Question

(12 pts.) A study examined the effectiveness of Botox injections in the corrugator supercilii muscles for the treatment of chronic migraines. Of the 26 migraine-prone patients participating in the study, 14 reported complete migraine elimination.

(a) Determine a plus four 99% confidence interval for the population proportion of migraine patients with complete migraine remission following such Botox injections.

(b) How large a sample n would you need to estimate p with margin of error 0.03 with 99% confidence? Use pˆ = 0.54 as an estimated value for p.

(c) How large a sample n would you need to estimate p with margin of error 0.03 with 99% confidence? Use pˆ = 0.50 as the estimated value for p.

Answer 1

a)

Solution :

Given that,

n = 26

x = 14

Point estimate = sample proportion = = x / n = 0.538

1 - = 0.462

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (((0.538 * 0.462) / 26)

= 0.252

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.538 - 0.252 < p < 0.538 + 0.252

0.286 < p < 0.790

b)( 0.286 , 0.790 )

Solution :

Given that,

= 0.54

1 - = 0.46

margin of error = E = 0.03

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= (2.576 / 0.03)² * 0.54 * 0.46

= 1831.47

sample size = 1832

c)

Solution :

Given that,

= 0.50

1 - = 0.50

margin of error = E = 0.03

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= (2.576 / 0.03)² * 0.5 * 0.5

= 1843.27

sample size = 1844