Researchers were interested in whether there is a difference in the time men and

Question

Researchers were interested in whether there is a difference in the time men and women take to get ready for a date. Use the SPSS output below to answer the following questions. Assume µ1 is the male participants’ time to get ready and µ2 is the female participants’ time to get ready. Based on the output, in regard to the hypothesis of no difference in average time between men and women, you should:

A) Reject null, because given a true null, the probability of this result is < 0.05.

B) Reject null, because the t statistic is relatively small.

C) Fail to reject null, because given a true null, the probability of this result is > 0.05.

D) Fail to reject null, because the confidence interval is negative.

Group Statistics Deviation Mean Mean Sex time to get read male 4 32.25 15.798 7.899 62.50 15.155 7.577 female Independent Samples Test Levene's Test for Equality of Variances t-test for Equality of Means 95% Confidence Interval of the Difference Sig. (2- Mean Std. Error tailed) Difference Difference Lower Upper time to get read Equal variances -30.250 .049 .033 10.946 -57.034 -3.466 .832 -2.764 assumed Equal variances -2.764 5.990 .033 -30.250 10.946 -57.045 -3.455 not assumed

Explanation / Answer

Given that,
mean(x)=32.25
standard deviation , s.d1=15.798
number(n1)=4
y(mean)=62.5
standard deviation, s.d2 =15.155
number(n2)=4
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =3.182
since our test is two-tailed
reject Ho, if to < -3.182 OR if to > 3.182
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =32.25-62.5/sqrt((249.5768/4)+(229.67403/4))
to =-2.764
| to | =2.764
critical value
the value of |t | with min (n1-1, n2-1) i.e 3 d.f is 3.182
we got |to| = 2.76359 & | t | = 3.182
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.7636 ) = 0.07
hence value of p0.05 < 0.07,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -2.764
critical value: -3.182 , 3.182
decision: do not reject Ho
p-value: 0.07

[ANSWER]
C) Fail to reject null, because given a true null, the probability of this result is > 0.05

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