Question

Of the 219 white GSS2008 respondents in their 20s, 63 of them claim the ability to speak a language other than English. With 99% confidence, what is the upper limit of the population proportion based on these statistics

Answer 1

The upper limit of the 99% confidence interval for the population proportion based on these statistics is 0.3665.

Explanation:

We are given that of the 219 white GSS 2008 respondents in their 20's, 63 of them claim the ability to speak a language other than English.

So, the sample proportion is :
`\hat p` = X/n = 63/219

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

P.Q. =
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = sample proportion =
`(63)/(219)`

n = sample of respondents = 219

p = population proportion

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the population​ proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at

0.5% level of significance are -2.5758 & 2.5758}

P(-2.5758 <
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` < 2.5758) = 0.99

P(
`-2.5758 * {\sqrt{(\hat p(1-\hat p))/(n) } }` <
`{\hat p-p}` <
`2.5758 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.99

P(
`\hat p-2.5758 * {\sqrt{(\hat p(1-\hat p))/(n) } }` < p <
`\hat p+2.5758 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.99

99% confidence interval for p = [
`\hat p-2.5758 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ,
`\hat p+2.5758 * {\sqrt{(\hat p(1-\hat p))/(n) } }`]

= [
`(63)/(219) -2.5758 * {\sqrt{((63)/(219)(1-(63)/(219)))/(219) } }` ,
`(63)/(219) +2.5758 * {\sqrt{((63)/(219)(1-(63)/(219)))/(219) } }` ]

= [0.2089 , 0.3665]

Therefore, 99% confidence interval for the population proportion based on these statistics is [0.2089 , 0.3665].

Hence, the upper limit of the population proportion based on these statistics is 0.3665.

Answer 2

The upper limit for population proportion is 0.3666

Explanation:

We are given the following in the question:

Sample size, n = 219

Number of people who have ability to speak a language other than English, x = 63

`\hat{p} = (x)/(n) = (63)/(219) = 0.2877`

99% Confidence interval:

`\hat{p}\pm z_(stat)\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`

`z_(critical)\text{ at}~\alpha_(0.05) = 2.58`

Putting the values, we get:

`0.2877\pm 2.58(\sqrt{( 0.2877(1- 0.2877))/(219)})\\\\ = 0.2877\pm 0.0789\\\\=(0.2088,0.3666)`

is the required 99% confidence interval for population proportion.

Thus, the upper limit for population proportion is 0.3666