Question

A winery claims class of students drink, on average 1.4 glasses of wine per night on nights after class. The professor randomly samples 34 students and finds the average to be 1.6 glasses consumed. The professor knows the population standard deviation to be equal to 0.5 glasses of wine. At 95% confidence level, is the winery's claims supported by this sample?

Answer 1

Solution:

mu = 1.4

n = 34

sigma = 0.5

x bar = 1.6

alpha = 0.05

Z = 1.96 ( From table )

Margin of Error = E = Z(sigma/sqrt(n)) = 0.1681

Lower Limit = x bar - E = 1.4319

Upper Limit = x bar + E = 1.7681

95% confidence interval is (1.4319 , 1.7681).

At 95% confidence level, is the winery's claims supported by this sample?

• The confidence interval contain the null hypothesis value ( mu = 1.6 ), hence the winery's claims supported by this sample.