Answer to Question #344634 in Statistics and Probability for geresu

Question

a coffe manufacturer is interested in whether the mean daily consumption of regular coffee drinkers decaffeinated coffee . 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 5.84 cups a day. use the .01 significance level. compute the p-value .

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1\ge\mu_2$

$H_a:\mu_1<\mu_2$

This corresponds to a left-tailed test, and a z-test for two means, with known population standard deviations will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ and the critical value for a left-tailed test is $z_c = -2.3263.$

The rejection region for this left-tailed test is$R = \{z: z < -2.3263\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{X}_1-\bar{X}_2}{\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}}$$

$$=\dfrac{4.35-5.84}{\sqrt{(1.2)^2/50+(1.36)^2/40}}=-5.4393$$

Using the P-value approach:

The p-value is $p=P(Z<-5.4393)=0.00000002,$ and since $p=0.00000002<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$

is less than $\mu_2,$ at the $\alpha = 0.01$ significance level.