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.

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  [tex]p-value  = 0.03682 [/tex]

Step-by-step explanation:

From the question we are told that

   The first sample size is  [tex]n_ 1 = 50[/tex]

   The first sample mean is [tex]\=x_1  =  3.84[/tex]

The first sample standard deviation is  [tex]s_1  = 1.26[/tex]

    The second sample size is  [tex]n_2  =  40[/tex]      

   The second  sample mean is   [tex]\=x_2  =  3.35 [/tex]

    The second sample standard deviation is [tex]s_2  =  1.36[/tex]

    The level of significance is  [tex]\alpha =  0.05[/tex]

The  null hypothesis is  [tex]H_o  :  \mu_1  = \mu_2[/tex]

The alternative hypothesis is  [tex]H_a  :  \mu_1  >  \mu_2[/tex]

Generally the test hypothesis is mathematically represented as

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

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

=>   [tex]   t =  1.789  [/tex]      

Generally the p-value is mathematically represented as

      [tex]p-value  =  P(t > 1.789)[/tex]

From the z table  

    [tex]\Phi (1.789) =  0.03682[/tex]

So

   [tex]p-value  =  P(t > 1.789) =  0.03682 [/tex]

From the value obtained we see that [tex]p-value < \alpha[/tex] 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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