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 11 nursing students from Group 1 resulted in a mean score of 51.1 with a standard deviation of 4.3 . A random sample of 18 nursing students from Group 2 resulted in a mean score of 63.1 with a standard deviation of 7 . Can you conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2? Let μ 1 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 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

Pooled Variance
sp = sqrt((((n1 - 1)*s1^2 + (n2 - 1)*s2^2)/(n1 + n2 - 2))*(1/n1 + 1/n2))
sp = sqrt((((11 - 1)*4.3^2 + (18 - 1)*7^2)/(11 + 18 - 2))*(1/11 + 1/18))
sp = 2.3498

Test statistic,
t = (x1bar - x2bar)/sp
t = (51.1 - 63.1)/2.3498
t = -5.107

Decision rule:
This is left tailed test, for α = 0.1 and df = n1 + n2 - 2 = 27
Critical value of t is -1.314.
Hence reject H0 if t < -1.314

Reject the null hypothesis