Question

A survey of 526 citizens found that 356 of them Favor a new bill introduced by the city.

We want to find a 95% confidence interval for the true proportion of the population who favor the bill.

What is the lower limit of the interval? (Round to 3 decimal digits)

Answer #1

Solution :

Given that,

n = 526

x = 356

= x / n = 356 /526 = 0.677

1 - = 1 - 0.677 = 0.323

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

= 0.05

Z_{}
= Z_{0.05} = 1.645

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

= 1.645 * (((0.677 * 0.323) / 900) = 0.034

A 95 % confidence interval for population proportion p is ,

- E

0.677 - 0.034

0.643

The 95% confidence interval for the population proportion p = lower limit = 0.643

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