Question

A coffee manufacturer is interested in whether the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers. A random sample of 50 regular-coffee drinkers showed a mean of 4.35 cups per day. A sample of 40 decaffeinated-coffee drinkers showed a mean of 5.12 cups per day. Assume the population standard deviation for those drinking regular coffee is 1.20 cups per day and 1.36 cups per day for those drinking decaffeinated coffee. Perform an appropriate test at the 1% level of significance. Use the critical value approach.Compute the p-value.

Answer 1

The P-Value ≅0 (zero).

Step-by-step explanation:

From the given data we have

Regular coffee drinkers sample size = n1 = 50

Decaffeinated-coffee drinkers sample size = n2= 40

Regular coffee drinkers sample mean= x1 = 4.35

Decaffeinated-coffee drinkers sample mean = x2= 5.12

Regular coffee drinkers population standard deviation = σ1 = 1.2

Decaffeinated-coffee drinkers population standard deviation = σ2= 1.36

1) Formulate null and alternate hypothesis

H0: u1≥ u2 Ha: u1 < u2

The null hypothesis is that the mean of the regular coffee drinkers is greater or equal to the mean of decaffeinated-coffee drinkers

against the claim

the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers.

2) The test statistic is

z= x1-x2/ sqrt( σ1 ²/n1 + σ2²/n2)

Putting the values

z = 4.35- 5.12/ sqrt( 1.44/50 + 1.8496/40)

z= -5.44

3) The significance level is 0.01

The critical region is Z < -2.33

4) Since the calculated value of z= -5.44 is less than the z ∝= -2.33 we reject H0.

5) the P-value can be calculated using the calculator.

The P-Value is < 0.00001.

P= 0

Which means that the claim is accepted that the mean of the regular coffee drinkers is less than the mean of decaffeinated-coffee drinkers.