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 17 nursing students from Group 1 resulted in a mean score of 41.1 with a standard deviation of 7.5 A random sample of 15 nursing students from Group 2 resulted in a mean score of 49.3 with a standard deviation of 8.9 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 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. (Reject or fail to reject null hypothesis)

Answer 1

Solution:

State the null and alternative hypotheses for the test.

Null hypothesis: H₀: the mean score for Group 1 is same as the mean score for Group 2.

Alternative hypothesis: H_a: the mean score for Group 1 is significantly lower than the mean score for Group 2.

H₀: µ₁ = µ₂ versus H_a: µ₁ < µ₂

Compute the value of the t test statistic.

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

We are given

X1bar = 41.1

X2bar = 49.3

S1 = 7.5

S2 = 8.9

n1 = 17

n2 = 15

(X1bar – X2bar) = 41.1 – 49.3 = -8.2

df = n1 + n2 – 2 = 17 + 15 – 2 = 30

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(17 – 1)*7.5^2 + (15 – 1)*8.9^2]/(17 + 15 – 2)

S_p² = 66.9647

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

t = -8.2/ sqrt[66.9647*((1/17)+(1/15))]

t = -8.2 / 2.8989

t = -2.8287

Test statistic = t = -2.829

Determine the decision rule for rejecting the null hypothesis H₀.

We have

df = 30

α = 0.01

Test is one tailed (lower tailed)

So, required critical t value by using t-table is given as below:

Critical value = -2.4573

Decision rule: Reject H₀ if test statistic t < -2.4573

State the test's conclusion.

Here, test statistic = t = -2.829 < critical value = -2.4573

So, we reject the null hypothesis H₀

There is sufficient evidence to conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2.