Question

An article in Journal of the American Statistical Association (1990, Vol. 85, pp. 972–985) measured weight of 30 rats under experiment controls. Suppose that there are 12 underweight rats. (a) Calculate a 90% two-sided confidence interval on the true proportion of rats that would show underweight from the experiment. Round your answers to 3 decimal places. Enter your answer; confidence interval, lower bound ≤p≤ Enter your answer; confidence interval, upper bound (b) Using the point estimate of p obtained from the preliminary sample, what sample size is needed to be 90% confident that the error in estimating the true value of p is no more than 0.02? n= Enter your answer in accordance to the item b) of the question statement (c) How large must the sample be if we wish to be at least 90% confident that the error in estimating p is less than 0.02, regardless of the true value of p? n= Enter your answer in accordance to the item c) of the question statement

Answer 1

Solution :

Given that,

n = 30

x = 12

Point estimate = sample proportion = = x / n = 12 / 30 = 0.40

1 - = 1 - 0.40 =0.60

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 (((0.40 * 0.60) / 30)

= 0.147

A 90% confidence interval for population proportion p is ,

   - E p + E

0.40 - 0.147 p 0.40 + 0.147

( 0.253 p 0.547 )

b) margin of error = E = 0.02

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.645 / 0.02)² * 0.40 * 0.60

= 1623.61

sample size = n = 1624

c) =  1 - =   0.5

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.645 / 0.02)² * 0.5 * 0.5

= 1691.26

sample size = n = 1692