Based on a sample of 80 recent Masters graduates (40 male and 40 female), the following information was made available regarding their annual salaries

Tyra 2021-03-06 Answered
Based on a sample of 80 recent Masters graduates (40 male and 40 female), the following information was made available regarding their annual salaries. The standard deviation of salaries for the male graduates was $40,000 and that for the female graduates was $25,000.
a) For the male graduates, what is the probability of obtaining a sample mean salary within $10,000 of the population mean?
b) Consider the same question in (a) but for the female graduates. In which case, males or females, do we have a higher probability of obtaining a sample estimate within $10,000 of the population mean? Why?
c) Suppose that the sample mean salary of females is $125,000. Develop a 95% confidence interval estimate for the mean salary of all female graduates
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Answered 2021-03-07 Author has 92 answers

a)
Mean Salary within 10,000 means within 40000/4=σ/4
=0.25σ
Lets assume the Standard normal where
μ=0
σ=1
To calculate probability for,
P(0.25<X<0.25)=P(0.25<Z<0.25)
=P((0.2501<Z<(0.2501))
=P(Z<0.25)P(Z<0.25)
=0.59870.4013
=0.1974
The probability of obtaining a sample mean salary within $10,000 of the population mean is 0.1974.
b)
Mean Salary within 10,000 means within 25000/2.5=σ/2.5
=0.4σ
Lets assume the Standard normal where
μ=0
σ=1
To calculate probability for,
P(0.4<X<0.4)=P(0.4<Z<0.4)
=P((0.401)<Z<(0.401))
=P(Z<0.4)P(Z<0.4)
=0.65540.3446
=0.3108
The probability of obtaining a sample mean salary within $10,000 of the population mean is 0.3108.
Females will have a higher probability of obtaining a sample estimate within $10,000
c)
Level of Significance ,a=0.05
Degree of freedom,
DF=n1=39
The value of 't-value' is,
ta/2=t5/2
=t2.5
=2.0227
Standard Error of the given data is,
SE=σn
=2500040
=3952.8471
Margin of error of the given data is,
E=t×SE
=2.0227×3952.84708
=7995.3879
Confidence interval is
Interval Lower Limit =xE=1250007995.387887=117004.612
Interval Upper Limit =x+E=1250007995.387887=132995.388
95% confidence interval is (117005<μ<132995)

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