Questions: A fitness center claims that the mean amount of time that a person spends at the gym per visit is at most 38 minutes.

A fitness center claims that the mean amount of time that a person spends at the gym per visit is at most 38 minutes.
Transcript text: A fitness center claims that the mean amount of time that a person spends at the gym per visit is at most 38 minutes.
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Solution

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Solution Steps

Step 1: Standard Error Calculation

The standard error \(SE\) is calculated using the formula:

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{30}} \approx 0.9129 \]

Step 2: Test Statistic Calculation

The test statistic \(Z_{\text{test}}\) is calculated using the formula:

\[ Z_{\text{test}} = \frac{\bar{x} - \mu_0}{SE} = \frac{37 - 38}{0.9129} \approx -1.0954 \]

Step 3: P-value Calculation

For a left-tailed test, the P-value is determined as follows:

\[ P = T(z) \approx 0.1367 \]

Step 4: Summary of Results
  • Test Statistic: \(Z_{\text{test}} \approx -1.0954\)
  • P-value: \(P \approx 0.1367\)
  • Standard Error: \(SE \approx 0.9129\)
  • Sample Mean: \(\bar{x} = 37\)
  • Sample Standard Deviation: \(\sigma = 5\)

Final Answer

The correct statement that results in the null hypothesis \(\mu \geq 38\) and the alternative hypothesis \(\mu < 38\) is:

\(\boxed{\text{A fitness center claims that the mean amount of time that a person spends at the gym per visit is at most 38 minutes.}}\)

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