A study was designed to compare the attitudes of two groups of nursing students

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 14 nursing students from Group 1 resulted in a mean score of 51.2 with a standard deviation of 2.9 . A random sample of 17 nursing students from Group 2 resulted in a mean score of 64.9 with a standard deviation of 2.7 . 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 = 0.01 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 HO. Round your answer to three decimal places. Step 4 of 4: State the test conculsion to reject the null hypothesis or fail to reject the null hypothesis.

Explanation / Answer

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

Null Hypothesis H0: mean score for Group 1 (µ1) - mean score for Group 2 (µ2) = 0

Alternative Hypothesis H1:  mean score for Group 1 (µ1) - mean score for Group 2 (µ2) < 0

Or, Null Hypothesis H0: 1 - 2 = 0

Alternative Hypothesis H1: 1 - 2 < 0

Step 2 of 4: Compute the value of the t test statistic

Standard error, SE = sqrt[(s12/n1) + (s22/n2)]

where s1 = 2.9, s2 = 2.7, n1=14, n2 = 17

SE = sqrt((2.9*2.9)/14 + (2.7*2.7)/17) = 1.014662

Degree of freedom, DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] } = 27.00

t = (x1 - x2) / SE where x1 = 51.2 and x2 = 64.9

t = (51.2-64.9) / 1.014662 = -13.502

Step 3 of 4: Determine the decision rule for rejecting the null hypothesis

Since we have a one-tailed test, the P-value is the probability that a t statistic having 27 degrees of freedom is more extreme than -13.502; that is, less than -13.502.

Using t-distribution table to find P(t < -13.502) for DF=27, we get P-value = 7.978e-14

Step 4 of 4: State the test conculsion to reject the null hypothesis or fail to reject the null hypothesis.

Since the P-value (7.978e-14) is less than the significance level ( = 0.01), we cannot accept the null hypothesis.

So, we reject the null hypothesis and accept the alternative hypothesis that the mean score for Group 1 (µ1) - mean score for Group 2 (µ2) < 0. Or, mean score for Group 1 (µ1) is significantly lower than mean score for Group 2 (µ2).

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