Answered

18. Use the data to identify the correct response.

The mean resting pulse rate for men is 72 beats per minute. A simple random sample of men who regularly work out at Mitch's Gym is obtained, and their resting pulse rates (in beats per minute) are listed below. Use a 0.05 significance level to test the claim that these sample pulse rates come from a population with a mean less than 72 beats per minute. Assume that the standard deviation of the resting pulse rates of all men who work out at Mitch's Gym is known to be 6.6 beats per minute. Use the traditional method of testing hypotheses.

[tex]\[
\begin{array}{cccccc}
56 & 59 & 69 & 84 & 74 & 64 \\
70 & 66 & 80 & 59 & 71 & 76 \\
63 & 63 \\
\end{array}
\][/tex]

Choose the correct null and alternative hypotheses associated with this experiment.

A. [tex]\( H_0: \mu = 72 \)[/tex] beats per minute [tex]\( H_1: \mu \ \textless \ 72 \)[/tex] beats per minute

B. [tex]\( H_0: \mu = 72 \)[/tex] beats per minute [tex]\( H_1: \mu \neq 72 \)[/tex] beats per minute

C. [tex]\( H_0: \mu \neq 72 \)[/tex] beats per minute [tex]\( H_1: \mu = 72 \)[/tex] beats per minute

D. [tex]\( H_0: \mu \leq 72 \)[/tex] beats per minute [tex]\( H_1: \mu \ \textgreater \ 72 \)[/tex] beats per minute



Answer :

GinnyAnswer
To tackle this problem, let's go through the steps to test the hypothesis using the traditional method. The goal is to test the claim that the sample pulse rates come from a population with a mean less than 72 beats per minute, given the sample data and the known population standard deviation.

### Step 1: State the Hypotheses

First, we need to set up our null and alternative hypotheses:
- [tex]\( H_0 \)[/tex]: The population mean pulse rate is equal to 72 beats per minute ([tex]\( \mu = 72 \)[/tex]).
- [tex]\( H_1 \)[/tex]: The population mean pulse rate is less than 72 beats per minute ([tex]\( \mu < 72 \)[/tex]).

Hence, the correct hypothesis from the provided choices is:
[tex]\[ H_0: \mu = 72 \text{ beats per minute} \][/tex]
[tex]\[ H_1: \mu < 72 \text{ beats per minute} \][/tex]

### Step 2: Gather Sample Data and Statistics

From the sample data provided, we have resting pulse rates:
[tex]\[ [56, 59, 69, 84, 74, 64, 69, 70, 66, 68, 59, 71, 76, 63]. \][/tex]

Given:
- Population standard deviation ([tex]\( \sigma \)[/tex]): 6.6 beats per minute
- Significance level ([tex]\( \alpha \)[/tex]): 0.05

For these data, we know:
- Sample size ([tex]\( n \)[/tex]): 14
- Sample mean ([tex]\( \bar{x} \)[/tex]): 67.71 beats per minute (approx.)

### Step 3: Calculate the Test Statistic

We calculate the standard error of the mean (SEM) using the formula:
[tex]\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{6.6}{\sqrt{14}} \approx 1.76. \][/tex]

Next, we calculate the z-score using the formula:
[tex]\[ z = \frac{\bar{x} - \mu_0}{\text{SEM}} = \frac{67.71 - 72}{1.76} \approx -2.43. \][/tex]

### Step 4: Determine the Critical Value

For a one-tailed test at the 0.05 significance level, the critical z-value is:
[tex]\[ z_{\alpha} = -1.645. \][/tex]

### Step 5: Make a Decision

To make a decision, we compare the calculated z-score with the critical z-value:
- If [tex]\( z \)[/tex] is less than [tex]\( z_{\alpha} \)[/tex], we reject [tex]\( H_0 \)[/tex].
- Observed z-score: -2.43
- Critical z-value: -1.645

Since [tex]\( -2.43 < -1.645 \)[/tex], we reject the null hypothesis.

### Conclusion

Given these results, we have sufficient evidence at the 0.05 significance level to support the claim that the sample pulse rates come from a population with a mean pulse rate less than 72 beats per minute.