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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.
Mathematics
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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