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 7 nursing students from Group 1 resulted in a mean score of 57.7 with a standard deviation of 2.7. A random sample of 15 nursing students from Group 2 resulted in a mean score of 68.5 with a standard deviation of 6.1. 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 2 of 4 : Compute the value of the t test statistic. Round your answer to three decimal places.

Answer 1

Solution:

Ho:μ1=μ2

Ha:μ1<μ2

alpha=0.05

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

=((7-1)*2.7^2+(15-1)*6.1^2)/(7+15-2)

=28.234

pooled standard devaition=Sp=sqrt(Sp^2)=sqrt(28.234)=5.313568

t=x1bar-x2bar/Sp*sqrt(1/n1+1/n2)

=(57.7-68.5)/5.313568*sqrt(1/7+1/15))

t= -4.440387

test statistic,t=-4.440

df=n1+n2-2=7+15-2=20

p value in excel

=T.DIST(-4.44,20,TRUE)

=0.000125805

P=0.0001

P<0.05

Reject Ho

There is sufficient statisitcal evidence at 5% level of significance to conclude that

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

ANSWER:

test statistic,t=-4.440