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 40.8 with a standard deviation of 5.4 A random sample of 13 nursing students from Group 2 resulted in a mean score of 54.5 with a standard deviation of 2.3. 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=0.01 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. Reject or fail to reject your hypothesis?

Answer 1

Answer:

Given,

Null hypothesis Ho : u1 = u2

Alternative hypothesis Ha : u1 < u2

sample n1 = 11 , n2 = 13

Mean x1 = 40.8 , x2 = 54.5

Standard deviation s1 = 5.4 , s2 = 2.3

consider,

test statistic t = (x1 - x2)/sqrt(Sp^2(1/n1 + 1/n2))

Sp^2 = [(n1-1)s1^2 + (n2-1)s2^2]/(n1+n2-2)

substitute values

= [(11-1)5.4^2 + (13-1)2.3^2]/(11+13-2)

= 16.14

substitute Sp^2 in t

t = (40.8 - 54.5)/sqrt(16.14(1/11 + 1/13))

t = - 8.324

degree of freedom = n1+n2-2

= 11 + 13 - 2

= 22

alpha = 0.01

Corresponding critical value t value is - 2.508

So reject Ho, if t < - 2.508

Here we observe that, test statistic < critical value, so we reject Ho.