Using this set of data comparing heart rates of males and fe

Using this set of data comparing heart rates of males and females: (using a level of significance of .05) data set is in comments

1.) State the null and alternative hypothesis

2.) Decide whether the z or t test is appropriate and why?

3.) Find the critical value

4.)State the decision

Solution

Set Up Hypothesis
Null, heart rates of males and females are equal Ho: u1 = u2
Alternate, n\'t Ho - H1: u1 != u2
Test Statistic
X(Mean)=72.1379
Standard Deviation(s.d1)=4.984 ; Number(n1)=29
Y(Mean)=71.414
Standard Deviation(s.d2)=8.5629; Number(n2)=29
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =72.1379-71.414/Sqrt((24.84026/29)+(73.32326/29))
to =0.393
| to | =0.393
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 28 d.f is 2.048
We got |to| = 0.39346 & | t | = 2.048
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != 0.3935 ) = 0.697
Hence Value of P0.05 < 0.697,Here We Do not Reject Ho

[ANSWERS]
1. Ho: u1 = u2, H1: u1 != u2
2. t - test , Population standard deviation is n\'t given & size<30
3.
Few professors use the crtical value to Min (n1-1, n2-1) i.e 28 d.f is -2.048,2.048
Few professors use the crtical value to (n1+n2-2) i.e 56 d.f is -2.003,2.003

4.to | < | t | and Here we Do not Reject Ho