Mathematics: Amarketing 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.

does he have a budget? If not, this doesnt make sense...

Amarketing research company desires to know the

Duriiee

12.01.2022

(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)

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