Question

10. Construct the 90% confidence interval for each of the following scenarios. a. You wish to estimate the proportion of automobiles that are purchased over the Internet. You took a random sample of 200 people and found that 38 purchased a car over the Internet. b. In a random sample of 17 bikers, the mean time spent riding a day was 2.813 hours and the standard deviation was 1.205. Assume the hours spent are normally distributed.

Answer 1

Solution :

Given that,

a)

n = 200

x = 38

Point estimate = sample proportion = = x / n = 38 / 200 = 0.19

1 - = 1 - 0.19 = 0.81

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z _0.05 = 1.645

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

= 1.645 * (((0.19*0.81) / 200)

= 0.046

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.19 - 0.046 < p < 0.19 + 0.046

0.144 < p < 0.236

The 90% confidence interval for the population proportion p is : (0.144 , 0.236)

b)

Solution :

Given that,

Point estimate = sample mean = = 2.813

sample standard deviation = s = 1.205

sample size = n = 17

Degrees of freedom = df = n - 1 = 16

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,16 = 1.746

Margin of error = E = t_/2,df * (s /n)

= 1.746 * (1.205 / 17)

= 0.510

The 90% confidence interval estimate of the population mean is,

- E < < + E

2.813 - 0.510 < < 2.813 + 0.510

2.303 < < 3.323

(2.303 , 3.323)