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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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Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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In this problem, we have that:

$\mu = 11, \sigma = 3$

a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 25, s = \frac{3}{\sqrt{25}} = 0.6$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

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b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

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$Z = \frac{11 - 11}{0.6}$

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Here we have that $n = 100, s = \frac{3}{\sqrt{100}} = 0.3$

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$Z = \frac{10.8 - 11}{0.3}$

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0.497, rounded to three decimal places.

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