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 12 nursing students from Group 1 resulted in a mean score of 54.7 with a standard deviation of 6. A random sample of 15 nursing students from Group 2 resulted in a mean score of 66.6 with a standard deviation of 3. Can you conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2? Let μ1 represent the mean score for Group 1 and μ2 represent the mean score for Group 2. Use a significance level of α=0.1 for the test. Assume that the population variances are equal and that the two populations are normally distributed. Step 1 : State the null and alternative hypotheses for the test. Step 2 : Compute the value of the t test statistic. Round your answer to three decimal places Step 3 : Determine the decision rule for rejecting the null hypothesis H0. Round your answer to three decimal places. Reject or fail to reject your hypothesis?

Answer 1

Step 1:

H0: Null Hypothesis: ( the mean score for Group 1 is not significantly lower than the mean score for Group 2)

HA: Alternative Hypothesis: ( the mean score for Group 1 is significantly lower than the mean score for Group 2) (Claim)

Step 2:

Pooled Standard Deviation is given by:

Test Statistic is given by:

the value of the t test statistic = - 6.724

Step 3:

= 0.10

df = 12+ 15 - 2 = 25

From Table, critical value of t = - 1.316

the decision rule for rejecting the null hypothesis H0 :

Reject H0 if t < - 1.316

Since calculated value of t = - 6.724 is less than critical value of t = - 1.316:

Correct option:

Reject our hypothesis