Answer:
Step-by-step explanation:
Hello!
The objective is to compare the mean rating on the client satisfaction survey of two consultants A and B.
Consultant A
X₁: Rating on the client satisfaction survey given to consultant A
n₁= 16
X[bar]₁= 6.82
S₁= 0.64
Consultant B
X₂: Rating on the client satisfaction survey given to consultant B
n₂= 10
X[bar]₂= 6.25
S₂= 0.75
a. The claim is that consultant A, which is more experienced, has a higher average service rate than consultant B, which has less experience, then the hypotheses are:
H₀: μ₁ ≤ μ₂
H₁: μ₁ > μ₂
b. If both variables have a normal distribution and the population's variances are unknown but equal, the statistic for this test is a t-test for independent samples with pooled sample variance.
[tex]t_{H_0}= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}[/tex]
[tex]Sa^2= \frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} = \frac{15*(0.64^2)+9*(0.75^2)}{16+10-2}= 0.47[/tex]
Sa= 0.68
[tex]t_{H_0}= \frac{6.82-6.25-0}{0.68*\sqrt{\frac{1}{16} +\frac{1}{10} } }= 2.079[/tex]
c. The p-value is the probability of obtaining a value as extreme as the value of the statistic. As the test it is one-tailed and has the same direction (right), symbolically:
P(t₂₄≥2.079)= 1 - P(t₂₄<2.079)= 1 - 0.9758= 0.0242
d. The p-value:0.0242 is less than α:0.05, then the decision is to reject the null hypothesis.
Using a significance level of 5%, there is significant evidence to say that the average service rate of consultant A is higher than the average service rate of consultant B.
I hope this helps!
