Question

The Chartered Financial Analyst (CFA^®) designation is fast becoming a requirement for serious investment professionals. It is an attractive alternative to getting an MBA for students wanting a career in investment. A student of finance is curious to know if a CFA designation is a more lucrative option than an MBA. He collects data on 38 recent CFAs with a mean salary of $138,000 and a standard deviation of $34,000. A sample of 80 MBAs results in a mean salary of $130,000 with a standard deviation of $46,000. Use Table 2.

  

μ₁ is the population mean for individuals with a CFA designation and μ₂ is the population mean of individuals with MBAs. Let CFAs and MBAs represent population 1 and population 2, respectively.

  

a-1.

Set up the hypotheses to test if a CFA designation is more lucrative than an MBA at the 5% significance level. Do not assume that the population variances are equal.

	H0: μ1 − μ2 = 0; HA: μ1 − μ2 ≠ 0
	H0: μ1 − μ2 ≥ 0; HA: μ1 − μ2 < 0
	H0: μ1 − μ2 ≤ 0; HA: μ1 − μ2 > 0

  

a-2.	Calculate the value of the test statistic. (Round all intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.)

  

Test statistic

  

a-3.

Approximate the p-value.

	p-value < 0.01
	0.01 ≤ p-value < 0.025
	0.025 ≤ p-value < 0.05
	0.05 ≤ p-value < 0.10
	p-value ≥ 0.10

  

a-4.

Do you reject the null hypothesis at the 5% level?

	Yes, since the p-value is less than α.
	No, since the p-value is less than α.
	Yes, since the p-value is more than α.
	No, since the p-value is more than α.

  

b.

Using the critical value approach, can we conclude that CFA is more lucrative?

	Yes, since the value of the test statistic is less than the critical value of 1.661.
	Yes, since the value of the test statistic is less than the critical value of 1.985.
	No, since the value of the test statistic is less than the critical value of 1.661.
	No, since the value of the test statistic is less than the critical value of 1.985.

Answer 1

H₀: μ₁ − μ₂ ≤ 0; H_A: μ₁ − μ₂ > 0

Test Statistic :-

t = 1.0608

Test Criteria :-
Reject null hypothesis if


DF = 95

No, since the value of the test statistic is less than the critical value of 1.661.

Result :- Fail to Reject Null Hypothesis

Decision based on P value
P - value = P ( t > 1.0608 ) = 0.1457

p-value ≥ 0.10

Reject null hypothesis if P value <    level of significance
P - value = 0.1457 > 0.05 ,hence we fail to reject null hypothesis

No, since the p-value is more than α.

Conclusion :- We Accept H0