Respuesta :
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
[tex]\hat{p} = \dfrac{x}{n} = \dfrac{63}{219} = 0.2877[/tex]
99% Confidence interval:
[tex]\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 2.58[/tex]
Putting the values, we get:
[tex] 0.2877\pm 2.58(\sqrt{\dfrac{ 0.2877(1- 0.2877)}{219}})\\\\ = 0.2877\pm 0.0789\\\\=(0.2088,0.3666)[/tex]
is the required 99% confidence interval for population proportion.
Thus, the upper limit for population proportion is 0.3666
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 : [tex]\hat p[/tex] = X/n = 63/219
Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;
P.Q. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion = [tex]\frac{63}{219}[/tex]
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 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 2.5758) = 0.99
P( [tex]-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]{\hat p-p}[/tex] < [tex]2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.99
P( [tex]\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < p < [tex]\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.99
99% confidence interval for p = [ [tex]\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] , [tex]\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex]]
= [ [tex]\frac{63}{219} -2.5758 \times {\sqrt{\frac{\frac{63}{219}(1-\frac{63}{219})}{219} } }[/tex] , [tex]\frac{63}{219} +2.5758 \times {\sqrt{\frac{\frac{63}{219}(1-\frac{63}{219})}{219} } }[/tex] ]
= [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.
