Question

# A study was designed to compare the attitudes of two groups of nursing students towards computers....

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.

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

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