Respuesta :
Answer:
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.
Step-by-step explanation:
To solve this problem, we run a hypothesis test about the difference of population means.
[tex]$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\\\\[/tex]
The appropriate hypothesis system for this situation is:
[tex]H_0: \mu_X-\mu_Y=0\\H_a: \mu_X-\mu_Y \leq 0\\\\[/tex]
Difference of means in the null hypothesis is:
[tex]\mu_X - \mu_Y = M_0=0\\\\[/tex]
[tex]$The test statistic is$ Z=\frac{( \bar X-\bar Y)-M_0}{\sqrt{\frac{\sigma^2_X}{n_X}+\frac{\sigma^2_Y}{n_Y}}}\\[/tex][tex]$The calculated statistic is Z_c=\frac{[(4.35-5.84)-0]}{\sqrt{\frac{1.20^2}{50}+\frac{1.36^2}{40}}}=-5.43926\\p-value = P(Z \leq Z_c)=0.0000\\\\[/tex]
Since, the calculated statistic [tex]Z_c[/tex] is less than critical [tex]Z_{1-\alpha}[/tex], 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.
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