Question

If n = 300 and X = 90, construct a 90% confidence interval estimate of the population proportion.

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Note: the "< " listed above are suppos to be underlined.  

Answer 1

Solution :

Given that,

n = 300

x = 90

= x / n = 90 / 300 = 0.300

1 - = 1 - 0.300 = 0.700

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 * (((0.300 * 0.700) / 300)

= 0.044

A 90% confidence interval for population proportion p is ,

- E < < + E

0300 - 0.044 < < 0.300 + 0.044

0.256 < < 0.344