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 $142,000 and a standard deviation of $45,000. A sample of 55 MBAs results in a mean salary of $126,000 with a standard deviation of $27,000. 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 10% 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. 0.025 Picture p-value < 0.050 0.050 Picture p-value < 0.100 0.010 Picture p-value < 0.025 p-value Picture 0.010 p-value Picture 0.100 a-4. Do you reject the null hypothesis at the 10% level? No, since the p-value is more than α. No, since the p-value is less than α. Yes, since the p-value is more than α. Yes, since the p-value is less than α. 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 1.673. 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 1.673.

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_CFA< u_MBA
Alternative hypothesis: u_CFA > u_MBA

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 8157.4518
DF = 91

t = [ (x₁ - x₂) - d ] / SE

t = 1.9614

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is thesize of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of 1.9614.

Therefore, the P-value in this analysis is 0.0264.

Interpret results. Since the P-value (0.0264) is less than the significance level (0.10), we have to reject the null hypothesis.

Reject null hypothesis.

a) Yes, since the p-value is less than α.

b) Yes, since the value of the test statistic is more than the critical value of 1.291.