Math homework geometry please help!

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maryam615

11.10.2020

Mathematics

95% confidence interval for the mean consumption of meat among people over age 30 is [2.9 pounds , 3.1 pounds].

Step-by-step explanation:

We are given that a sample of 2092 people over age 30 was drawn and the mean meat consumption was 3 pounds. Assume that the population standard deviation is known to be 1.4 pounds.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean meat consumption = 3 pounds

s = population standard deviation = 1.4 pounds

n = sample of people = 2092

$\mu$ = population mean consumption of meat

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $3-1.96 \times {\frac{1.4}{\sqrt{2092} } }$ , $3+1.96 \times {\frac{1.4}{\sqrt{2092} } }$ ]

= [2.9 pounds , 3.1 pounds]

Therefore, 95% confidence interval for the mean consumption of meat among people over age 30 is [2.9 pounds , 3.1 pounds].

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