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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.

56 59 69 84 74 64 69
70 66 80 59 71 76 63

Choose the correct test statistic and critical-value(s) associated with this experiment.

A) Test statistic: z = –1.94 Critical-value: z = –1.645
B) Test statistic: z = 1.94 Critical-value: z = 1.645
C) Test statistic: z = –1.94 Critical-value: z = ±1.645
D) Test statistic: z = 1.94 Critical-value: z = –1.645

1 answer

To conduct the hypothesis test, we need to follow these steps:

  1. State the hypotheses:

    • Null hypothesis (\(H_0\)): \(\mu = 72\) (the mean resting pulse rate for men is equal to 72 beats per minute).
    • Alternative hypothesis (\(H_a\)): \(\mu < 72\) (the mean resting pulse rate for men is less than 72 beats per minute).

  2. Calculate the sample mean (\(\bar{x}\)): Given the sample data: \[ 56, 59, 69, 84, 74, 64, 69, 70, 66, 80, 59, 71, 76, 63 \] The number of observations (n) is 14. Now, let’s compute the sample mean. \[ \bar{x} = \frac{56 + 59 + 69 + 84 + 74 + 64 + 69 + 70 + 66 + 80 + 59 + 71 + 76 + 63}{14} = \frac{ 63 + 54 + 69 + 84 + 74 + 64 + 69 + 70 + 66 + 80 + 59 + 71 + 76 + 63 }{14} \] \[ \bar{x} = \frac{ 64 + 84 + 69 + 64 + 69 + 70 + 66 + 80 + 59 + 76 + 63 }{14} = \frac{935}{14} = 66.7857 \approx 69 \]

  3. Calculate the test statistic (z): The formula for the z test statistic is: \[ z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} \] Where:

    • \(\bar{x}\) = sample mean
    • \(\mu_0\) = population mean (72)
    • \(\sigma\) = population standard deviation (6.6)
    • \(n\) = sample size (14)

    Calculation: \[ z = \frac{69 - 72}{6.6 / \sqrt{14}} = \frac{-3}{6.6 / 3.7417} = \frac{-3}{1.76} \approx -1.70 \]

  4. Determine the critical value: At a significance level of 0.05 for a one-tailed test, the critical value (z) can be found from the z-table: \[ z_{critical} = -1.645 \]

  5. Compare the test statistic to the critical value: If \(z < z_{critical}\), we reject the null hypothesis.

Based on the calculations:

  • Test statistic: approximately \(z = -1.70\)
  • Critical value: \(z = -1.645\)

Now, let's evaluate the options:

A) Test statistic: z = –1.94 Critical-value: z = –1.645
B) Test statistic: z = 1.94 Critical-value: z = 1.645
C) Test statistic: z = –1.94 Critical-value: z = ±1.645
D) Test statistic: z = 1.94 Critical-value: z = –1.645

None of the given answer choices match our calculated test statistic. It seems that there might be a mistake in the test statistic answers provided. However, since our findings indicate that the test statistic is around -1.70 and our critical-value is -1.645, the correct conclusion is to reject the null hypothesis since -1.70 < -1.645.

In summary, we should select option A as the closest representation:

A) Test statistic: z = –1.94 Critical-value: z = –1.645

The reason being both values align with the hypothesis test indicating a statistically significant result.