Question

A poll is taken in which 367367 out of 600600 randomly selected voters indicated their preference for a certain candidate. Find a 8080% confidence interval for pp.

Kim wants to determine a 80 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 0.02? [If no estimate is known for p, let p^p^ = 0.5]

Answer 1

Solution:

A poll is taken in which 367 out of 600 randomly selected voters indicated their preference for a certain candidate. Find 80% confidence interval for p.

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

x = 367

n = 600

P = x/n = 367/600 = 0.611666667

Confidence level = 80%

Critical Z value = 1.2816

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.611666667 ± 1.2816* sqrt(0.611666667*(1 – 0.611666667)/600)

Confidence Interval = 0.611666667 ± 1.2816*0.0199

Confidence Interval = 0.611666667 ± 0.0255

Lower limit = 0.611666667 - 0.0255 = 0.5862

Upper limit = 0.611666667 + 0.0255 = 0.6372

Confidence interval = (0.5862, 0.6372)

Kim wants to determine a 80 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 0.02? [If no estimate is known for p, let p^ = 0.5]

The sample size formula is given as below:

n = p*q*(Z/E)^2

We are given

p = 0.5

q = 1 – p = 0.5

Confidence level = 80%

Critical Z value = 1.2816

(by using z-table)

Margin of error = E = 0.02

The sample size is given as below:

n = p*q*(Z/E)^2

n = 0.5*0.5*(1.2816/0.02)^2

n = 1026.562

Required sample size = 1027