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 1010 nursing students from Group 1 resulted in a mean score of 45.345.3 with a standard deviation of 4.54.5. A random sample of 1717 nursing students from Group 2 resulted in a mean score of 55.855.8 with a standard deviation of 6.96.9. 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=0.1α=0.1 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.

Answer 1

n1 = 10

= 45.3

s1 = 4.5

n2 = 17

= 55.8

s2 = 6.9

Claim: The mean score for Group 1 is significantly lower than the mean score for Group 2.

The null and alternative hypothesis is

Level of significance = α=0.1

The population variances are equal.

So we have to use here pooled variance.

Test statistic is

Degrees of freedom = n1 + n2 - 2 = 10 + 17 - 2 = 25

Critical value = 1.316     ( Using t table)

| t | > critical value we reject null hypothesis.

Conclusion:

The mean score for Group 1 is significantly lower than the mean score for Group 2.