The mean height of American males is 69.5 inches.

The mean height of American males is 69.5 inches. The heights of the 43 male U.S. presidents (Washington to Obama) have a mean of 70.78 inches and a standard deviation of 2.77 inches. Treating the 43 presidents as a simple random sample, determine if there is evidence to suggest that U.S. presidents are taller than the average American male. Use α=0.05 level of significance and clearly label each step. (This question was written before Trump came into office) (5pts) Step 1: State the null and alternative hypothesis (H0 and Ha) and state if it’s Two, Left, or Right Tailed. Step 2: Decide on the significance level α. Step 3: Compute the value of the test statistic: t0 = (xÌ„ – μ0) / (s/√n). Now choose if you are doing a Critical Value Approach or a P-Value Approach. Step 4 (Critical Value): The critical values are (a) Two-Tailed test: (b) Left-Tailed test (c) Right-Tailed test. Step 5 (Critical Value): If the value of the test statistic falls in the rejection region, reject H0. Otherwise, do not reject H0. Step 6: Interpret the results of the Hypothesis Test.

The Correct Answer and Explanation is:

Here is the full solution to the hypothesis test to determine if U.S. presidents are taller than the average American male:

Step 1: State the Hypotheses

• Null Hypothesis (H₀): μ = 69.5 inches (U.S. presidents are not taller than the average American male)
• Alternative Hypothesis (Hₐ): μ > 69.5 inches (U.S. presidents are taller than the average American male)

This is a Right-Tailed test because we are testing whether the mean is greater than 69.5.

Step 2: Significance Level

• The significance level is α = 0.05

Step 3: Compute the Test Statistic

Given:

• Sample mean (x̄) = 70.78 inches
• Population mean (μ₀) = 69.5 inches
• Sample standard deviation (s) = 2.77 inches
• Sample size (n) = 43

Using the t-test formula: t0=xˉ−μ0s/n=70.78−69.52.77/43≈1.280.4224≈3.03t₀ = \frac{x̄ – μ₀}{s / \sqrt{n}} = \frac{70.78 – 69.5}{2.77 / \sqrt{43}} \approx \frac{1.28}{0.4224} \approx 3.03t0​=s/n​xˉ−μ0​​=2.77/43​70.78−69.5​≈0.42241.28​≈3.03

We will use the Critical Value Approach.

Step 4: Critical Value (Right-Tailed Test)

Degrees of freedom (df) = n – 1 = 42
At α = 0.05 and df = 42, the critical value from the t-distribution table is approximately 1.682.

Step 5: Decision Rule

Since the calculated t-value (3.03) is greater than the critical value (1.682), it falls in the rejection region.

Therefore, reject the null hypothesis H₀.

Step 6: Conclusion

There is sufficient evidence at the 0.05 significance level to conclude that U.S. presidents (from Washington to Obama) are significantly taller than the average American male.

Explanation (300 words)

This problem involves testing whether U.S. presidents are taller than the average American male. The average male height in the U.S. is 69.5 inches, and the average height of the 43 presidents is 70.78 inches. We are using a one-sample t-test because the population standard deviation is unknown, and the sample size is relatively small.

We start by setting the null and alternative hypotheses. The null hypothesis claims there is no difference, while the alternative suggests that presidents are taller. Since we are looking for whether they are specifically taller (and not simply different), this is a one-tailed test in the right direction.

We calculate the test statistic using the t-formula for a sample: t=xˉ−μ0s/nt = \frac{x̄ – μ₀}{s / \sqrt{n}}t=s/n​xˉ−μ0​​. After computing the t-value, we compare it with the critical value for 42 degrees of freedom and α = 0.05. The result is a t-statistic of 3.03, which exceeds the critical value of 1.682.

Since our test statistic falls in the rejection region, we reject the null hypothesis. This tells us that the observed sample provides enough statistical evidence to support the idea that U.S. presidents are, on average, taller than typical American males. This conclusion could reflect sociopolitical preferences that subtly favor taller individuals for leadership roles.