question archive A study was designed to compare the attitudes of two groups of nursing students towards computers
A study was designed to compare the attitudes of two groups of nursing students towards computers
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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 18 nursing students from Group 1 resulted in a mean score of 48.6 with a standard deviation of 6.8. A random sample of 12 nursing students from Group 2 resulted in a mean score of 61.9 with a standard deviation of 4.4. 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 statistic..Rond 3 decimal places
Step 3 of 4: Determine the decision rule for rejecting the null hypothesis Ho. Round your answer to three decimal places
Step 4 of 4: State the test conclusion
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Answer:
(a) Ho: u1 = u2 ; H1: u1 < u2
(B) test statistic: -5.975
(c) reject Ho, if to < -1.701
(d) decision: reject Ho, conclude that the mean score for group 1 is significantly lower than the mean score for group 2
Step-by-step explanation
Given that,
mean(x)=48.6
standard deviation , s.d1=6.8
number(n1)=18
y(mean)=61.9
standard deviation, s.d2 =4.4
number(n2)=12
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.7011
since our test is left-tailed
reject Ho, if to < -1.7011
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (17*46.24 + 11*19.36) / (30- 2 )
s^2 = 35.68
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=48.6-61.9/sqrt((35.68( 1 /18+ 1/12 ))
to=-13.3/2.2261
to=-5.9746
| to | =5.9746
critical value
the value of |t α| with (n1+n2-2) i.e 28 d.f is 1.7011
we got |to| = 5.9746 & | t α | = 1.7011
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: left tail - ha : ( p < -5.9746 ) = 0
hence value of p0.05 > 0,here we reject Ho
