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 1111 nursing students from Group 1 resulted in a mean score of 40.840.8 with a standard deviation of 5.45.4. A random sample of 1313 nursing students from Group 2 resulted in a mean score of 54.554.5 with a standard deviation of 2.32.3. Can you conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2? Let μ1μ1 represent the mean score for Group 1 and μ2μ2 represent the mean score for Group 2. Use a significance level of α=0.01α=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 t test statistic. Round your answer to three decimal places.

Step 3 of 4:

Determine the decision rule for rejecting the null hypothesis H0H0. Round your answer to three decimal places.

Step 4 of 4:

State the test's conclusion.

Answer 1

Date: 15/12/2019 Answer Step 1 To test the hypothesis is that the mean score for Group 1 is significantly lower than the mean score for Group 2 at 1% significance level. The null and alternative hypothesis is, Ho :4 = H2 Et > 'rt:°H Step 2 The t test-statistics is, First, compute the pooled variance then find the va of the t test-statistics.

The pooled variance is, (n, – 1)s + (n, – 1)s (n - 1) + (n, – 1) (11–1) x 5.4? + (13 – 1) × 2.32 (11-1) + (13 – 1) = 16.14 %3D The t test-statistics is, (X - x2) - (4 - 4) d. п, (40.80 – 54.50)-0 16.14 , 16.14 11 13 = -8.324 The t test-statistics is -8.324

Step 3 The t-critical value is, The sample size is small and left-tailed test. Look in the column headed a = 0.01 and the row headed in the t distribution table by using degree of freedom is, d.f.=n, +n, -2 = 11+13 – 2 = 22 The t-critical value is - 2.508 The decision rule is, Reject the null hypothesis, if the t test-statistics is less than the t-critical value which is 2.508. Step 4

The conclusion is that the t value corresponds to sample statistics is fall in the critical region, so the null hypothesis is rejected at 1% level of significance. There is sufficient evidence to indicate that the mean score for Group 1 is significantly lower than the mean score for Group 2. The result is statistically significant.