Question

To test if the participation in a “communication training” will result in a higher student evaluations, a randomly selected group of 20 faculty members participate in the training and the difference of their ratings (on a 4-point scale) are recorded to test The following are the data with summary statistics Faculty before training after training difference 1 1.94 2.26 0.32 2 3.19 3.62 0.43 3 2.96 3.33 0.37 4 2.99 3.46 0.47 5 2.73 3.1 0.37 6 2.62 2.97 0.35 7 2.87 3.2 0.33 8 3.21 3.44 0.23 9 3.4 3.8 0.4 10 3.11 3.45 0.34 11 2.77 3.1 0.33 12 2.8 3.25 0.45 13 1.87 2.15 0.28 14 2.47 2.78 0.31 15 3.44 3.82 0.38 Atul N Roy                                                                                                                                                          pg. 3 16 2.77 3.12 0.35 17 2.4 2.68 0.28 18 2.12 2.49 0.37 19 2.34 2.79 0.45 20 2.84 3.09 0.25 Variable N Mean StDev before training 20 2.7420 0.4456 after training 20 3.095 0.465 difference 20 0.3530 0.0660 Use the above information to compute the test statistic and state if you would reject or not reject the null hypothesis at 5% level of significance.

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d< 0

Alternative hypothesis: u_d > 0

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

Before After d A B d - dbar) 2 1.94 3.19 2.96 2.99 2.26 3.62 3.33 3.46 0.32 0.001089 0.43 0.005929 2.73 2.62 2.87 3.21 0.37 0.000289 0.47 0.0136889 0.37 0.000289 0.35 0.0000089 0.33 0.0005289 0.230.015129 2.97 3.2 3.44 3 3.11 2.77 2.8 1.87 2.47 3.45 3.25 2.15 2.78 0.34 0.00016899 0.33 0.0005289 0.45 0.0094 0.28 0.005329 0.31 0.00184 2899 2.77 3.12 0.35 0.0000089 2.4 2.68 0.28 0.0053289 2.34 2.84 54.84 2.742 2.79 3.09 61.9 3.095 0.45 0.00940 0.250.010609 7.06 0.08282 8 Sum Mean 0.353 0.004141

$s = \sqrt{\frac{\sum (d-\bar{d})^{2}}{n-1}}$

s = 0.066022

SE = s / sqrt(n)

S.E = 0.01476

DF = n - 1 = 20 -1

D.F = 19

t = [ (x₁ - x₂) - D ] / SE

t = 23.92

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 19 degrees of freedom is greater than 23.92.

Thus, the P-value = less than 0.0001.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

Reject H₀. from the above test we have sufficient evidence in the favor of the claim that the participation in a “communication training” will result in a higher student evaluation.