A teacher would like to determine if quiz scores improve after completion of a worksheet. Thestudents take a pre-quiz before the worksheet and then another quiz after the worksheet. (Pre-quizscore - Post-quiz score) Assume quiz scores are normally distributed. The grades for each quiz aregiven below. Use a significance level of a = 0.05.H₂: Hd = 0Ha: Hd <0pre-quiz1216171611151271513post-quiz1316151614151391415What is the test statistic?test statistic =(Report answer accurate to 2 decimal places.)What is the p-value for this sample?p-value=(Report answer accurate to 3 decimal places.)The correct decision is to Select an answerConclusion: Select an answerthat the worksheet improved quiz scores.

Suppose μd is the mean value of the differences for the pre quiz score and post quiz scores.That is, μd = μPre-quiz – μPost-quiz. The observations are taken on same students; one for pre quiz and another for post quiz. Hence, the observations are related. Moreover, the sample is a random sample and the population is normally distributed. Hence, paired t-test for can be used. It is needed to whether the worksheet improved quiz scores. The null and alternative hypotheses are: H0: μd = 0 Ha: μd < 0 The test is left tailed. The sample size is n = 10. Consider the below table. The test statistic is: The test statistic is t = –1.26. The degrees of freedom is, n – 1 = 10 – 1 = 9. Using the EXCEL formula =T.DIST(-1.26,9,TRUE), the p-value is approximately 0.120. The excel output is, The p-value is 0.120. The level of significance is α = 0.05. Decision rule: If p-value is less than or equal to the level of significance, reject the null hypothesis. Otherwise fail to reject the null hypothesis. Conclusion: The p-value of 0.120 is greater than α = 0.05. Thus, the correct decision is to fail to reject the null hypothesis. Conclusion: There is not enough evidence to conclude that the worksheet improved quiz scores.