A reading comprehension program claims that it can significantly increase your r

Question

A reading comprehension program claims that it can significantly increase your recall memory after only one 3-hour class. 50 students sign up for the course, and each student takes a 100 point reading comprehension test at the beginning of the course and again at the end. The expectation is that scores at the end of the course should be higher. The difference between each student's before and after score is calculated, yielding the following information: Difference (D) = (After Score - Before Score) Mean difference (D^- over bar) = -6.00 Standard Deviation (s_D) = 17.50) You decide to test whether this program significantly improves reading comprehension. What would be the value of your t-statistic here? (a) -0.343 (b) -0.357 (c) -2.400 (d) -7.500 (e) -16.800 (f) +16.800 The critical value for a one-tail t-test with 49 degrees of freedom and alpha equal to 0.05 is +1.677. Based on this test and its critical value, what would your decision and conclusion be? (a) Reject H_0. The course improves reading comprehension. (b) Reject H_0. The course leads to a decline in reading comprehension. (c) Do not reject H_0. The course does not improve reading comprehension. (d) None of the above. There is not enough information to make a decision. When comparing two independent samples, under what circumstances would a nonparametric test be more appropriate to do than a t-test? (a) when the sampled populations are normally distributed. (b) when the sampled populations have homogeneous variances (c) when the sampled populations have badly skewed distributions (d) when the sampled observations are not statistically independent of one another

Explanation / Answer

22)

t-stastic = ( Difference in means) / [Standard deviation/sqrt(n-1)] = -6/[17.5/sqrt(49)] = -2.400

Hence, correct answer is C

23) SInce absolute t-value of 2.4 is greater than 1.677, we reject the null hypothesis and can state than the course leads to a decline in reading comprehension. Hence, B is the correct answer

24) When the population has outliers which cannot be removed, we use a nonparametric test. In badly skewed distributions, where there are outliers, we should use a non parametric test. hence, correct answer is C

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