Mathematics, 04.04.2020 11:04 donnamorgan32

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

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Answer from: rouchedavisin4

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 from: Sanchezj4655

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.

Answer from: Quest

Ithink you're supposed to multiply 12.50 which is how much he makes per hour and then multiply it by each number then add whats left togrther and there's your answer

Answer from: Quest

it’s supposed to be -4+45=41

step-by-step explanation: