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 10 nursing students from Group 1 resulted in a mean score of 65.8 with a standard deviation of 5.1. A random sample of 16 nursing students from Group 2 resulted in a mean score of 70.4 with a standard deviation of 7.6. 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=0.05 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 H0. Round your answer to three decimal places.

Step 4 of 4: State the test's conclusion.

Answer 1

The samples given are independent since they are from two different nursing students. We want to test the difference of the population means. The population SD are unknown and are assumed to be equal. Also the data is normally distributed so we will conduct an independent samples' t-test for equal variances.

We want to test if mean of group (1) is lower than mean of group 2: (1) < (2) : (1) - (2) < 0. So this is a left tailed one sided test.

Step 1 of 4: State the null and alternative hypotheses for the test.

H_0: μ1 - μ2 = 0

H_1: μ1 - μ2 < 0

	Group 1	Group 2
n	10	16
Mean	65.8	70.4
SD	5.1	7.6
Var(SD2)	26.01	57.76

Step 2 of 4: Compute the value of the t test statistic. Round your answer to three decimal places.

Pooled variance =

= 45.8538

Test Stat =

Where the null difference = 0

Test stat =-1.6852

Step 3 of 4: Determine the decision rule for rejecting the null hypothesis H0. Round your answer to three decimal places.

p-value approach

We reject the null hypothesis if p-value < level of significance

Critical value approach

We reject the null hypothesis if |Test Stat| > C.V. at level of significance

Step 4 of 4: State the test's conclusion.

Level of significance

p-value = P ( > |T.S. |)

= P(t₂₄ > 1.69) ........using t-dist tables

p-value = 0.052

Since p-value >0.05

We do not reejct the null hypothesis.

C.V. =

=t_{24, 0.05}

C.V. =1.7109

Since |test Stat| < C.V.

We do not reject the null hypothesis.

There is insufficient evidence to support the claim that mean of group (1) is significantly lower than mean of group 2.