Question

Of 1000 randomly selected cases of lung cancer, 833 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. sps i (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

Solution :

Given that,

a) Point estimate = sample proportion = $\hat p$ = x / n = 833 / 1000 = 0.833

1 - $\hat p$ = 1 - 0.833 = 0.167

Z$\alpha$/2 = Z_0.025 = 1.96

Margin of error = E = Z_{$\alpha$ / 2} * (($\hat p$ * (1 - $\hat p$ ))$\sqrt$ / n)

= 1.96 ($\sqrt$((0.833 * 0.167) / 1000)

= 0.023

A 95% confidence interval for population proportion p is ,

$\hat p$ - E $\leq$ p $\leq$ $\hat p$ + E

0.833 - 0.023 $\leq$ ​​​​​​​ p $\leq$ ​​​​​​​ 0.833 + 0.023

( 0.810 $\leq$ ​​​​​​​ p $\leq$ ​​​​​​​ 0.856 )

b) margin of error = E = 0.03

sample size = n = (Z_{$\alpha$ / 2} / E )² * $\hat p$ * (1 - $\hat p$ )

= (1.96 / 0.03 )² * 0.833 * 0.167

= 593.78

sample size = n = 594

c) $\hat p$ =  1 - $\hat p$ = 0.5

sample size = n = (Z_{$\alpha$ / 2} / E )² * $\hat p$ * (1 - $\hat p$ )

= (1.96 / 0.03)² * 0.5 * 0.5

= 1067.11

sample size = n = 1068