Mathematics, 21.04.2020 02:26 sainijasdeep27

Time magazine reported the result of a telephone poll of randomly selected adult Americans. The poll asked these adult Americans: "Should the federal tax on cigarettes be raised to pay for health care reform?" The results of the survey showed that among 620 non-smokers, 318 had said "yes" and among 195 smokers, 35 had said "yes".
We want to estimate the actual difference between the proportions of non-smokers and smokers who said "yes" to federal tax increase. Notation: 1=non-smokers and 2-smokers. Based on this data, what is the upper bound for a 95% confidence interval for the difference in the population proportions, Pi - P2?

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Answer from: pthalia

The 95% confidence interval for the difference in the population proportions( Pi - P2)

(0.2674 ,0.4055)

the upper bound for a 95% confidence interval for the difference in the population proportions, Pi - P2

0.4055

Step-by-step explanation:

Step :- (1)

Given data the results of the survey showed that among 620 non-smokers, 318 had said "yes"

The first proportion $p_{1} = \frac{318}{620} =0.5129$

q₁ = 1- p₁ = 1-0.5129 =0.4871

Given data the results of the survey showed that among 195 smokers, 35 had said "yes".

The second proportion $p_{2} = \frac{35}{195} =0.179$

q₂ = 1- p₂ = 1-0.179 =0.821

Step :-(2)

The 95% confidence interval for the difference in the population proportions( Pi - P2)

(p₁-p₂ ± z₀.₉₅ se(p₁-p₂))

The standard error (p₁-p₂) is defined by

= $\sqrt{\frac{p_{1}q_{1} }{n_{1} }+\frac{p_{2}q_{2} }{n_{2} } }$

= $\sqrt{\frac{0.5129 X 0.4871 }{620 }+\frac{0.179 X 0.821 }{195 } }$

= 0.0339

The 95% confidence interval for the difference in the population proportions( Pi - P2)

(p₁-p₂ ± z₀.₉₅ se(p₁-p₂))

(p₁-p₂ - z₀.₉₅ se(p₁-p₂) , p₁-p₂ + z₀.₉₅ se(p₁-p₂) )

(0.5129-0.179) - 1.96 × 0.0339 , 0.5129 -0.179) - 1.96 × 0.0339)

(0.339 -0.0665 , 0.339 +0.0665 )

(0.2674 ,0.4055)

Conclusion:-

The 95% confidence interval for the difference in the population proportions( Pi - P2)

(0.2674 ,0.4055)

Answer from: cuhh

The 95% confidence interval for difference in proportion of non-smokers and smokers who said "yes" to federal tax increase is (0.26, 0.40).

Step-by-step explanation:

The confidence interval for difference in proportion formula is,

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

The given information is:

$n_{1}=620\\n_{2}=195\\X_{1}=318\\X_{2}=35$

Compute the sample proportions as follows:

$\hat p_{1}=\frac{X_{1}}{n_{1}}=\frac{318}{620}=0.51\\\\\hat p_{2}=\frac{X_{2}}{n_{2}}=\frac{35}{195}=0.18$

Compute the critical value of z for the 95% confidence level as follows:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the standard normal table.

Compute the 95% confidence interval for difference between proportions as follows:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

$=(0.51-0.18)\pm 1.96\times \sqrt{\frac{0.51(1-0.51)}{620}+\frac{0.18(1-0.18)}{195}}$

$=0.33\pm 0.067\\=(0.263, 0.397)\\\approx (0.26, 0.40)$

Thus, the 95% confidence interval for difference in proportion of non-smokers and smokers who said "yes" to federal tax increase is (0.26, 0.40).

Answer from: Quest

i think the answer is true

step-by-step explanation:

Answer from: Quest

answerim :

step-by-step explanation:

im not sure but my dad knows

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