Question

Question 1 0/3 pt You wish to test the following claim (Ha) at a significance level of a=0.01. Ho:p1=p2 Ha:p1#u2 You obtain the following two samples of data. -1.40 Test statistic three decimal places .19 P value four decimal places. Accept or reject the null Sample 1 101.8 54.7 76.6 26.8 91.5 83.8 84.1 96.4 80.7 83.2 94.4 89.6 71.2 81.9 87.3 89.6 93.1 60.2 91.1 103.9 85.4 82.3 90 74.3 98.7 100.9 96.4 74.7 95.8 98.1 67.6 81.3 72.6 96.9 68.9 92.3

Answer 1

first, we have to find the mean and standard deviation of the two given samples

Sample 1:

T1 = 83.88, S1 = 12.45, n1 = 63

Sample 2:

12 = 79.04, S2 = 20.51, n1 = 45

The following null and alternative hypotheses need to be tested: Ho: Mi = 12 Ha: 11 + H2 This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Based on the information provided, we have assumed that the population variances are unequal, then the number of degrees of freedom is computed as follows:

df = (si/nı + s/n2) (8/n)? _ (s/n2) 1+1 + 2 +1

Based on the information provided, the significance level is a = 0.01, and the degrees of freedom are df = 66.919. 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 two-tailed test is te = 2.651, for a = 0.01 and df = 66.919. The rejection region for this two-tailed test is R = {t: 1t| > 2.651}.

Since it is assumed that the population variances are unequal, the t-statistic is computed as follows: X1 - 2 + Vn 83.88 - 79.04 = 12.452 120.512 = 1.408 T 45

Since it is observed that t = 1.408 <tc = 2.651, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=0.1636, and since p= 0.1636 > 0.01, it is concluded that the null hypothesis is not rejected.

TEST Statistic = 1.408

P value = 0.1636

Accept the null hypothesis.

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