The Chartered Financial Analyst (CFA®) designation is fast becoming a requiremen
ID: 3314118 • Letter: T
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 36 recent CFAs with a mean salary of $145,000 and a standard deviation of $36,000. A sample of 47 MBAs results in a mean salary of $135,000 with a standard deviation of $24,000. Use Table 2. 1 is the population mean for individuals with a CFA designation and 2 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 10% significance level. Do not assume that the population variances are equal. H0: 1 2 = 0; HA: 1 2 0 H0: 1 2 0; HA: 1 2 < 0 H0: 1 2 0; HA: 1 2 > 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 Picture 0.100 p-value Picture 0.010 0.010 Picture p-value < 0.025 0.025 Picture p-value < 0.050 0.050 Picture p-value < 0.100 a-4. Do you reject the null hypothesis at the 10% level? No, since the p-value is more than . No, since the p-value is less than . Yes, since the p-value is more than . Yes, since the p-value is less than . b. Using the critical value approach, can we conclude that CFA is more lucrative? No, since the value of the test statistic is more than the critical value of 1.297. No, since the value of the test statistic is more than the critical value of 1.672. Yes, since the value of the test statistic is more than the critical value of 1.297. Yes, since the value of the test statistic is more than the critical value of 1.672.
Explanation / Answer
Ans:
H0: 1 2 0;
HA: 1 2 > 0
Population variances are unequal.
So,
standard error of mean=sqrt((36000^2/36)+(24000^2/47))=6946.6049
Test statistic:
t=(145000-135000)/6946.6=1.44
df=36-1=35 (as we consider smaller of n1-1 or n2-1 to be conservative )
p-value(right tailed)=tdist(1.440,35,1)=0.0794
0.05<p-value<0.1
significance level=0.1
As,p-value<0.1,we reject null hypothesis.
Yes, since the p-value is less than.
critical value approach:
Yes, since the value of the test statistic is more than the critical value of 1.297.