Questions: Chapter 7 Discussion Population birth weights of boys and girls are estimated with a 95% confidence interval (measured in hectograms - hg) A 95% confidence interval for the population mean birth weight for boys is: 31.8 hg < μ < 33.6 hg A 95% confidence interval for the population mean birth weight for girls is: 29.3 hg < μ < 32.5 hg These intervals overlap...Explain how each of the following conclusions could be true...(use the limits of the confidence intervals, use correct units of measure, include statistical values in each explanation, and use statistical vocabulary, and correct spelling and punctuation and sentence structure): 1. The population mean birth weight of boys is greater 2. The population mean birth weight of girls is greater 3. The population mean birth weight of boys and girls are the same

Solution Steps

The mean birth weight for boys and girls is calculated as follows:

\[ \bar{x}_{boys} = \frac{31.8 + 33.6}{2} = 32.7 \, \text{hg} \]

\[ \bar{x}_{girls} = \frac{29.3 + 32.5}{2} = 30.9 \, \text{hg} \]

From the calculated means, we observe:

\[ \bar{x}_{boys} > \bar{x}_{girls} \quad \text{(i.e., } 32.7 \, \text{hg} > 30.9 \, \text{hg}\text{)} \]

This indicates that the population mean birth weight of boys is greater than that of girls.

The confidence intervals for boys and girls are:

• Boys: \( (31.8 \, \text{hg}, 33.6 \, \text{hg}) \)
• Girls: \( (29.3 \, \text{hg}, 32.5 \, \text{hg}) \)

Since the intervals overlap, it is possible that the population mean birth weights of boys and girls are the same.

To statistically compare the means, we calculate the standard error \( (SE) \):

\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{1.62}{2} + \frac{5.12}{2}} = 1.8358 \]

Next, we compute the test statistic \( (t) \):

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{32.7 - 30.9}{1.8358} = 0.9805 \]

The degrees of freedom \( (df) \) is calculated as:

\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} = \frac{11.3569}{7.2097} = 1.5752 \]

Finally, we find the p-value \( (P) \):

\[ P = 2(1 - T(|t|)) = 2(1 - T(0.9805)) = 0.4531 \]

Final Answer