Question

Proctored Nonproctored ?2 30 84.06 A study was done on proctored and nonproctored tests. The results are shown in the table ssume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal Complete parts (a) and (b) below Use a 0 05 significance level for both parts X 77.73 s 11761883 a. Test the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests What are the null and alternative hypotheses? The test statistic, t, is(Round to two decimal places as needed) The P.value is(Round to three decimal places as needed ) State the conclusion for the test OA. Fal to reject Ho There is sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests Reject Ho There is sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests O B. C Fail tn reiert Ha There is nnt sufficient avirdenre to siinnort the rlaim that sturdents taking nonnrortnred tacts net a Click to select your answer(s).

Answer 1

a.)

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested This corresponds to a left-tailed test, for which a z-test for two population means, with known population standard deviations will be used (2) Rejection Region Based on the information provided, the significance level is α-0.05, and the critical value for a leit- tailed test is zc1.64. The rejection region for this left-tailed test is R-{: z <-1.64) The z-statistic is computed as follows: 77.73 -84.06 1.588 Since it is observed that z-1.588> z1.64, it is then concluded that the null hypothesis is not rejected.

O C. Fail to reject Ho. There is not sufficient evidence to support the claim that students taking nonproctored tests get a higher mean scare than those taking proctored tests

b. confidence interval

using this formula confidence interval is

Confidence Interval The 95% confidence interval for μι-μ2 is-14.142 < μι-μ2 < 1.482

c). yes, because the confidence interval does not contain $\small \mu 1-\mu 2$ value .