Question

1. A retail store runs an advertising campaign on a radio station. They decide to measure the effectiveness of the campaign by measuring the increase in customers compared to previous days. They choose 35 days at random, and find the average number of customers to have increased by 83.3 customers per day. Historically, the number of customers per day has a standard deviation of 17.5 customers. What is the 95% confidence interval for the population mean increase in customers?

Select one:

a. 79.51 to 87.09

b. 79.54 to 87.06

c. 77.50 to 89.10

d. 55.70 to 110.90

e. 78.42 to 88.18

2. A movie theater in a tourist destination notices that their attendance improves when it is rainy outside. For the past year, the standard deviation of movie attendance has been 6.1 people. The theater looks at the attendance of 42 movies while it was raining last month, and the average attendance was 91.2 people per showing. What is the 80% confidence interval?

Select one:

a. 89.65 to 92.75

b. 73.19 to 109.21

c. 90.00 to 92.40

d. 89.36 to 93.04

e. 91.01 to 91.39

Answer 1

Solution :

1) Given that,

Point estimate = sample mean = = 83.3

Population standard deviation =    = 17.5

Sample size = n = 35

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 = 0.05

/2 = 0.025

Z/2 = Z_0.025 = 1.96

Margin of error = E = Z/2 * ( /n)

= 1.96 * ( 17.5 /  35 )

= 5.80

At 95% confidence interval estimate of the population mean is,

  ± E

83.3 ± 5.80   

( 77.50 to 89.10 )  

correct option is = c

2) Given that,

Point estimate = sample mean = = 91.2

Population standard deviation =    = 6.1

Sample size = n = 42

At 80% confidence level

= 1 - 80%  

= 1 - 0.80 =0.20

/2 = 0.10

Z/2 = Z_0.10 = 1.282

Margin of error = E = Z/2 * ( /n)

= 1.28 * ( 6.1 /  42 )

= 1.20

At 80% confidence interval estimate of the population mean is,

  ± E

91.2 ± 1.20

( 90.00 to 92.40 )  

correct option is = c