Question

Suppose that many large samples are taken from a population and that the

sample proportions are normally distributed, with a mean of 0.22 and a

standard error of 0.029. What is the 68% confidence interval?

O

O

O

O

A. (0.162 to 0.278)

B. (0.104 to 0.336)

C. (0.133 to 0.307)

D. (0.191 to 0.249)

Answer 1

Answer:

$0.22 -1 *0.029 =0.191$

$0.22 +1 *0.029 =0.249$

And the best option would be:

D. (0.191 to 0.249)

Step-by-step explanation:

For this case we know that the mean is:

$\bar X = 0.22$

And the standard error is given by:

$SE = 0.029$

We want to construct a 68% confidence interval so then the significance level would be :

$\alpha=1-0.68 = 0.32$ and $\alpha/2 =0.16$. The confidence interval is given by:

$\bar X \pm z_{\alpha/2} SE$

Now we can find the critical value using the normal standard distribution and we got looking for a quantile who accumulate 0.16 of the area on each tail and we got:

$z_{\alpha/2}= 1$

And replacing we got:

$0.22 -1 *0.029 =0.191$

$0.22 +1 *0.029 =0.249$

And the best option would be:

D. (0.191 to 0.249)