Question

A random sample of 700 voters in a particular city found 266 voters who voted yes on proposition 200. Find a 95% confidence interval for the true percent of voters in this city who voted yes on proposition 200. Express your results to the nearest hundredth of a percent.

Answer #1

Solution :

Given that,

n = 700

x = 266

Point estimate = sample proportion = = x / n = 266/700=0.38

1 - = 1-0.38=0.62

At 95% confidence level

= 1 - 95%

= 1 - 0.95 =0.05

/2
= 0.025

Z/2
= Z_{0.025 = 1.96} ( Using z table )

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

= 1.96 (((0.38*0.62) /700 )

E = 0.0360

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.38 -0.0360 < p <0.38+ 0.0360

0.3440< p <0.4160

The 95% confidence interval for the population proportion p is : 0.34,0.42

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HW 25
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Q2
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Please show work so I
can understand how to solve thanks
A random sample of 800
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