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In a survey of high school seniors, 72% of those surveyed stated they were planning on attendingcollege the following year. The margin of error was 4%. Find the likely interval that contains thepercentage of high school seniors that were planning on attending college the following year.

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Solution

For this case we have the proportion from the sample given

phat=0.72

And we have also the margin of error ME=0.04

And we want to find the likely interval so we can calculate the 95% confidence interval with the following formula:

[tex]p\pm ME[/tex]

And replacing we got:

0.72 -0.04 =0.68

0.72 +0.04= 0.76

Then the confidence interval would be given by

0.68 < p < 0.76

For the second case we have:

For the second case we have the following:

[tex]p=0.18,n=350[/tex]

Then we can replace in the margin of error formula and we got:

[tex]ME=1.96\cdot\sqrt[]{\frac{0.18\cdot(1-0.18)}{350}}=0.04[/tex]

Then the interval would be:

0.18-0.04 =0.14

0.18+0.04=0.22

Then the answer is

B

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