Question

4. In 2017, chicken consumption in pounds consumed for 100 randomly selected people hasa mean * = 55.2 pounds and a standard deviation s = 23 pounds. Construct a 90%confidence interval the mean weight of chicken consumption in 2017.

Answer 1

(a)

The given parameters are:

$\begin{gathered} \text{Mean}=\bar{X}=55.2 \\ \text{Standard deviation}=\sigma=23 \\ Sample\text{ size}=n=100 \\ z=1.644854\text{ (90\% confidence ineterval)} \end{gathered}$

(b)

The formula to find the margin of error for a 90% confidence interval is given below:

$E=z\times\frac{\sigma}{\sqrt[]{n}}$

Substitute the value from part (a), to get

$\begin{gathered} E=1.644854\times\frac{23}{\sqrt[]{100}} \\ =1.644854\times\frac{23}{10} \\ =3.7832 \end{gathered}$

Thus, the margin of error is 3.7832.

(d)

The given sample's confidence interval is,

$55.2\pm3.7832$

So, the confidence interval is (51.42 to 58.98).

(d)

For 90% confidence interval, the mean weight of chicken consumption is between 51.42 pounds and 58.98 pounds.