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 1313 nursing students from Group 1 resulted in a mean score of 68.668.6 with a standard deviation of 4.24.2. A random sample of 1818 nursing students from Group 2 resulted in a mean score of 74.174.1 with a standard deviation of 88. 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. Ho: μ1 (=,≠,<,>,≤,≥) μ2 Ha: μ1 (=,≠,<,>,≤,≥) μ2

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 H0. Round your answer to three decimal places. Reject Ho if (t, I t I) (<,>) ____

Step 4 of 4: State the test's conclusion. A. Reject Null Hypothesis B. Fail to Reject Null Hypothesis

Answer 1

=68.6, =74.1

s1​=4.2, s2​=8

n1​=13, n2​=18

= 0.01

Step 1:

Ho :

Ha:

Step 2:

Compute the value of the t test statistic

formula is

t test statistic = −2.257

Step 3:

df = n1 + n2 -2 = 13 + 18 -2 = 29

now calculate critical value for left tailed test with df = 29

to determine decesion rule.

we get critical value as

critical value = −2.462

Reject Ho if t < −2.462

step 4:

since (t test statistic = −2.257 ) > (critical value = −2.462 )

Fail to Reject Null Hypothesis