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A coffee manufacturer is interested in the mean daily consumpiton of regular-coffee drinkers and decaffeinated-coffee drinkers. A random sample of 50 regular-coffee drinkers showed a mean of 3.84 cups per day. A sample of 40 decaffeinated-coffee drinkers shoed a mean of 3.35 cups per day. The sample stnadard deviation for those drinking regular coffee is 1.20 cups per day an 1.36 cups per day for those drinking decaffeinated coffee. Assume the population standard deviations are euql. At the 5% significance level, is the mean daily consumption of reglar-coffee drinkers greater than that of decaffeinated-cofee? Find the p-value for this hypothesis test.

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Answer:

The decision rule is

Reject the null hypothesis

The conclusion

There is sufficient evidence to conclude that mean daily consumption of regular-coffee drinkers greater than that of decaffeinated-coffee.

The p-value is
p-value  = 0.03682

Explanation:

From the question we are told that

The first sample size is
n_ 1 = 50

The first sample mean is
\=x_1  =  3.84

The first sample standard deviation is
s_1  = 1.26

The second sample size is
n_2  =  40

The second sample mean is
\=x_2  =  3.35

The second sample standard deviation is
s_2  =  1.36

The level of significance is
\alpha =  0.05

The null hypothesis is
H_o  :  \mu_1  = \mu_2

The alternative hypothesis is
H_a  :  \mu_1  >  \mu_2

Generally the test hypothesis is mathematically represented as


t =  \frac{( \= x_1 - \= x_2 ) - 0}{ \sqrt{ (s^2_1 )/(n_1)  + (s^2_2)/(n_2) } }

=>
t =  \frac{( 3.84 - 3.35  ) - 0}{ \sqrt{ (1.26^2 )/(50)  + (1.36^2)/(40) } }

=>
t =  1.789

Generally the p-value is mathematically represented as


p-value  =  P(t > 1.789)

From the z table


\Phi (1.789) =  0.03682

So


p-value  =  P(t > 1.789) =  0.03682

From the value obtained we see that
p-value < \alpha hence

The decision rule is

Reject the null hypothesis

The conclusion

There is sufficient evidence to conclude that mean daily consumption of regular-coffee drinkers greater than that of decaffeinated-coffee

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