Question

Researchers are trying to estimate the proportion of patients trying a new experimental drug who experience a skin rash as a side-effect. 245 patients took this drug as part of early research, and these 245 patients are considered representative of the population who would take this drug in the future. 2.86% of these patients reported a skin rash. We would like to construct a 90% confidence interval for the proportion of potential drug users who would develop a skin rash.

What would be the standard error for our sample proportion, using the information we have from our sample to estimate that? Round to 3 decimal places

a 90% confidence interval will go 1.65 standard errors out in either direction. Calculate the lower bound of this 90% confidence interval for the true proportion of drug users who experience a skin rash as a side effect

Answer 1

(a)Solution :

Given that,

n = 245

Point estimate = sample proportion = 0.286

1 - = 1 - 0.286=0.714

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 (((0.286 *0.714) /245 )

= 0.0289 0.047

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.286 - 0.047 < p < 0.286 + 0.047

0.239< p < 0.333

The 90% confidence interval for the population proportion p is : (0.239 , 0.333)

(b)standard error = (( * (1 - )) / n) = (((0.286 *0.714) /245 ) = 0.029

(c) standard error 1.65 it not the proportion