A researcher wants to know the proportion of students at the university who live in residence....

A researcher wants to know the proportion of students at the university who live in residence. A sample of 50 students was taken and the researcher found 25 of them lived in residence. What is the 99% confidence interval for the proportion of students who live in residence?

Solutions

Expert Solution

Solution :

Given that,

n = 50

x = 25

Point estimate = sample proportion = = x / n = 25/50=0.5

1 - = 1- 0.5 =0.5

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576 ( Using z table )

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

= 2.576* ( $\sqrt$ ((0.5*0.5) /50 )

= 0.1822

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.5-0.1822 < p <0.5+ 0.1822

0.3178< p < 0.6822

The 99% confidence interval for the population proportion p is : 0.3178, 0.6822

orchestra answered 1 year ago

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