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1. To test whether the mean resting pulse rates of men and women differ, the pul

ID: 3314276 • Letter: 1

Question

1. To test whether the mean resting pulse rates of men and women differ, the pulse rates of random samples of men and women were measured, with the following results:

men n1 = 360 ¯x1 = 73.5 s1 = 13.0 women n2 = 380 ¯x2 = 76.3 s2 = 12.8

The test is performed at the 1% level of significance.

(a) State the null and alternative hypotheses for the test.

(b) State the formula for the test statistic and compute its value. Justify your answer.

(c) Construct the rejection region and make a decision.

(d) State a conclusion about the mean resting pulse rates of men and women, based on the test you performed.

Explanation / Answer

Given that,
mean(x)=73.5
standard deviation , s.d1=13
number(n1)=360
y(mean)=76.3
standard deviation, s.d2 =12.8
number(n2)=380
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.01
from standard normal table, two tailed t /2 =2.59
since our test is two-tailed
reject Ho, if to < -2.59 OR if to > 2.59
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =73.5-76.3/sqrt((169/360)+(163.84/380))
to =-2.9505
| to | =2.9505
critical value
the value of |t | with min (n1-1, n2-1) i.e 359 d.f is 2.59
we got |to| = 2.95047 & | t | = 2.59
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.9505 ) = 0.003
hence value of p0.01 > 0.003,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -2.9505
critical value: -2.59 , 2.59
decision: reject Ho
p-value: 0.003
mean resting pulse rates of men and women differ