Question

A company randomly chose 200 US households and found that in 120 of them, the women made the majority of the purchasing decisions. We want to know the population proportion of US housholds where women make the majority of purchasing decisions. Construct a 96% confidence interval to find p.  

a) what is p hat?

b) what is alpha?

c) what is the value of the test statistic

d) what is the value of the standard errror

e) what is the value of the margin of error

f) what is the lower boundary of the confidence interval

g) what is the upper boundary of the confidence interval

Answer 1

Solution :

Given that,

n = 200

x = 120

a) Point estimate = sample proportion = = x / n = 120/200 = 0.600

1 - = 1-0.600 = 0.400

At 96% confidence level

b) = 1-0.96% =1-0.96 =0.04

/2 =0.04/ 2= 0.020

Z/2 = Z_{0.020 = 2.054}

c) Z_/2 = 2.054

d) standard error =(( * (1 - )) / n) = 0.0346

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

= 2.054 * ((0.600*(0.400) /200 )

= 0.071

A 96% confidence interval for population proportion p is ,

- E < p < + E

0.600-0.071 < p < 0.600+0.071

(0.529,0.671 )

f) Lower boundary of the confidence interval =0.529

g) Upper boundary of the confidence interval = 0.671