Hypothesis Test: Independent Groups (t-test, unequal variance) PE Protocol PT Protocol 149.28 212.16 mean 185.89 190.58 std. dev. 853 821 n 1665 df -62.883 difference (PE Protocol - PT Protocol) 9.206 standard error of difference 0 hypothesized difference -6.831 t 5.91E-12 p-value (one-tailed, lower) -80.940 confidence interval 95.% lower -44.826 confidence interval 95.% upper 18.057 margin of error Figure 3: Hypothesis Test of Independent Groups (t-test, unequal variance) The tal = -6.8 falls in the rejection region. We conclude that the new protocol (PE) results in a shorter average service time than the traditional protocol (PT) based on available data.Hypothesis Test: Mean vs. Hypothesized Value 150.000 hypothesized value 147.922 mean QueueTime 137.971 std. dev. 3.372 std. error 1674 n -062 z 2689 p-value (one-tailed, lower) 141.313 confidence interval 95.% lower 154 532 confidence interval 95.% upper 6.609 margin of error Figure 2: Results of TiQ, Hypothesis Test: Mean versus Hypothesized Value Average Service Time Test The team performed a test of hypothesis to determine whether the service time (ST) with new service protocol PE is lower than with the current PT protocol. A significance level of a=0.05 was used. This is a test of means for two independent samples with unknown variances assumed unequal. Sample 1 is the data from the PT (current) protocol. Sample 2 is the data from the PE (new) protocol. We tested whether the mean ST with protocol PE is smaller than the mean with protocol PT. The null and alternate hypotheses are: Ho :pt=p2 H1:p1=p2 This is a right-tailed test. Because of the very large samples, there is no real difference between finding the critical value with a normal distribution or the { distribution. The critical value with a significance level a=0.05ist=1.645 The decision rule becomes: Reject Ho if tgaz > 1.645. Time in Queue Test The team performed a test of hypothesis to determine whether the average JiQ is lower than the industry standard of 2.5 minutes (150 seconds). A significance level a=0.05 was used. This was a test of mean against a hypothesized value of 150 seconds. Because the sample size was large, we assumed knowledge of the population's variance. The null and alternate hypotheses is: Ho : p > 150 H1 : p