Do one of the following, as appropriate: (a) Find the critical value$ z_{?/2}$, (b) find the critical value $ t_{?/2}$, (c) state that neither the normal nor the t distribution applies
-A $99 \%$ confidence interval (in inches) for the mean height of a population is $65.64

Question

Do one of the following, as appropriate: (a) Find the critical value$ z_{?/2}$, (b) find the critical value $ t_{?/2}$, (c) state that neither the normal nor the t distribution applies
-A $99 \%$ confidence interval (in inches) for the mean height of a population is $65.64 < \mu < 67.16$. This result is pased on a sample of size 144 . If the confidence interval $65.85 < \mu < 66.95$ is obtained from the same sample data, what is the degree of confidence? (Hint: you will first need to find the sample mean and sample standard deviation).
A) $94 \%$
B) $92 \%$
C) $93 \%$
D) $95 \%$

Quizplus · Accepted Answer

The degree of confidence is $94 \%$. To find this, first calculate the sample mean ($\mu$) as the midpoint of the initial confidence interval: $\mu = \frac{65.64 + 67.16}{2} = 66.4$. The margin of error (E) for the 99% confidence interval is half the width of the interval: $E = \frac{67.16 - 65.64}{2} = 0.76$. Using the formula for the margin of error $E = z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$, where $n = 144$, solve for the sample standard deviation ($\sigma$). Knowing $z_{0.005} = 2.576$ for a 99% confidence level, $\sigma = E \sqrt{n} / z_{\alpha/2} = 0.76 \sqrt{144} / 2.576 = 4.32$. For the second confidence interval, the margin of error is $E = \frac{66.95 - 65.85}{2} = 0.55$. To find the corresponding z-value, use $E = z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$, which gives $z_{\alpha/2} = E \sqrt{n} / \sigma = 0.55 \sqrt{144} / 4.32$. This z-value corresponds to a confidence level of 94%, hence the correct answer is 94%.

Do One of the Following, as Appropriate: (A) Find the Critical