Question

Kim wants to determine a 90 percent confidence interval for the
true proportion of high school students in the area who attend
their home basketball games. How large of a sample must she have to
get a margin of error less than 0.03?

HINT: To find n, since no previous study has been done, use the
value p = 0.5 for the proportion and one of the values (1.282,
1.645, 1.96, 2.576) for the critical value depending on the
confidence level. Don't forget to round your value of n up.

Answer #1

Solution,

Given that,

= 1 - = 0.5

margin of error = E = 0.03

At 90% confidence level

= 1 - 90%

= 1 - 0.90 =0.10

/2
= 0.05

Z_{/2}
= Z_{0.05} = 1.645

sample size = n = (Z_{ / 2} / E )^{2} *
* (1 -
)

= (1.645 / 0.03 )^{2} * 0.5 * 0.5

= 751.67

sample size = n = 752

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