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 16 nursing students from Group 1 resulted in a mean score of 41.3 with a standard deviation of 2.5. A random sample of 11 nursing students from Group 2 resulted in a mean score of 46.7 with a standard deviation of 4.8. 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=0.01 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 3 decimals. )

Step 3. Determine the decision rule for rejecting the null hypothesis H0. Round your answer to three decimal places.

Step 4. State the test's conclusion. A) Reject Null Hypothesis B) Fail to Reject Null Hypothesis

Answer 1

Let μ1 represent the mean score for Group 1 and μ2 represent the mean score for Group 2

Now , we want to test the claim that the mean score for Group 1 is significantly lower than the mean score for Group 2.

Step 1 ) Hypothesis :
Hopi = pz
Vs
HQ: M1 <H2

Since , the population variances are equal and that the two populations are normally distributed.

Therefore , use the t-test for two samples.

The pooled estimate is ,

(n1 - 1)si + (n2 – 1)s n1 + n2 - 2 /(16 – 1)2.52 + (11 – 1)4.82 -= 3.6008 16+ 11 - 2

Step 2 ) The test statistic is ,

t= X7-X, = t +n2+2 SVITA

41.3-46.7 = -3.8289 3.6008.

Step 3 ) Critical Value : ; From t-table

tanı+ng+2 = 10.01,16+11-2 = to.01.25 = -2.485

Rejection rule : If   then reject Ho , accept otherwise

ESTAT <-tanı +ng+2

Decision : If

USTAT = -3.8289 < tani +12+2 = -2.485

Therefore , reject Ho

Step 4 )

Conclusion :

A) Reject Null Hypothesis

Hence , there is sufficient evidence to support the claim that the mean score for Group 1 is significantly lower than the mean score for Group 2.