Question

question 12
a study was conducted to estimate μ, the mean commute distance that all employed u.s. adults travel to work. suppose a random sample of 49 employed u.s. adults gives a mean commute distance of 22 miles and that from prior studies, the population standard deviation is assumed to be σ = 8.4 miles.
how large a sample of u.s. adults is needed in order to estimate μ with a 95% confidence interval of length 2.4 miles?
a. 784
b. 111
c. 196
d. 49

Explanation:

Step1: Recall the formula for the margin of error

The formula for the margin of error $E$ of a confidence - interval for the population mean when the population standard deviation $\sigma$ is known is $E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, and the length of the confidence interval $L = 2E$. Given $L = 2.4$, then $E=\frac{L}{2}=1.2$. For a 95% confidence interval, the critical value $z_{\alpha/2}=1.96$. The population standard deviation $\sigma = 8.4$.

Step2: Rearrange the margin - of - error formula to solve for $n$

Starting from $E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, we can solve for $n$. First, square both sides of the equation: $E^{2}=z_{\alpha/2}^{2}\frac{\sigma^{2}}{n}$. Then, cross - multiply to get $n=\frac{z_{\alpha/2}^{2}\sigma^{2}}{E^{2}}$.

Step3: Substitute the known values into the formula for $n$

Substitute $z_{\alpha/2}=1.96$, $\sigma = 8.4$, and $E = 1.2$ into the formula $n=\frac{z_{\alpha/2}^{2}\sigma^{2}}{E^{2}}$.
\[

$$\begin{align*} n&=\frac{(1.96)^{2}\times(8.4)^{2}}{(1.2)^{2}}\\ &=\frac{3.8416\times70.56}{1.44}\\ &=\frac{271.485296}{1.44}\\ & = 196 \end{align*}$$

\]

Answer:

C. 196