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 10 nursing students from Group 1 resulted in a mean score of 55.9 with a standard deviation of 5.4. A random sample of 14 nursing students from Group 2 resulted in a mean score of 64.5 with a standard deviation of 5.7. 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.

Reject H0 if (t, ItI) (<,>)_____________

Step 4 of 4: State the test's conclusion. (Reject or Fail to Reject Null Hypothesis)

Answer 1

a)

Given,

The null and alternative hypothesis for the test is given below

Null hypothesis

Ho : 1 = 2

Alternative hypothesis

H1 : 1 < 2

b)

To give the t test statistic

n1 = 10

n2 = 14

x1 = 55.9

x2 = 64.5

s1 = 5.4

s2 = 5.7

Now to give the test statistic

Sp = sqrt(((n1-1)*s1^2+(n2-1)*s2^2)/(n1+n2-2))

substitute the values given'

Sp = sqrt(((10-1)*5.4^2+(14-1)*5.7^2)/(10+14-2))

Sp = 5.5792

for test statistic

t = (x1- x2)/Sp *sqrt(1/n1+1/n2)

substitute values

t = (55.9 - 64.5) / 5.5792*sqrt(1/10 + 1/14)

= - 8.6 / 2.3100

t = - 3.723 [Rounded to three decimal places]

c)

Now decision rule

Since from the critical value calculator

Here reject null hypothesis Ho if t < 1.717

d)

Test's conclusion :

we have sufficient evidence to conclude that the mean score for group1 is lower than group2

i.e.,

Reject the null hypothesis Ho.