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3 votes
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)

User Ian Cook
by
6.6k points

1 Answer

2 votes

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)

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)

User Carrier
by
6.7k points
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