Answer:
We Reject H₀ if t calculated > t tabulated
But in this case,
0.83 is not greater than 2.056
Therefore, we failed to reject H₀
There is no difference between the mean pulse rate of people who do not exercise and the mean pulse rate of people who do exercise.
Step-by-step explanation:
Refer to the attached data.
The Null and Alternate hypothesis is given by
Null hypotheses = H₀: μ₁ = μ₂
Alternate hypotheses = H₁: μ₁ ≠ μ₂
The test statistic is given by
[tex]$ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } } $[/tex]
Where [tex]\bar{x}_1[/tex] is the sample mean of people who do not exercise regularly.
Where [tex]\bar{x}_2[/tex] is the sample mean of people who do exercise regularly.
Where [tex]s_1[/tex] is the sample standard deviation of people who do not exercise regularly.
Where [tex]s_2[/tex] is the sample standard deviation of people who do exercise regularly.
Where [tex]n_1[/tex] is the sample size of people who do not exercise regularly.
Where [tex]n_2[/tex] is the sample size of people who do exercise regularly.
[tex]$ t = \frac{72.7 - 69.7}{\sqrt{\frac{10.9^2}{16} + \frac{8.2^2}{12} } } $[/tex]
[tex]t = 0.83[/tex]
The given level of significance is
1 - 0.95 = 0.05
The degree of freedom is
df = 16 + 12 - 2 = 26
From the t-table, df = 26 and significance level 0.05,
t = 2.056 (two-tailed)
Conclusion:
We Reject H₀ if t calculated > t tabulated
But in this case,
0.83 is not greater than 2.056
Therefore, We failed to reject H₀
There is no difference between the mean pulse rate of people who do not exercise and the mean pulse rate of people who do exercise.
