Question

Periodically, customers of a financial services company are asked to evaluate the company's financial consultants and services. Higher ratings on the client satisfaction survey indicate better service, with 7 the maximum service rating. Independent samples of service ratings for two financial consultants are summarized here. Consultant A has 10 years of experience, whereas consultant B has 1 year of experience. Use

α = 0.05

and test to see whether the consultant with more experience has the higher population mean service rating.

Consultant A	Consultant B
n1 = 16	n2 = 10
x1 = 6.85	x2 = 6.25
s1 = 0.68	s2 = 0.75

State the null and alternative hypotheses.

H₀:

μ₁ − μ₂ = 0

H_a:

μ₁ − μ₂ ≠ 0

H₀:

μ₁ − μ₂ > 0

H_a:

μ₁ − μ₂ ≤ 0

    

H₀:

μ₁ − μ₂ ≠ 0

H_a:

μ₁ − μ₂ = 0

H₀:

μ₁ − μ₂ ≤ 0

H_a:

μ₁ − μ₂ = 0

H₀:

μ₁ − μ₂ ≤ 0

H_a:

μ₁ − μ₂ > 0

(b)

Compute the value of the test statistic. (Round your answer to three decimal places.)

(c)

What is the p-value? (Round your answer to four decimal places.)

p-value =

What is your conclusion?

Reject H₀. There is insufficient evidence to conclude that the consultant with more experience has a higher population mean rating.

Reject H₀. There is sufficient evidence to conclude that the consultant with more experience has a higher population mean rating.    

Do not Reject H₀. There is sufficient evidence to conclude that the consultant with more experience has a higher population mean rating.

Do not reject H₀. There is insufficient evidence to conclude that the consultant with more experience has a higher population mean rating.

Answer 1

Solution :-
Given That :-
	X1 =	6.85	X2 =	6.25
	n1 =	16	n2 =	10
	S1 =	0.68	S2 =	0.75
	alpha =	0.05
	S1^2 =	0.4624	S2^2 =	0.5625
Hypothesis :-
Ho: μ1​ = μ2​		This is a Right tailed Test
Ha: μ1​ > μ2​
Test Statistic :-

				DF =	n1 + n2 - 2
t =	0.6			DF =	24
	0.285026
				S^p =	11.9985
t =	2.105				24
				S^p =	0.500
Critical value :-
By using t value Calculator,
t critical (tc) for Right tailed test at alpha = 0.05 is 1.711
Since t stat = 2.105 > t critical = 1.711
It is Concluded that Null Hypothesis is rejected.
P - Value :-
By using p value Calculator,
P - Value at t stat = 2.105 is 0.0229
P - Value = 0.0229 < Significance level 0.05
Conclusion :-
Reject H0. - There is sufficient evidence to conclude that the consultant with more
Experience has a higher population mean rating.