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artcher [175]
3 years ago
10

A research company desires to know the mean consumption of meat per week among people over age 40. A sample of 610 people over a

ge 40 was drawn and the mean meat consumption was 3.1 pounds. Assume that the population standard deviation is known to be 1.1 pounds. Construct the 85% confidence interval for the mean consumption of meat among people over age 40. Round your answers to one decimal place.
Mathematics
1 answer:
Papessa [141]3 years ago
5 0

Answer: (3.0,\ 3.2)

Step-by-step explanation:

Confidence interval for population mean is given by :-

\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}  (1)

, where \overline{x} = Sample mean

z_{\alpha/2}= Critical z-value

\sigma = Population standard deviation.

n= Sample size.

As per given , we have

n= 610

\sigma=1.1

\overline{x}=3.1

Significance level for 85% confidence : \alpha=1-0.85=0.15

By z-table critical two tailed z-value : z_{\alpha/2}=z_{0.075}=1.44

Put all values in (1) , we get

3.1\pm 1.44\dfrac{1.1}{\sqrt{610}}

3.1\pm 1.44\dfrac{1.1}{24.698}

3.1\pm 0.064

=(3.1-0.064,\ 3.1+0.064)=(3.036,\ 3.164)\approx(3.0,\ 3.2)

Hence, the 85% confidence interval for the mean consumption of meat among people over age 40.  = (3.0,\ 3.2)

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