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

Answer:

$157.3-1.96\frac{15.6}{\sqrt{234}}=155.301$    

$157.3+1.96\frac{15.6}{\sqrt{234}}=159.299$    

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)

Step-by-step 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:

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

$157.3-1.96\frac{15.6}{\sqrt{234}}=155.301$    

$157.3+1.96\frac{15.6}{\sqrt{234}}=159.299$    

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)