#3-Effect of Exercise on Resting Heart Rate A physiologist resting heart rate. T
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Question
#3-Effect of Exercise on Resting Heart Rate A physiologist resting heart rate. The heart rates were then put on the exercise later. The wants to determine whether a particular exercise program has an 8 es of 8 randomly selected people were measured. Those people program and their heart rates were measured again one year data is summarized in the table below. Heart Rate Before 66 Heart Rate-After 79 75 76 75 75 73 75 53 52 101 93 a) this exercise State the null and alternative hypothesis for testing whether or not program had an effect on resting heart rate. ch test should we use to test these hypotheses? You may assume that the data comes from normally distributed populations. Apply the test you chose in part b to the data provided above. Report the test statistic or the p-value for this test. c) d) At the 5% significance level, what should we conclude based on the results from this test?Explanation / Answer
Sample 1 :- Heart rate after , Sample 2 :- Heart rate before
The following table is obtained:
From the sample data, it is found that the corresponding sample means are:
X¯1=78 , X¯2=72.625
Also, the provided sample standard deviations are:
s1=12.784 , s2=16.758
and the sample size is n = 8. For the score differences we have
Dbar=5.375 SD=19.985
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: D = 0
Ha: D 0
This corresponds to a two-tailed test, for which a t-test for two paired samples be used.
(2) Rejection Region
Based on the information provided, the significance level is =0.05, and the degrees of freedom are df=7.
Hence, it is found that the critical value for this two-tailed test is tc=2.365, for =0.05 and df=7.
The rejection region for this two-tailed test is R={t:t>2.365}.
(3) Test Statistics
The t-statistic is computed as shown in the following formula:
t = [Dbar / ( SD/n) ] = [ 5.375/ ( 19.985/8) ] = 0.761
(4) Decision about the null hypothesis
Since it is observed that |t| =0.761tc=2.365, it is then concluded that the null hypothesis is not rejected.
Using the P-value approach: The p-value is p=0.4717, and since p =0.47170.05, 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 1 is different than 2, at the 0.05 significance level.
Sample 1 Sample 2 Difference = Sample 1 - Sample 2 77 66 11 75 79 -4 75 76 -1 63 75 -12 101 52 49 67 53 14 73 75 -2 93 105 -12 Average 78 72.625 5.375 St. Dev. 12.784 16.758 19.985 n 8 8 8Related Questions
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