Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis).   The test statistic in a left-tailed test is z = -2.05

Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the nullhypothesis (reject the null hypothesis or fail to reject the null hypothesis). The test statistic in a right-tailed test is z = 0.52.

A teacher would like to determine if quiz scores improve after completion of a worksheet. The students take a pre-quiz before the worksheet and then another quiz after the worksheet. Assume that the quiz scores are normally distributed. The grades for each quiz are given below. Use a significance level of α=0.05. List 1 has the post-quiz scores. List 2 has the pre-quiz scores. The hypothesis test is to see if the Post-Quiz Scores are > the Pre-Quiz Scores So,        H0:   μd=0         H1:  μd > 0  post-quiz    pre-quiz11    137    1317    1418    1713    136    616    1221    1814    718    14 What is the test statistic?test statistic =  (Report answer accurate to 3 decimal places.)  What is the p-value for this sample? p-value =  (Report answer accurate to 3 decimal places.)  The correct decision is to  .  The correct summary would be:  that the worksheet did improve the average quiz scores

Use the given information to find the P-value. Also, use a 0.05 significance level and state theconclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis).With H1: p (does not equal) 0.377 the test statistic is z = 3.06.