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We know that the mean weight of men is greater than the mean weight of women, and the mean height of men is greater than the mean height of women. A person's body mass index (BMI) is computed by dividing weight (kg) by the square of height \((\mathrm{m})\). Given below are the BMI statistics for random samples of females and males taken from Data Set 1 "Body Data" in Appendix B. a. Use a \(0.05\) significance level to test the claim that females and males have the same mean BMI. b. Construct the confidence interval that is appropriate for testing the claim in part (a). c. Do females and males appear to have the same mean BMI? $$ \begin{aligned} \text { Female BMI: } & n=70, \bar{x}=29.10, s=7.39 \\ \text { Male BMI: } & n=80, \bar{x}=28.38, s=5.37 \end{aligned} $$

Step by step solution

01

Define the Hypotheses

Formulate the null and alternative hypotheses. Null hypothesis (H_0): 渭_f = 渭_m, the mean BMI of females (渭_f) is equal to the mean BMI of males (渭_m). Alternative hypothesis (H_1): 渭_f 鈮� 渭_m, the mean BMI of females is not equal to the mean BMI of males.

02

Identify the Significance Level

The given significance level (伪) is 0.05.

03

Calculate the Test Statistic

To test the hypothesis about the difference between two means, use the following formula for the test statistic ( t ): \[ t = \frac{(\bar{x}_f - \bar{x}_m)}{ \sqrt{ \frac{s_f^2}{n_f} + \frac{s_m^2}{n_m}} } \]Where: \( \bar{x}_f = 29.10 \) (mean BMI of females) \( \bar{x}_m = 28.38 \) (mean BMI of males) \( s_f = 7.39 \) (standard deviation of female BMI) \( s_m = 5.37 \) (standard deviation of male BMI) \( n_f = 70 \) (sample size for females) \( n_m = 80 \) (sample size for males).

04

Compute the Pooled Standard Error

Calculate the standard error (SE): \[ SE = \sqrt{ \frac{s_f^2}{n_f} + \frac{s_m^2}{n_m}} \]Substitute the values: \[ SE = \sqrt{ \frac{(7.39)^2}{70} + \frac{(5.37)^2}{80} } = \sqrt{ \frac{54.6121}{70} + \frac{28.8369}{80} } = \sqrt{ 0.7802 + 0.3605} = \sqrt{1.1407} \approx 1.0688 \]

05

Calculate the Test Statistic

Now substitute the values into the test statistic formula. \[ t = \frac{(29.10 - 28.38)}{1.0688} = \frac{0.72}{1.0688} \approx 0.6739 \]

06

Find the Critical Value and Make a Decision

With a significance level of 0.05 and degrees of freedom calculated using min(n_f - 1, n_m - 1) = min(69, 79) = 69, the critical value for a two-tailed test can be found in a t-distribution table or using a calculator. The critical value is approximately 卤2.00. Since 0.6739 < 2.00, do not reject the null hypothesis (H_0).

07

Construct the Confidence Interval

Use the following formula to construct a 95% confidence interval for the difference in means: \[ (\bar{x}_f - \bar{x}_m) \pm t_{critical} \times SE \]Substitute the values: \[ (29.10 - 28.38) \pm 2.00 \times 1.0688 = 0.72 \pm 2.1376 \]This results in the interval: \[ (-1.4176, 2.8576) \]

08

Conclusion

Since the confidence interval includes 0, and the test statistic did not lead to the rejection of the null hypothesis, there isn't enough evidence to conclude that the means are different. Thus, females and males do not appear to have different mean BMIs.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis

In hypothesis testing, the null hypothesis (denoted as H鈧�) represents a statement of no effect or no difference. In this exercise, the null hypothesis is: \(H_0: \mu_f = \mu_m \), which means that the mean BMI of females is equal to the mean BMI of males.

This hypothesis serves as the starting point of any statistical test. You assume it is true until you have enough evidence against it.

The null hypothesis is important because it provides a baseline to compare your data against. If the data significantly deviate from this baseline, you might reject the null hypothesis.

Alternative Hypothesis

The alternative hypothesis (denoted as H鈧� or H_a) is a statement that contradicts the null hypothesis. For our problem, the alternative hypothesis is: \(H_1: \mu_f 鈮� \mu_m\), which means that there is a difference between the mean BMI of females and males.

This hypothesis is what you might believe to be true or wish to prove. When the null hypothesis is rejected, it suggests that the evidence supports the alternative hypothesis.

The alternative hypothesis can be one-tailed (one-directional) or two-tailed (bidirectional). Here, it is two-tailed because we are testing for any difference, not just that one mean is greater than the other.

Confidence Interval

A confidence interval provides a range of values within which the true population parameter is expected to fall, with a certain level of confidence鈥攊n this case, 95%. For the example given in the exercise, we calculated the confidence interval for the difference in mean BMI between females and males as: \((-1.4176, 2.8576) \).

This means we are 95% confident that the true difference in mean BMI between females and males lies within this interval.

If the interval includes zero, which it does in our case, it indicates there is no significant difference between the two means at the specified level of confidence.

Significance Level

The significance level, denoted as \(\alpha \), is the threshold for determining whether a result is statistically significant. In this problem, we use \(\alpha = 0.05 \).

This means there is a 5% risk of concluding that a difference exists when there is no actual difference. Essentially, it sets the criteria for how extreme the observed data must be under the null hypothesis for you to reject it.

A common choice for \alpha is 0.05, but the level can be adjusted depending on the context and how much risk the researcher is willing to accept.