Question

A research company desires to know the mean consumption of meat per week among people over age 29. A sample of 2092 people over age 29 was drawn and the mean meet consumption was 2.9 pounds. Assume that the standard deviation is known to be 1.4 pounds. Construct a 95% confidence interval for the mean consumption of meat among people over age 29. Round your answer to one decimal place.

Answer 1

Answer:

The 95% confidence interval for the mean consumption of meat among people over age 29 is between 2.8 pounds and 3 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:

$\alpha = \frac{1-0.95}{2} = 0.025$

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 $1-0.025 = 0.975$, so $z = 1.96$

Now, find the margin of error 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.

$M = 1.96*\frac{1.4}{\sqrt{2092}} = 0.1$

The lower end of the interval is the sample mean subtracted by M. So it is 2.9 - 0.1 = 2.8 pounds

The upper end of the interval is the sample mean added to M. So it is 2.9 + 0.1 = 3 pounds.

The 95% confidence interval for the mean consumption of meat among people over age 29 is between 2.8 pounds and 3 pounds.

Answer 2

Answer:

$2.9-1.96\frac{1.4}{\sqrt{2092}}=2.84$    

$2.9+1.96\frac{1.4}{\sqrt{2092}}=2.96$    

The confidence interval is given by $2.84 \leq \mu \leq 2.96$

Step-by-step explanation:

Information given

$\bar X=2.9$ represent the sample mean

$\mu$ population mean

$\sigma =1.4$ represent the population standard deviation

n=2092 represent the sample size  

Confidence interval

The confidence interval is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Since the Confidence interval is 0.95 or 95%, the significance is $\alpha=0.05$ and $\alpha/2 =0.025$, and the critical value would be $z_{\alpha/2}=1.96$

Replacing the info we got:

$2.9-1.96\frac{1.4}{\sqrt{2092}}=2.84$    

$2.9+1.96\frac{1.4}{\sqrt{2092}}=2.96$    

The confidence interval is given by $2.84 \leq \mu \leq 2.96$