Question

Website 1 Website 2 n 80 n2=95 A market analyst wants to know if the new website he designed is showing increased page views per visit and calculates the summary statistics in the table to the right. Assume that the new website is website 1 and the old website is website 2. Test the hypothesis that the mean number of page views from the two websites is the same.. You may assume that the number of page views from each website follow a Normal distribution. Complete parts a through c below. y16.6 2 6.4 $2 =2.9 t 0.438 S 3.1 a) Calculate the P-value of the statistic given that the approximation formula gives df 163.6. The P-value is 0.331 (Round to three decimal places as needed.) b) Calculate the P-value of the statistic using the rule that df min(n, 1, n2-1). The P-value is 0.331 (Round to three decimal places as needed.) c) What do you conclude at o 0.05? Fail to reject Ha for both cases. There is not sufficient evidence to reject the claim that the mean number of page visits is the same for the two websites, in favor of the claim of increased page visits.

Answer 1

The provided sample means are shown below: X6.6 X2 6.4 Also, the provided sample standard deviations are 81 3.1 82 2.9 and the sample sizes are ni 80 and n2 95 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested Ho: 12 Ha: 2 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) Rejection Region Based on the information provided, the significance level is a =0.05, and the degrees of freedom are df 173. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: Hence, it is found that the critical value for this right-tailed test is te 1.654, for a 0.05 and df = 173. The rejection region for this right-tailed test is R = {t :t> 1.654}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X, — Хз (n;-1)+(ng- nn-2 6.6 6.4 = 0.44 (80-1)3.1+(95-1)2.92 80+95 2 (4) Decision about the null hypothesis 0.44 < t = 1.654, it is then concluded that the null Since it is observed that t hypothesis is not rejected. Using the P-value approach: The p-value is p 0.3301, and since p 0.3301 2 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean is greater than 2, at the 0.05 significance level

(a) p value will be 0.330259 corresponds to df= 163.5

(b) p value = 0.330569 corresponds to df=79

(c) fail to reject H0.it it not sufficient ......

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