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 6 nursing students from Group 1 resulted in a mean score of 40.1 with a standard deviation of 5.9. A random sample of 10 nursing students from Group 2 resulted in a mean score of 45.6 with a standard deviation of 6.8. 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=0.05 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 three decimal places Step 3: Determine the decision rule for rejecting the null hypothesis H0. Round your answer to three decimal places. Step 4: Reject or fail to reject your hypothesis?

Answer #2

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((((6 - 1)*5.9^2 + (10 - 1)*6.8^2)/(6 + 10 - 2))*(1/6 +
1/10))

sp = 3.3529

Test statistic,

t = (x1bar - x2bar)/sp

t = (40.1 - 45.6)/3.3529

t = -1.640

3)

Rejection Region

This is left tailed test, for α = 0.05 and df = n1 + n2 - 2 =
14

Critical value of t is -1.761.

Hence reject H0 if t < -1.761

4)

fail to reject null hypothesis.

answered by: anonymous

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