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 41 recent CFAs with a mean salary of $139,000 and a standard deviation of $60,000.
A sample of 53 MBAs results in a mean salary of $132,000 with a standard deviation of $23,000. (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) 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.
(b) Calculate the value of the test statistic. (Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)

Answer 1

Answer:

Step-by-step explanation:

We would set up the hypothesis test.

For the null hypothesis,

H0: µ1 ≤ µ2

For the alternative hypothesis,

Ha: µ1 > µ2

This is a test of two means

Random variable = x1 - x2 = difference in the sample mean amount for individuals with a CFA designation and sample mean amount of individuals with MBAs

This is a right tailed test

Since the population standard deviation is not given, the t test would be used to determine the test statistic. The formula is

(x1 - x2)/√(s1²)/n1 + (s2²)/n2

x1 and x2 = sample means

s1 and s2 = sample standard deviations

n1 and n2 = number of samples

From the information given,

x1 = 139000

x2 = 132000

s1 = 60000

s2 = 23000

n1 = 41

n2 = 53

test statistic = (139000 - 132000)/√(60000²)/41 + (23000²)/53

= 7000/√(87804878.049 + 9981132.075471697)

Test statistic = 0.71

Answer 2

In this exercise we have to analyze the hypotheses and calculate the value of the statistical test, in this way we will find that:

A) $(x_1 - x_2)/\sqrt{(S_1)^2/n_1+(S_2^2)/n2}$

B)$0.71$

We would set up the hypothesis test. For the null hypothesis:

$H_0: \mu_1 \leq \mu_2$

For the alternative hypothesis, we know that:  

$H_a: \mu_1 > \mu_2$

This is a test of two means, that a  random variable: $x_1 - x_2$ representes the difference in the sample mean amount for individuals with a CFA designation and sample mean amount of individuals with MBAs. This is a right tailed test.  Since the population standard deviation is not given, the t test would be used to determine the test statistic. The formula is:  

$(x_1 - x_2)/\sqrt{(S_1)^2/n_1+(S_2^2)/n2}$

• Sample means: $x_1, x_2$

• Sample standard deviations: $s_1, s_2$

• Number of samples: $n_1, n_2$

So from the information given in the base text, we have to:  

• $x_1 = 139000$

• $x_2 = 132000$

• $S_1 = 60000$

• $S_2 = 23000$

• $n_1 = 41$

• $n_2 = 53$

Are performing the calculations we have to:

$(139000 - 132000)/\sqrt{(60000^2)/41} + (23000^2)/53 \\= 7000/\sqrt{(87804878.049 + 9981132.075471697)} \\= 0.71$

See more about hypothesis at brainly.com/question/2695653