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 12 nursing students from Group 1 resulted in a mean score of 59.7 with a standard deviation of 2.8. A random sample of 15 nursing students from Group 2 resulted in a mean score of 64.7 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 = 0.1 for the test. Assume that the population variances are equal and that the two populations are normally distributed.Step 1: State the null and alternative hypotheses for the test.Step 2: Compute the value of the t test statistic. Round your answer to three decimal placesStep 3: 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:

Kindly check explanation

Step-by-step explanation:

Given the following :

Group 1:

μ1 = 59.7

s1 = 2.8

n1 = sample size = 12

Group 2:

μ2 = 64.7

s2 = 8.3

n2 = sample size = 15

α = 0.1

Assume normal distribution and equ sample variance

A.)

Null and alternative hypothesis

Null : μ1 = μ2

Alternative : μ1 < μ2

B.)

USing the t test

Test statistic :

t = (m1 - m2) / S(√1/n1 + 1/n2)

S = √(((n1 - 1)s²1 + (n2 - 1)s²2) / (n1 + n2 - 2))

S = √(((12 - 1)2.8^2 + (15 - 1)8.3^2) / (12 + 15 - 2))

S = 6.4829005

t = (59.7 - 64.7) / 6.4829005(√1/12 + 1/15)

t = - 5 / 2.5108165

tstat = −1.991384

Decision rule :

If tstat < - tα, (n1+n2-2) ; reject the Null

tstat < t0.1,25

From t table :

-t0.1, 25 = - 1.3163

tstat = - 1.9913

-1.9913 < - 1.3163 ; Hence reject the Null