A 99% confidence interval (in inches) for the mean height of a population is $65.7

Question

A  99% confidence interval (in inches) for the mean height of a population is $65.7 < \mu < 67.3$
This result is based on a sample of size 144. Construct the  95% confidence interval. (Hint: you will first need to find the sample mean and sample standard deviation).
A) $65.7 \text { in. } < \mu < 67.3 \text { in. }$
B) $65.9 \text { in. } < \mu < 67.1 \text { in. }$
C) $65.6 \text { in. } < \mu < 67.4 \text { in. }$
D) $66.0 \text { in. } < \mu < 67.0 \text { in. }$

Quizplus · Accepted Answer

To find the 95% confidence interval, we first calculate the sample mean ($\bar{x}$) and the standard error (SE) using the 99% confidence interval provided. The sample mean is the midpoint of the 99% confidence interval: $\bar{x} = \frac{65.7 + 67.3}{2} = 66.5$ inches. The margin of error (MOE) for the 99% confidence interval can be calculated as half the width of the interval: $MOE_{99\%} = \frac{67.3 - 65.7}{2} = 0.8$ inches. The z-score for a 99% confidence level is approximately 2.576, and for a 95% confidence level, it is approximately 1.96. We can find the standard deviation ($\sigma$) of the sample using the formula for the margin of error: $MOE = z \times \frac{\sigma}{\sqrt{n}}$, solving for $\sigma$ gives $\sigma = MOE_{99\%} \times \frac{\sqrt{n}}{z_{99\%}}$. Substituting the values, we get $\sigma = 0.8 \times \frac{\sqrt{144}}{2.576} = 4.48$. To find the 95% confidence interval, we use the formula $\bar{x} \pm z_{95\%} \times \frac{\sigma}{\sqrt{n}}$, which gives $66.5 \pm 1.96 \times \frac{4.48}{\sqrt{144}}$, resulting in a margin of error of $0.6$ inches for the 95% confidence interval. Thus, the 95% confidence interval is $66.5 \pm 0.6$, which is $65.9 < \mu < 67.1$.

A 99% Confidence Interval (In Inches) for the Mean 65.7<μ<67.365.7 < \mu < 67.365.7<μ<67.3

A 99% Confidence Interval (In Inches) for the Mean $65.7 < \mu < 67.3$