The heights of 20- to 29-year-old females are known to have a population standard deviation σ = 2.7 inches.A simple random sample of n = 15 females 20 to 29 years old results in the following data: $$\begin{array} { l l l l l }
63.1 & 67.9 & 64.8 & 62.2 & 65.4 \
\hline 63.3 & 66.2 & 68.2 & 69.7 & 64.1 \
\hline 68.4 & 69.9 & 67.3 & 64.5 & 70.2
\end{array}$$
A) (64.98, 67.72); we are 95% confident that the mean height of 20- to 29-year-old females is between 64.98 and 67.72 inches.
B) (64.85, 67.85); we are 95% confident that the mean height of 20- to 29-year-old females is between 64.85 and 67.85 inches.
C) (65.12, 67.58); we are 95% confident that the mean height of 20- to 29-year-old females is between 65.12 and 67.58 inches.
D) (65.20, 67.50); we are 95% confident that the mean height of 20- to 29-year-old females is between 65.20 and 67.50 inches.

Question

The heights of 20- to 29-year-old females are known to have a population standard deviation σ = 2.7 inches.A simple random sample of n = 15 females 20 to 29 years old results in the following data: $$\begin{array} { l l l l l } 
63.1 & 67.9 & 64.8 & 62.2 & 65.4 \
\hline 63.3 & 66.2 & 68.2 & 69.7 & 64.1 \
\hline 68.4 & 69.9 & 67.3 & 64.5 & 70.2
\end{array}$$
A) (64.98, 67.72); we are 95% confident that the mean height of 20- to 29-year-old females is between 64.98 and 67.72 inches.
B) (64.85, 67.85); we are 95% confident that the mean height of 20- to 29-year-old females is between 64.85 and 67.85 inches.
C) (65.12, 67.58); we are 95% confident that the mean height of 20- to 29-year-old females is between 65.12 and 67.58 inches.
D) (65.20, 67.50); we are 95% confident that the mean height of 20- to 29-year-old females is between 65.20 and 67.50 inches.

Quizplus · Accepted Answer

To find the correct confidence interval, we first calculate the sample mean ($\bar{x}$) and then use the formula for a confidence interval when the population standard deviation ($\sigma$) is known, which is $\bar{x} \pm z \frac{\sigma}{\sqrt{n}}$, where $z$ is the z-score corresponding to the desired confidence level (for 95% confidence, $z = 1.96$), $\sigma$ is the population standard deviation, and $n$ is the sample size.First, calculate the sample mean ($\bar{x}$):$\bar{x} = \frac{63.1 + 67.9 + 64.8 + 62.2 + 65.4 + 63.3 + 66.2 + 68.2 + 69.7 + 64.1 + 68.4 + 69.9 + 67.3 + 64.5 + 70.2}{15} = 66.35$Then, calculate the margin of error (ME):$ME = z \frac{\sigma}{\sqrt{n}} = 1.96 \frac{2.7}{\sqrt{15}} \approx 1.37$Finally, calculate the confidence interval:Lower limit = $\bar{x} - ME = 66.35 - 1.37 = 64.98$Upper limit = $\bar{x} + ME = 66.35 + 1.37 = 67.72$Therefore, the 95% confidence interval is (64.98, 67.72).

The Heights of 20- to 29-Year-Old Females Are Known to Have