A coffee manufacturer is interested in whether the mean daily consumption of reg

Question

A coffee manufacturer is interested in whether the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers. Assume the population standard deviation for those drinking regular coffee is 1.36 cups per day and 1.46 cups per day for those drinking decaffeinated coffee. A random sample of 52 regular-coffee drinkers showed a mean of 4.53 cups per day. A sample of 48 decaffeinated-coffee drinkers showed a mean of 5.25 cups per day. The decision rule is to reject H0: r d if z < The test statistic is z = The p-value is

Explanation / Answer

Compute test statistic Z,

Pooled estimate, Sigmaxbar-xbar=sqrt [s1^2/n1+s^2/n2], where s1 and s2 are standard deviations of regular coffee drinkers and decaffeniated coffee drinkers.

=sqrt[1.36^2/52+1.46^2/48]

=0.28

Z(obtained)=(x1bar-x2bar)/Sigmaxbar-xbar=(4.53-5.25)/0.28=-2.57

The critical Z is -1.96.

Decision rule, reject H0 if test statistic>-1.96.

The test statistic falls in critical region. Therefore, reject null hypothesis.

P value is 0.005.

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