Question

A researcher was interested in comparing the resting pulse rates of people who exercise regularly and people who do not exercise regularly. Independent simple random samples of 16 people ages 30 dash 40 who do not exercise regularly and 12 people ages 30 dash 40 who do exercise regularly were​ selected, and the resting pulse rate​ (in beats per​ minute) of each person was measured. The summary statistics are to the right. Apply the nonpooled​ t-interval procedure to obtain a​ 95% confidence interval for the​ difference, mu 1 minus mu 2​, between the mean pulse rate of people who do not exercise and the mean pulse rate of people who do exercise. Assume that the requirements for using the procedure are satisfied and round to two decimal places.

Answer 1

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.

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

`$ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{(s_1^2)/(n_1) + (s_2^2)/(n_2) } } $`

Where
`\bar{x}_1` is the sample mean of people who do not exercise regularly.

Where
`\bar{x}_2` is the sample mean of people who do exercise regularly.

Where
`s_1` is the sample standard deviation of people who do not exercise regularly.

Where
`s_2` is the sample standard deviation of people who do exercise regularly.

Where
`n_1` is the sample size of people who do not exercise regularly.

Where
`n_2` is the sample size of people who do exercise regularly.

`$ t = \frac{72.7 - 69.7}{\sqrt{(10.9^2)/(16) + (8.2^2)/(12) } } $`

`t = 0.83`

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.