Respuesta :
Answer:
So the p value is higher than any significance level given, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is higher or equal than 0. So we can't conclude that jogging leads to a reduction in one's pulse rate.
Step-by-step explanation:
A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.
Let put some notation
x=test value before , y = test value after
x: 74 86 84 79 70 78 79 70
y: 70 85 90 110 71 80 69 74
The system of hypothesis for this case are:
Null hypothesis: [tex]\mu_y- \mu_x \geq 0[/tex]
Alternative hypothesis: [tex]\mu_y -\mu_x <0[/tex]
The first step is calculate the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:
d: -4, -1, 6, 31, 1, 2, -10, 4
The second step is calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{29}{8}=3.625[/tex]
The third step would be calculate the standard deviation for the differences, and we got:
[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =12.130[/tex]
The 4 step is calculate the statistic given by :
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{3.625 -0}{\frac{12.130}{\sqrt{8}}}=0.845[/tex]
The next step is calculate the degrees of freedom given by:
[tex]df=n-1=8-1=7[/tex]
Now we can calculate the p value, since we have a left tailed test the p value is given by:
[tex]p_v =P(t_{(7)}<0.845) =0.787[/tex]
So the p value is higher than any significance level given, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is higher or equal than 0. So we can't conclude that jogging leads to a reduction in one's pulse rate.
The conclusion about the hypothesis is that; there is no evidence that shows that jogging leads to a reduction in one's pulse rate.
What is the Hypothesis Conclusion?
We are given pulse rates before and after. The, let x represent pulse rates before and let y represent pulse rates after. Thus, we have;
x: 74 86 84 79 70 78 79 70
y: 70 85 90 110 71 80 69 74
Let us define the hypothesis now as;
Null Hypothesis; H₀: μ_y - μ_x ≥ 0
Alternative Hypothesis; Hₐ: μ_y - μ_x < 0
The difference between the y and x values for respective terms which is values for (y - x) are; d: -4, -1, 6, 31, 1, 2, -10, 4
Mean difference is;
d' = ∑d/n
d' = (-4 + -1 + 6 + 31 + 1 + 2 + -10 + 4)/8
d' = 29/8
d' = 3.625
From online standard deviation calculator, the standard deviation of the mean difference is; s = 12.13
The test statistic is gotten from the formula;
t = (d' - (μ_y - μ_x))/(s/√n)
t = (3.625 - 0)/(12.13/√8)
t = 0.845
From online t-tables with a t-value of 0.845, df = 8 - 1 = 7, we have a p-value of; p - value = 0.787
Significance level could be 0.1, 0.05, 0.01 our p-value is higher than any of the significance levels and so we fail to reject the null hypothesis and conclude that there is no evidence that shows that jogging leads to a reduction in one's pulse rate.
Read more about hypothesis conclusion at; https://brainly.com/question/15980493
