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 47.5 with a standard deviation of 7.8 . A random sample of 15 nursing students from Group 2 resulted in a mean score of 54.7 with a standard deviation of 5.4 . 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 μ 2  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 2 of 4 :

Compute the value of the t test statistic. Round your answer to three decimal places.

Answer 1

The hypotheses are:

Rejection region:

Reject Ho if Tobs<-t_0.1,25=-1.317, since degree of freedom =n1+n2-2=12+15-2=25

Test statistic:

t=-2.83

P- value:

P-value calculated using t value and degree offreedom as

0.0045< P-value<0.005

Conclusion:

since t obs =-2.83< -t_{0.1, 25}=-1.317 and p value is less than 0.1 , we reject the null hypothesis and conclude that we havw enough evidence to suuport the claim.

The p-value and t0.1,25 are calculated using the table shown below: