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). 1. A random sample of 11 nursing students from Group 1 resulted in a mean score of 63.3 with a standard deviation of 3.7. A random sample of 13 nursing students from Group 2 resulted in a mean score of 70.2 with a standard deviation of 6.6. Can you conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2? 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.A: State the null and alternative hypotheses for the test.
B: Compute the value of the t test statistic. Round your answer to three decimal places
C: 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:

a. H0:μ1≥μ2

Ha:μ1<μ2

b. t=-3.076

c. Rejection region=[tcalculated<−1.717]

Reject H0

Step-by-step explanation:

a)

As the score for group 1 is lower than group 2,

Null hypothesis: H0:μ1≥μ2

Alternative hypothesis: H1:μ1<μ2

b) t test statistic for equal variances

t=(xbar1-xbar2)-(μ1-μ2)/sqrt[{1/n1+1/n2}*{((n1-1)s1²+(n2-1)s2²)/n1+n2-2}

t=63.3-70.2/sqrt[{1/11+1/13}*{((11-1)3.7²+(13-1)6.6²)/11+13-2}

t=-6.9/sqrt[{0.091+0.077}{136.9+522.72/22}]

t=-3.076

c. α=0.05, df=22

t(0.05,22)=-1.717

The rejection region is t calculated<t critical value

t<-1.717

We can see that the calculated value of t-statistic falls in rejection region and so we reject the null hypothesis at 5% significance level.