Question

In this assignment, you will test for a significant difference between the average resting heart rate of males and the average resting heart rate of females in your heart rate data. You have observed that the mean rates are not exactly the same but are they significantly different? You may use either of the two methods for testing a hypothesis illustrated in Realizeit: compare the two confidence intervals or use the data analysis tool to run a two-sample test with unequal variances as shown in the topic of testing two-samples. Steps Write the null hypotheses being tested Run the analysis either by using data analysis and the two-sample test or by comparing the two confidence intervals Interpret your data to determine if the resting male heart rate is the same as the resting female heart rate. Remember we are looking for whether the difference is a significant one, not just whether they are not the same. Heart rate before and after exercise M=0 F=1 Resting After Exercise 0 60.1 78.0 0 67.7 79.4 0 80.3 93.4 0 85.2 97.7 0 86.3 99.7 0 76.6 83.7 0 94.4 101.9 0 86.4 100.6 0 83.4 97.4 0 89.8 97.4 0 88.7 97.1 0 78.4 87.2 1 66.6 88.2 1 79.2 90.4 1 80.5 101.3 1 75.4 93.1 1 83.7 90.5 1 73.9 89.1 1 76.0 90.8 1 85.2 93.5 1 82.1 93.5 1 76.3 87.0 1 97.0 104.5 1 81.5 86.5 1 65.3 86.3 1 60.8 86.7 1 78.5 89.9 1 60.4 97.6 1 89.8 92.9 1 87.8 98.5

Answer 1

R code:

x=matrix(c(0, 60.1, 78.0, 0, 67.7, 79.4, 0, 80.3, 93.4, 0, 85.2, 97.7, 0, 86.3, 99.7,
0, 76.6, 83.7, 0, 94.4, 101.9, 0, 86.4, 100.6, 0, 83.4, 97.4, 0, 89.8, 97.4,
0, 88.7, 97.1, 0, 78.4, 87.2, 1, 66.6, 88.2, 1, 79.2, 90.4, 1, 80.5, 101.3,
1, 75.4, 93.1, 1, 83.7, 90.5, 1, 73.9, 89.1, 1, 76.0, 90.8, 1, 85.2, 93.5,
1, 82.1, 93.5, 1, 76.3, 87.0, 1, 97.0, 104.5, 1, 81.5, 86.5, 1, 65.3, 86.3,
1, 60.8, 86.7, 1, 78.5, 89.9, 1, 60.4, 97.6, 1, 89.8, 92.9, 1, 87.8, 98.5),
,nrow=30,ncol=3,byrow=TRUE)
Gender=x[,1];before=x[,2];after=x[,3]
t.test(before~Gender,alternative ="two.sided",paired = FALSE, var.equal = FALSE,
conf.level = 0.95)
t.test(after~Gender,alternative ="two.sided",paired = FALSE, var.equal = FALSE,
conf.level = 0.95)
t.test(after[1:12],before[1:12],alternative ="two.sided",paired = FALSE, var.equal = FALSE,
conf.level = 0.95)
t.test(after[13:30],before[13:30],alternative ="two.sided",paired = FALSE, var.equal = FALSE,
conf.level = 0.95)

Output 1:

Welch Two Sample t-test

data: before by Gender
t = 1.008, df = 23.961, p-value = 0.3235
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-3.838931 11.166709
sample estimates:
mean in group 0 mean in group 1
81.44167 77.77778

Since p-value= 0.3235>0.05 so there is insignificant difference of mean heart rates before exercise between male and female.

Output 2:

data: after by Gender
t = 0.2016, df = 16.68, p-value = 0.8427
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-5.240852 6.346407
sample estimates:
mean in group 0 mean in group 1
92.79167 92.23889

Since p-value= 0.8427>0.05 so there is insignificant difference of mean heart rates after exercise between male and female.

Hence there is insignificant effect of gender on heart rates.

Output 3:

Welch Two Sample t-test

data: after[1:12] and before[1:12]
t = 3.0532, df = 21.608, p-value = 0.005908
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
3.632462 19.067538
sample estimates:
mean of x mean of y
92.79167 81.44167

Since p-value=0.005908<0.05 so there is significant difference in mean female heart rate between before and after exercise.

Welch Two Sample t-test

data: after[13:30] and before[13:30]
t = 5.5003, df = 26.005, p-value = 9.015e-06
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
9.056851 19.865371
sample estimates:
mean of x mean of y
92.23889 77.77778

Since p-value=9.015e-06<0.05 so there is significant difference in mean male heart rate between before and after exercise.