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sammy [17]
3 years ago
11

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?
Mathematics
2 answers:
Yuki888 [10]3 years ago
8 0

Answer:

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

<u>Step :-(2)</u>

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

(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

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

(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)

<u>Conclusion</u>:-

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

(0.2674 ,0.4055)

Thepotemich [5.8K]3 years ago
7 0

Answer:

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 <em>z</em> 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).

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