Question

A marketing research company desires to know the mean consumption of meat per week among males over age 50. A sample of 217 males over age 50 was drawn and the mean meat consumption was 3.4 pounds. Assume that the population standard deviation is known to be 0.5 pounds. Construct the 95% confidence interval for the mean consumption of meat among males over age 5050. Round your answers to one decimal place.

Answer 1

Answer: (3.3, 3.5)

Step-by-step explanation:

The formula to calculate the confidence interval is given by :-

$\overline{x}\pm z^* \dfrac{\sigma}{\sqrt{n}}$

, where $\overline{x}$ = sample mean.

z* = critical value.

$\sigma$ = Population standard deviation.

n= Sample size.

As per given , we have

$\overline{x}=3.4$

$\sigma=0.5$

n= 217

By using z-table , the critical value for 95% confidence level : z* = 1.96

Now, the  95% confidence interval for the mean consumption of meat among males over age 50 will be :

$3.4\pm (1.96) \dfrac{0.5}{\sqrt{217}}$

$3.4\pm (1.96) \dfrac{0.5}{14.73091986}$

$3.4\pm (1.96) 0.0339422$

$3.4\pm 0.066526712\approx3.4\pm 0.1=(3.4-0.1,\ 3.4+0.1)=(3.3,\ 3.5)$

Hence, the 95% confidence interval for the mean consumption of meat among males over age 50 =(3.3, 3.5)