 21 11.9 Click the icon to view the Student distribution table a) Perform a hypothesis test. Determine the null and alternative hypotheses O A HOM > B. Hy: H2 OB HM, H, H2 + C Họ P = H1 H1 * PR OD. H.H2 Determine the test statistic -Round to two decimal places as needed.) Determine the critical values). Select the correct choice below and fill in the answer boxes) within your choice (Round to three decimal places as needed.) O A The urtical values

Answer 1

a)

The following null and alternative hypotheses need to be tested: Ho: Mi M2 Ha: Hi > H2

test statistic value is

X X2 t= sa 721 + 12 = 50.8 - 43 17:12 + = 1.979 11.92 11

Based on the information provided, the significance level is a = 0.01, and the degrees of freedom are df = 14.211. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal: Hence, it is found that the critical value for this right-tailed test is te = 2.62, for a = 0.01 and df = 14.211.

critccal value = 2.62

Since it is observed that t = 1.979 <te=2.62, it is then concluded that the null hypothesis is not rejected.

b)

df = (si/nı + s/n2) (s/nı) (s/na + n1+1 ng+1 – 2 = 15.022 The critical value for a = 0.01 and df = 15.022 degrees of freedom is to = tı-a/2;1–1 = 2.946. The corresponding confidence interval is computed as shown below: Since we assume that the population variances are unequal, the standard error is computed as follows: se = + n2 7.12 19 + 11.92 11 = 3.94 Now, we finally compute the confidence interval: CI (X1 - X- te x se, X1 - X2 +te x se) = (50.8 – 43 – 2.946 x 3.94,50.8 – 43 +2.946 x 3.94) = (-3.809, 19.409)

we are allowed to solve one question only.