In: Statistics and Probability
5. (Excel exercise) According to a recent report, consumption of alcoholic beverages by young women of drinking age has been increasing in the U.S. and Europe. Annual alcohol consumption data for a random sample of 20 young women in the United States is reported in the following table.
X = Annual alcohol consumption (liters)
X
266
170
164
93
82
222
102
0
199
115
113
93
174
130
171
110
97
169
0
130
a. Assuming that the population is approximately bell-shaped and symmetric, construct and interpret a 95% confidence interval for the mean annual consumption of alcoholic beverages by young women in the United States. What is the margin of error?
b. Now, construct and interpret a 99% confidence interval for the mean annual consumption of alcoholic beverages by young women in the United States. What happens to the margin of error when we change the confidence level from 95% to 99%?
From the given data,
Sample mean xbar = 130
Sample std.dev s = 65.3911
a)
The t value at 95% confidence interval is,
df = n -1 = 20-1 = 19
alpha = 1 - 0.95 = 0.05
alpha/2 = 0.05/2 = 0.025
t(alpha/2,df) = t(0.025,19) = 2.0930
Margin of error = E =z *(s/sqrt(n))
= 2.0930 *(65.3911/sqrt(20))
=30.6036
The 95% confidence interval is
mean -E < mu < mean +E
130 - 30.6036 < mu < 653.5 + 30.6036
99.3964 < mu < 160.6036
WE are 95% confident that the mean annual consumption of alcoholic beverages by young women in the United States is between 99.3464 and 160.6036
b)
The t value at 99% confidence interval is,
df = n -1 = 20-1 = 19
alpha = 1 - 0.99 = 0.01
alpha/2 = 0.01/2 = 0.02
t(alpha/2,df) = t(0.02,19) =2.8609
Margin of error = E =z *(s/sqrt(n))
= 2.8609 *(65.3911/sqrt(20))
=41.8318
The 99% confidence interval is
mean -E < mu < mean +E
130 - 41.8318 < mu < 653.5 + 41.8318
88.1682< mu < 171.8318
WE are 95% confident that the mean annual consumption of
alcoholic beverages by young
women in the United States is between 88.1682 and 171.8318
AS the confidence interval increses teh margin of error
increases