Question

In: Statistics and Probability

5. (Excel exercise) According to a recent report, consumption of alcoholic beverages by young women of...

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)

266

170

164

222

102

199

115

113

174

130

171

110

169

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%?

Solutions

Expert Solution

From the given data,

Sample mean xbar = 130
Sample std.dev s = 65.3911

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

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

orchestra answered 1 year ago

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