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 = 6.82
x2 = 6.28
s1 = 0.65
s2 = 0.75
(a)
State the null and alternative hypotheses.
H0:
μ1 − μ2 ≤ 0
Ha:
μ1 − μ2 = 0
H0:
μ1 − μ2 > 0
Ha:
μ1 − μ2 ≤ 0
H0:
μ1 − μ2 ≠ 0
Ha:
μ1 − μ2 = 0
H0:
μ1 − μ2 ≤ 0
Ha:
μ1 − μ2 > 0
H0:
μ1 − μ2 = 0
Ha:
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 =
(d)
What is your conclusion?
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. Do not Reject H0. There is sufficient evidence to conclude that the consultant with more experience has a higher population mean rating.Reject H0. There is insufficient evidence to conclude that the consultant with more experience has a higher population mean rating.

Answer 1

Answer:

A) Null hypothesis; H0: μ1 − μ2 ≤ 0

Alternative hypothesis; Ha: μ1 − μ2 > 0

B) Test statistic = t = 1.878

C) p-value is 0.038823.

D) Reject the Null hypothesis H0

Step-by-step explanation:

We are given;

α = 0.05

n1 = 16

n2 = 10

bar x1 = 6.82

bar x2 = 6.28

s1 = 0.65

s2 = 0.75

A) The hypothesis is as follows;

Null hypothesis; H0: μ1 − μ2 ≤ 0

Alternative hypothesis; Ha: μ1 − μ2 > 0

B) Formula yo determine the test statistic is;

t = ((bar x1) - (bar x))/√((s1)²/n1) + (s2)²/n2))

Plugging in the relevant values, we have;

t = (6.82 - 6.28)/√((0.65)²/16) + (0.75)²/10))

t = 0.54/√(0.02640625 + 0.05625)

t = 0.54/0.2875

t = 1.878

C) The formula for the degree of freedom is;

Δ = [(s1)²/n1) + (s2)²/n2))]²/[((s1²/n1)²/(n1 - 1)) + ((s2²/n2)²/(n1 - 1))  

Plugging in the relevant values, we have;

Δ = [(0.65)²/16) + (0.75)²/10))]²/[((0.65²/16)²/(16 - 1)) + ((0.75²/10)²/(10 - 1))  

Δ = 0.00683205566/(0.000046486 + 0.0003515625)

Δ ≈ 17

Thus, the P-value;

From online p-value calculator from t-score and DF which i attached, we have the p-value as;

The p-value is 0.038823.

D) The p-value result is significant at p < 0.05

Thus, we reject the Null hypothesis H0