Question

Of 1000 randomly selected cases of lung cancer, 820 resulted in death within 10 years. Construct a 95% two-sided confidence interval on the death rate from lung cancer. (a) Construct a 95% two-sided confidence interval on the death rate from lung cancer. Round your answers to 3 decimal places. (b) Using the point estimate of p obtained from the preliminary sample, what sample size is needed to be 95% confident that the error in estimating the true value of p is less than 0.03? (c) How large must the sample if we wish to be at least 95% confident that the error in estimating p is less than 0.03, regardless of the true value of p?

Answer 1

Part a

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

x = 820

n = 1000

P = x/n = 820/1000 = 0.82

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.82 ± 1.96* sqrt(0.82*(1 – 0.82)/1000)

Confidence Interval = 0.82 ± 1.96*0.0121

Confidence Interval = 0.82 ± 0.0238

Lower limit = 0.82 - 0.0238 =0.7962

Upper limit = 0.82 + 0.0238 = 0.8438

Confidence interval = (0.7962, 0.8438)

Part b

The sample size formula is given as below:

n = p*q*(Z/E)^2

We are given

p = 0.82

q = 1 – p = 1 – 0.82 = 0.18

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Margin of error = E = 0.03

The sample size is given as below:

n = p*q*(Z/E)^2

n = 0.82*0.18*(1.96/0.03)^2

n = 630.0224

Required sample size = 631

Part c

The sample size formula is given as below:

n = p*q*(Z/E)^2

Estimate for proportion is not given, so we take

p = 0.5

q = 1 – p = 0.5

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Margin of error = E = 0.03

The sample size is given as below:

n = p*q*(Z/E)^2

n = 0.5*0.5*(1.96/0.03)^2

n = 1067.111

Required sample size =1068