Question

David Marciano

1. A medical researcher suspects that the pulse rate of smokers is higher than the pulse rates of non-smokers. Use the sample statistics to test the researcher’s suspicion at α=0.05

Smokers	Nonsmokers
n1= 100	n2 = 100
1 = 86 bpm	2= 83 bpm
S1=4.8 bpm	S2=5.3 bpm

1 H₀:___   H_a:_____                      5.Decision:{Circle one}Reject H₀ or Fail to Reject H₀

2 α=______                                                    6. P-value _________

3 Critical Value   _________                                 7.Statement:___________________________________

                                                                                          ___________________________________

4Test stat______                                                                   

Answer 1

Given sample means are,

X₁ = 86

X₂ = 83

Given sample standard deviations are,

s₂​=4.8

s₂​=5.3

sample sizes are

n₁​=100

n₂=100.

(1) Null and Alternative Hypothesis

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ > μ2​

This corresponds to a right-tailed test, for which a t-test wiil be performed.

Testing for Equality of Variances

A F-test is used to test for the equality of variances.

H0 : s12 = s22

Ha : s12 ≠ s22

The critical values are

F_l=0.673 using Excel formula =F.INV(0.025,99,99)

F_u​=1.486, using Excel formula =F.INV(0.975,99,99)

So rejection region will be F<0.673 and F>1.486.

The following F-ratio is obtained:

and since F = 0.82, then the null hypothesis of equal variances is not rejected.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df = 198. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal as we have calculated above from F test

Hence, it is found that the critical value for this right-tailed test is tc​=1.653 using Excel formula =T.INV(0.95,198)

The rejection region for this right-tailed test is R = >1.653

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

test stat = t = 0.83

(4) Decision about the null hypothesis

Since it is observed that t = 0.83 <tc​=1.653, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.2038, using Excel formula =T.DIST.RT(0.83,198) and since p = 0.2038 >0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​(smokers) is greater than μ2​ (non smokers), at the 0.05 significance level.

Note- I have highlighted answer in above explaination .