Questions: A B C 1 Women's heights Men's heights 2 Mean 157.88 171.46 3 SE 0.50 0.53 4 Mean-SE 157.38 170.93 5 Mean + SE 158.38 171.99 SD 2.46 2.57

Solution Steps

The standard error for the difference between two means is calculated using the formula:

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

Given:

• \( s_1 = 2.46 \), \( n_1 = 100 \)
• \( s_2 = 2.57 \), \( n_2 = 100 \)

Substitute the values:

\[ SE = \sqrt{\frac{2.46^2}{100} + \frac{2.57^2}{100}} = \sqrt{\frac{6.0516}{100} + \frac{6.6049}{100}} = \sqrt{0.060516 + 0.066049} = \sqrt{0.126565} \approx 0.3558 \]

The test statistic \( z \) is calculated using the formula:

\[ z = \frac{\bar{x}_1 - \bar{x}_2}{SE} \]

Given:

• \(\bar{x}_1 = 157.88\)
• \(\bar{x}_2 = 171.46\)

Substitute the values:

\[ z = \frac{157.88 - 171.46}{0.3558} = \frac{-13.58}{0.3558} \approx -38.1718 \]

The p-value is determined from the standard normal distribution for the calculated \( z \)-value. Since the \( z \)-value is very large in magnitude and negative, the p-value is essentially 0:

\[ P(Z > -38.1718) = 0.0 \]

For a two-tailed test at a significance level of \(\alpha = 0.05\), the critical \( z \)-value is 1.96. Since the calculated \( z \)-value of \(-38.1718\) is far less than \(-1.96\), we reject the null hypothesis.

Final Answer