Question

A study of women’s weights found that a randomly selected sample of 234 women had a mean weight of 157.3 lb. Assuming that the population standard deviation is 15.6 lb., construct a 95% confidence interval estimate of the mean weight of all women.

A. (145.3, 160.5)
B. (155.3, 159,3)
C. (165.5, 173.5)
D. (185.7, 199.3)

Answer 1

`157.3-1.96(15.6)/(√(234))=155.301`

`157.3+1.96(15.6)/(√(234))=159.299`

So on this case the 95% confidence interval would be given by (155.301;159.299)

And the best option would be:

B. (155.3, 159,3)

Explanation:

Information given

`\bar X=157.3` represent the sample mean

`\mu` population mean (variable of interest)

`\sigma =15.6` represent the population standard deviation

n=234 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

`\bar X \pm z_(\alpha/2)(\sigma)/(√(n))` (1)

The Confidence level is is 0.95 or 95%, the significance is
`\alpha=0.05` and
`\alpha/2 =0.025`, the critical value for this case would be
`z_(\alpha/2)=1.96`

And replacing we got:

`157.3-1.96(15.6)/(√(234))=155.301`

`157.3+1.96(15.6)/(√(234))=159.299`

So on this case the 95% confidence interval would be given by (155.301;159.299)

And the best option would be:

B. (155.3, 159,3)