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KonstantinChe [14]

3 years ago

5

A marketing research company desires to know the mean consumption of meat per week among males over age 50. A sample of 217 male

s 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.

1 answer:

guajiro [1.7K]3 years ago

4 0

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)

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