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12345 [234]
3 years ago
15

A research company desires to know the mean consumption of meat per week among people over age 23. A sample of 164 people over a

ge 23 was drawn and the mean meat consumption was 4.1 pounds. Assume that the population standard deviation is known to be 0.7 pounds. Construct the 80% confidence interval for the mean consumption of meat among people over age 23. Round your answers to one decimal place.
Mathematics
1 answer:
Elan Coil [88]3 years ago
8 0

Answer:

The 80% confidence interval for the mean consumption of meat among people over age 23 is between 4 and 4.2 pounds.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.8}{2} = 0.1

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.1 = 0.9, so z = 1.28

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.28*\frac{0.7}{\sqrt{164}} = 0.07

The lower end of the interval is the mean subtracted by M. So it is 4.1 - 0.07 = 4.03 pounds

The upper end of the interval is the mean added to M. So it is 4.1 + 0.07 = 4.17 pounds

Rounded to one decimal place

The 80% confidence interval for the mean consumption of meat among people over age 23 is between 4 and 4.2 pounds.

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