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

Answer:

The upper limit for population proportion is 0.3666

Step-by-step 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

$\hat{p} = \dfrac{x}{n} = \dfrac{63}{219} = 0.2877$

99% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 2.58$

Putting the values, we get:

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

is the required 99% confidence interval for population proportion.

Thus, the upper limit for population proportion is 0.3666

Answer 2

Answer:

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

Step-by-step 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. =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion = $\frac{63}{219}$

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

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

P( $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$]

   = [ $\frac{63}{219} -2.5758 \times {\sqrt{\frac{\frac{63}{219}(1-\frac{63}{219})}{219} } }$ , $\frac{63}{219} +2.5758 \times {\sqrt{\frac{\frac{63}{219}(1-\frac{63}{219})}{219} } }$ ]

   = [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.