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 52.7 with a standard deviation of 8.5. A random sample of 12 nursing students from Group 2 resulted in a mean score of 63.2 with a standard deviation of 2.5. 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: Compute the value of the t test statistic. Round your answer to three decimal places.

Step 4 of 4: Determine the decision rule for rejecting the null hypothesis H0H0. Round your answer to three decimal places. State the test's conclusion.

Answer 1

1)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ1 = μ2
Alternative Hypothesis, Ha: μ1 < μ2

2)

Pooled Variance
sp = sqrt((((n1 - 1)*s1^2 + (n2 - 1)*s2^2)/(n1 + n2 - 2))*(1/n1 + 1/n2))
sp = sqrt((((16 - 1)*8.5^2 + (12 - 1)*2.5^2)/(16 + 12 - 2))*(1/16 + 1/12))
sp = 2.5425

Test statistic,
t = (x1bar - x2bar)/sp
t = (52.7 - 63.2)/2.5425
t = -4.130

3)

Rejection Region
This is left tailed test, for α = 0.05 and df = n1 + n2 - 2 = 26
Critical value of t is -1.706.
Hence reject H0 if t < -1.706

4)

reject the null hypothesis.