Question

"4. 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 1.20 cups per day and 1.36 cups per day for those drinking 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 per day. Use the .01 significance level."(1) This a (ONE TAILED or TWO-TAILED) test??

Answer 1

Answer and Step-by-step explanation

Given:

Sample size = n1 = 50

Sample mean = x1 = 3.84

Population standard deviation= ∂1 = 1.20

Sample size = n2 = 40

Sample mean = x2 = 3.35

Population standard deviation = ∂2 =1.36

Significance level = a = 0.05

Daily consumption of regular-coffee is greater than decaffeinated-coffee.

The hypothesis:

H0 : µ1 ≤ µ2

Ha : µ1 > µ2

If alternative hypothesis Ha contains <, then test is left tailed.

If the alternative hypothesis Ha contains >, then test is right tailed.

If the alternative hypothesis Ha contains ≠, then test is two tailed.

Right tailed test.

P ( z > z0) = 0.05

Let determine the z score that corresponds with probability of 1 – 0.5 – 0.05 =0.45

Determine the value of test statistic:

Z = x1 – x2 / √(∂1/n1+∂2/n2)

The p value is the probability of obtaining a value more extreme or equ

Answer 2

With a 0.01 significance level and samples of 50 and 40 cofee drinkers, there is enough statistical evidence to state that the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers.

The test is a one-tailed test.

Explanation:

To solve this problem, we run a hypothesis test about the difference of population means.

`$$Sample mean $\bar X: \bar X=4.35$\\Sample mean $\bar Y: \bar Y=5.84$\\Population variance $\sigma^2_X: \sigma^2_X=1.2^2$\\Population variance $\sigma^2_Y: \sigma^2_Y=1.36^2$\\Sample size $n_X=50$\\Sample size $n_Y=40$\\Significance level $\alpha=0.01$\\Z criticals values (for a 0.01 significance)\\(Left tail test) Z_(1-\alpha)=Z_(0.990)=-2.32635\\($Right tail test) Z_(\alpha)=Z_(0.010)=2.32636\\($Two-tailed test) $Z_(1-\alpha/2)=Z_(0.995)=-2.57583$ and $ Z_(\alpha/2)=Z_(0.005)=2.57583\\\\`

The appropriate hypothesis system for this situation is:

`H_0: \mu_X-\mu_Y=0\\H_a: \mu_X-\mu_Y  \leq 0\\\\`

Difference of means in the null hypothesis is:

`\mu_X - \mu_Y = M_0=0\\\\`

`$$The test statistic is$ Z=\frac{( \bar X-\bar Y)-M_0}{\sqrt{(\sigma^2_X)/(n_X)+(\sigma^2_Y)/(n_Y)}}\\`
`$$The calculated statistic is Z_c=\frac{[(4.35-5.84)-0]}{\sqrt{(1.20^2)/(50)+(1.36^2)/(40)}}=-5.43926\\p-value = P(Z \leq Z_c)=0.0000\\\\`

Since, the calculated statistic
`Z_c` is less than critical
`Z_(1-\alpha)`, the null hypothesis should be rejected. There is enough statistical evidence to state that the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers.