Question

A study was designed to compare the attitudes of two groups of nursing students towards computers. Group 1 had previously taken a statistical methods course that involved significant computer interaction. Group 2 had taken a statistic methods course that did not use computers. The students' attitudes were measured by administering the Computer Anxiety Rating Scale (CARS). A random sample of 15 nursing students from Group 1 resulted in a mean score of 59.659.6 with a standard deviation of 8.1. A random sample of 12 nursing students from Group 2 resulted in a mean score of 65.5with a standard deviation of 5.2. Can you conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2? Let μ1represent the mean score for Group 1 and μ2 represent the mean score for Group 2. Use a significance level of α=0.01 for the test. Assume that the population variances are equal and that the two populations are normally distributed.

Step 1 of 4: State the null and alternative hypotheses for the test.

Step 2 of 4: Compute the value of the test statistic. Round your answer to two decimal places.

Step 3 of 4: Determine the decision rule for rejecting the null hypothesis H0H0. Round the numerical portion of your answer to two decimal places.

Step 4 of 4: Make the decision for the hypothesis test.

Answer 1

Step 1:

The null and alternative hypotheses for the test are,

H0: μ1 = μ2

H1: μ1 < μ2

Step 2:

Since it is assumed that the population variances are equal and that the two populations are normally distributed, we use pooled t-test.

Pooled variances, Sp² = [(n1 - 1) s1² + (n2 - 1) s2² ] / (n1 + n2 - 2)

= [(15 - 1) * 8.1² + (12 - 1) * 5.2² ] / (15 + 12 - 2)

= 48.6392

Standard error of mean difference, SE =

= 2.701089

Test Statistic, t = (x1 - x2) / SE = (59.6 - 65.5) / 2.701089

= -2.18

Step 3:

Degree of freedom = n1 + n2 - 2 = 15 + 12 - 2 = 25

For left tail test, Critical value of t at α=0.01 and df = 25 is -2.79

Decision rule: Reject H0 if t < -2.79

Step 4:

Since test Statistic t is greater than the critical value, we fail to reject H0 and conclude that there is no significant evidence that the mean score for Group 1 is significantly lower than the mean score for Group 2 (μ1 < μ2)