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

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 $95\mathrm{%}$ confidence interval estimate for the mean salary of all female graduates You can still ask an expert for help Expert Community at Your Service • Live experts 24/7 • Questions are typically answered in as fast as 30 minutes • Personalized clear answers Solve your problem for the price of one coffee • Available 24/7 • Math expert for every subject • Pay only if we can solve it ## Expert Answer tafzijdeq Answered 2021-03-07 Author has 92 answers a) Mean Salary within 10,000 means within $40000\text{/}4=\sigma \text{/}4$ $=0.25\sigma$ Lets assume the Standard normal where $\mu =0$ $\sigma =1$ To calculate probability for, $P\left(-0.25 $=P\left(\left(\frac{-0.25-0}{1} $=P\left(Z<0.25\right)-P\left(Z<-0.25\right)$ $=0.5987-0.4013$ $=0.1974$ 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 $25000\text{/}2.5=\sigma \text{/}2.5$
$=0.4\sigma$
Lets assume the Standard normal where
$\mu =0$
$\sigma =1$
To calculate probability for,
$P\left(-0.4
$=P\left(\left(\frac{-0.4-0}{1}\right)
$=P\left(Z<0.4\right)-P\left(Z<-0.4\right)$
$=0.6554-0.3446$
$=0.3108$
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,
$DF=n-1=39$
The value of 't-value' is,
${t}_{a\text{/}2}={t}_{5\text{/}2}$
$={t}_{2.5}$
$=2.0227$
Standard Error of the given data is,
$SE=\frac{\sigma }{\sqrt{n}}$
$=\frac{25000}{\sqrt{40}}$
$=3952.8471$
Margin of error of the given data is,
$E=t×SE$
$=2.0227×3952.84708$
$=7995.3879$
Confidence interval is
Interval Lower Limit $=\stackrel{―}{x}-E=125000-7995.387887=117004.612$
Interval Upper Limit $=\stackrel{―}{x}+E=125000-7995.387887=132995.388$
$95\mathrm{%}$ confidence interval is $\left(117005<\mu <132995\right)$

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