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 1111 nursing students from Group 1 resulted in a mean score of 58.358.3 with a standard deviation of 8.98.9. A random sample of 1717 nursing students from Group 2 resulted in a mean score of 66.766.7 with a standard deviation of 5.15.1. Can you conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2? Let μ1μ1 represent the mean score for Group 1 and μ2μ2 represent the mean score for Group 2. Use a significance level of α=0.01α=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 H0H0. Round your answer to three decimal places.

Step 4 of 4:

State the test's conclusion.

Answer 1

T-test for two Means – Unknown Population Standard Deviations - Equal Variance

The following information about the samples has been provided: a. Sample Means : Xˉ1​=58.3 and Xˉ2​=66.7 b. Sample Standard deviation: s1=8.9 and s2=5.1 c. Sample size: n1=11 and n2=17 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ1 =μ2 Ha: μ1 <μ2 This corresponds to a Left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) The degrees of freedom Assuming that the population variances are equal, the degrees of freedom are given by n1+n2-2=11+17-2=26. (3a) Critical Value Based on the information provided, the significance level is α=0.01, and the degree of freedom is 26. Therefore the critical value for this Left-tailed test is tc​=-2.4786. This can be found by either using excel or the t distribution table. (3b) Rejection Region The rejection region for this Left-tailed test is t<-2.4786 (4)Test Statistics The t-statistic is computed as follows: (5) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0.0019 (6) The Decision about the null hypothesis (a) Using traditional method Since it is observed that t=-3.1844 < tc​=-2.4786, it is then concluded that the null hypothesis is rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.0019, and since p=0.0019≤0.01, it is concluded that the null hypothesis is rejected. (7) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ1​ is less than μ2, at the 0.01 significance level.

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