Question

According to a​ poll, 704 out of 1068 randomly selected smokers polled believed they are discriminated...

According to a​ poll, 704 out of 1068 randomly selected smokers polled believed they are discriminated against in public life or in employment because of their smoking. a. What percentage of the smokers polled believed they are discriminated against because of their​ smoking? b. Check the conditions to determine whether the CLT can be used to find a confidence interval. c. Find a​ 95% confidence interval for the population proportion of smokers who believe they are discriminated against because of their smoking. d. Can this confidence interval be used to conclude that the majority of smokers believe they are discriminated against because of their​ smoking? Why or why​ not?

a. The percentage of those taking the poll believed they are discriminated against because of their smoking is

b. Check the conditions to determine whether you can apply the CLT to find a confidence interval.

The Random and Independent condition

cannot

can

reasonably be assumed to hold.

The Large Sample condition

does not hold.

holds.

The Big Population condition

does not hold.

holds.

c. The​ 95% confidence interval is

​(___ ​ , ___).

​(Round to three decimal places as​ needed.)

Can this confidence interval be used to conclude that the majority of smokers believe they are discriminated against because of their​ smoking? Why or why​ not?

No,

Yes,

because the confidence interval

includes values above and below

only includes values above

only includes values below

50%.

75%.

100%.

25%.

0 0
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Answer #1

a)
pcap = 704/1068 = 0.6592
Hence percentage = 65.92%

b)
The Random and Independent condition can reasonably be assumed to hold.

The Large Sample condition holds.

The Big Population condition holds.

c)
sample size, n = 1068
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.6592 * (1 - 0.6592)/1068) = 0.0145

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.6592 - 1.96 * 0.0145 , 0.6592 + 1.96 * 0.0145)
CI = (0.631 , 0.688)

Yes,

because the confidence interval only includes values above 50%.

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