A Fitness Club Claims That The Average Age Of Its Members Is

1 A Fitness Club Claims That The Average Age Of Its Members Is 42 As

A fitness club claims that the average age of its members is 42. Assume that the ages are normally distributed. You believe that the average age is less than 42, therefore, you ask the ages of a few of the fitness club's members and get the following values: 40, 55, 50, 33, 38, 35, 30, and 45. Do you have any reason to think the fitness club's claim is incorrect? Using your knowledge on hypothesis testing, explain your conclusion carefully with detailed procedure. Find the p-value of your test, and interpret the results. Researchers will state the hypothesis they would like to confirm as the alternative hypothesis, why?

Paper For Above instruction

The scenario presented involves testing the claim of a fitness club regarding the average age of its members. The null hypothesis (H₀) states that the true mean age of members is 42, aligning with the club's claim, while the alternative hypothesis (H₁) suggests that the actual mean age is less than 42. This is a classic one-tailed hypothesis test designed to evaluate whether there is sufficient evidence to refute the club's claim based on sample data.

Given the sample ages—40, 55, 50, 33, 38, 35, 30, and 45—we first need to compute the sample mean and standard deviation to proceed with the hypothesis test. The sample mean (x̄) can be calculated as:

x̄ = (40 + 55 + 50 + 33 + 38 + 35 + 30 + 45) / 8 = 326 / 8 = 40.75

The sample standard deviation (s) is computed using the formula:

s = √[Σ(xi - x̄)² / (n - 1)]

Calculating the deviations:

• (40 - 40.75)² = 0.5625
• (55 - 40.75)² = 201.56
• (50 - 40.75)² = 87.56
• (33 - 40.75)² = 60.06
• (38 - 40.75)² = 7.56
• (35 - 40.75)² = 33.06
• (30 - 40.75)² = 114.06
• (45 - 40.75)² = 18.06

Sum of squared deviations ≈ 522.42

Thus,

s ≈ √(522.42 / 7) ≈ √74.6 ≈ 8.63

Next, we execute the hypothesis test. The test statistic (t) is calculated as:

t = (x̄ - μ₀) / (s / √n) = (40.75 - 42) / (8.63 / √8) ≈ (-1.25) / (8.63 / 2.828) ≈ (-1.25) / 3.052 ≈ -0.409

The degrees of freedom (df) are n - 1 = 7. Looking up the p-value corresponding to t ≈ -0.409 with df = 7 in the t-distribution table or using statistical software yields a p-value significantly higher than common significance levels (e.g., 0.05), indicating weak evidence against H₀.

Specifically, the p-value is approximately 0.34, which suggests that we do not have enough evidence to reject the null hypothesis that the average age of members is 42. This means that based on the sample, we cannot conclude that the average age is less than 42.

Researchers state their hypothesis they wish to confirm as the alternative hypothesis because doing so aligns with scientific inquiry to demonstrate the presence of an effect or difference. The alternative hypothesis (H₁) reflects the researcher's hypothesis of interest—in this case, that the average age is less than 42—and guides the analysis. If the evidence supports H₁ beyond a certain threshold (e.g., a p-value below 0.05), the researcher can confidently conclude that the mean is less than 42.

Regarding the second scenario

In the case of testing whether the mean miles per gallon increase is at least 5 mpg, the null hypothesis (H₀) is specified as μ ≥ 5, and the alternative hypothesis (H₁) as μ

Even if the null hypothesis is expressed as μ ≥ c or μ ≤ c, for a one-sided test, the critical value is calculated based on the specific value c in H₀ because the key question is whether the observed data sufficiently deviate in the direction that supports H₁. This approach streamlines decision-making by focusing solely on the boundary parameter c and the distribution of the test statistic under the null hypothesis.

In the example, testing whether the mean increase is less than 5 mpg involves calculating the test statistic relative to μ = 5, then comparing it with the critical value for a specified significance level (e.g., α = 0.05). Since our observed mean (4.5 mpg) is below 5, the critical value guides whether this difference is statistically significant or could have occurred by random chance, considering the variability in the data.

In conclusion, focusing only on the critical value based on the null hypothesis parameter c simplifies hypothesis testing for one-sided tests, regardless of whether the null includes equality or inequalities. It ensures clarity and efficiency in decision-making regarding statistical significance.

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