A researcher randomly sampled 30 graduates of an MBA program and recorded data c
ID: 3156145 • Letter: A
Question
A researcher randomly sampled 30 graduates of an MBA program and recorded data concerning their starting salaries. Of primary interest to the researcher was the effect of gender on starting salaries. Analysis of the mean salaries of the females and males in the sample is given below. Suppose the researcher was attempting to show statistically that the female MBA graduates have a significantly lower mean starting salary than the male MBA graduates.
Hypothesized Difference
0
Level of Significance
0.05
Population 1 Sample
Sample Size
18
Sample Mean
48266.7
Sample Standard Deviation
13577.63
Population 2 Sample
Sample Size
12
Sample Mean
55000
Sample Standard Deviation
11741.29
Difference in Sample Means
-6733.3
t-Test Statistic
-1.40193
Lower-Tail Test
Lower Critical Value
-1.70113
p-Value
0.085962
a) According to the test run, state an appropriate alternative hypothesis.
b) From the analysis in Table 10-2, what is the correct test statistic? Explain how you obtain your answer.
c) What is the 95% confidence interval estimate for the difference between two means? Show how you obtain your answer.
d) What conclusion do you reach for this test? Explain how you obtain your answer.
Hypothesized Difference
0
Level of Significance
0.05
Population 1 Sample
Sample Size
18
Sample Mean
48266.7
Sample Standard Deviation
13577.63
Population 2 Sample
Sample Size
12
Sample Mean
55000
Sample Standard Deviation
11741.29
Difference in Sample Means
-6733.3
t-Test Statistic
-1.40193
Lower-Tail Test
Lower Critical Value
-1.70113
p-Value
0.085962
Explanation / Answer
A)
Ha: u1 - u2 < 0 [ANSWER]
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b)
From the table, under "t test statistic",
t = -1.40193 [ANSWER]
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c)
Calculating the means of each group,
X1 = 48266.7
X2 = 55000
Calculating the standard deviations of each group,
s1 = 13577.63
s2 = 11741.29
Thus, the pooled standard deviation is given by
S = sqrt[((n1 - 1)s1^2 + (n2 - 1)(s2^2))/(n1 + n2 - 2)]
As n1 = 18 , n2 = 12
Then
S = 12887.45433
Thus, the standard error of the difference is
Sd = S sqrt (1/n1 + 1/n2) = 4802.870658
df = n1 + n2 - 2 = 28
For the 0.95 confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.025
t(alpha/2) = 2.048407142
lower bound = [X1 - X2] - t(alpha/2) * Sd = -19104.67498
upper bound = [X1 - X2] + t(alpha/2) * Sd = 5638.074983
Thus, the confidence interval is
( -19104.67498 , 5638.074983 ) [ANSWER]
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d)
As P = 0.085962 > 0.05, we fail to reject Ho.
There is no significant evidence at 0.05 level that female MBA graduates have a significantly lower mean starting salary than the male MBA graduates. [CONCLUSION]
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