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 16 nursing students from Group 1 resulted in a mean score of 55.4 with a standard deviation of 4.5 . A random sample of 8 nursing students from Group 2 resulted in a mean score of 66 with a standard deviation of 8.3 . 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=0.05 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 H0 . Round your answer to three decimal places.

Step 4 of 4:

State the test's conclusion.

Answer 1

n 1= 16

n 2=8

2.1 = 55.4

T266

s1 = 4.5

s2 = 8.3

claim: mean score for Group 1 is significantly lower than the mean score for Group 2

Step 1 of 4:

Null and alternative hypothesis is

Vs

Level of significance = 0.05

Step 2 of 4:

Here Population variances are equal.

So we have to use pooled variance.

Formula

$t=\frac{\bar{x1}-\bar{x2}}{Sp*\sqrt{\frac{1}{n1}+\frac{1}{n2}}}$

nln2-2

(16-1)20.258 (82 1)68 89-V 3013 752 48223-5.9772 (16 - 1)20.25 (8 1)68.89 3 303.75 482.23 16+8-2

55.4 - 66 251086 -40055 5,9772+V 5.97722.58824.0955

t一一4.096

Step 3 of 4:

Decision rule : p-value < α , Reject H0

d.f = n1 + n2 – 2 = 16 + 8 - 2 = 22

a-0.05

p-value = 0.0002 ( using t table )

p-value < α , Reject H0

Step 4 of 4:

conclusion : There is a sufficient evidence to support the claim that the mean score for Group 1 is significantly lower than the mean score for Group 2