CareerBank.com annual salary survey. CareerBank.comconducts an annual salary survey of accounting, finance, and banking professionals. For one survey, data were collected for 2,800 responses submitted online by professionals across the country who voluntarily responded to CareerBank.com’s Web-based survey. Salary comparisons were made by gender, education, and marital status. Some of the results are shown in the accompanying table.

Males Females

Mean Salary \(69,848 \)52,012

Number of Respondents 1,400 1,400

a. Suppose you want to make an inference about the difference between the mean salaries of male and femaleaccounting/finance/banking professionals at a 95% level of confidence. Why is this impossible to do using the information in the table?

Short Answer

Expert verified

Due to distinct sample variances and dependent variables, it is impossible to draw any conclusions from the data in the table.

Step by step solution

01

Given information

The sample sizes and means for the two samples were provided.

\(\begin{aligned}{l}n1 &= 1,400\\n2 &= 1,400\end{aligned}\)

\(\begin{aligned}{l}\bar x1 &= 69,848\\\bar x2 &= 52,012\end{aligned}\)

02

Explaining the inference

The process of analyzing the outcome and drawing conclusions from data with random variation is known as statistical inference. Additionally known as inferential statistics. Applications of statistical inference include hypothesis testing and confidence intervals. Statistical inference is a technique for determining a population's characteristics based on a random sample.

Analyzing the correlation between the dependent and independent variables is helpful. Estimating uncertainty or sample-to-sample variation is the goal of statistical inference. It enables us to offer a likely range of values for an item's actual values in the population.

The following elements are included in statistical inference:

  • Actual relationship between X and Y
  • Independence
  • Normal
  • Equal variance
  • Random
03

Explaining the reason for not drawing inference

\(\begin{aligned}{l}n1 &= 1,400\\n2 &= 1,400\end{aligned}\)

\(\begin{aligned}{l}\bar x1 &= 69,848\\\bar x2 &= 52,012\end{aligned}\)

\(\begin{aligned}{l}{\sigma ^2}1 &= Variance\,of\,population1\\{\sigma ^2}2 &= \,Variance\,of\,population2\end{aligned}\)

The test hypothesis is as follows.

Denominator\({S^2}2\) and numerator\({S^2}1\) both

\(\begin{aligned}{l}df &= n1 - 1\\df = 1,400 - 1\\df &= 1,399\end{aligned}\)

\(v1 = n1 - 1\)

\(df = n2 - 1\)

\(v2 = n2 - 1\)

\(\begin{aligned}{l}df &= 1,400 - 1\\df &= 1,399\end{aligned}\)

Data\(\left( x \right)\)

\(xi - \bar x\)

\({\left( {xi - \bar x} \right)^2}\)

Data\(\left( y \right)\)

\(yi - \bar y\)

\({\left( {yi - \bar y} \right)^2}\)

1,400

-68,448

4,685,128,704

1,400

-50,612

2,561,574,544

Total:1,400

Total:4,685,128,704

Total:1,400

Total:-50,612

Total: 2,561,574,544

\(\begin{aligned}{l}S{1^2} &= \frac{{\sum {{\left( {xi - \bar x} \right)}^2}}}{{n - 1}}\\S{1^2} = \frac{{4,685,128,704}}{{1,399}}\\S{1^2} &= 3,348,912.5832737\end{aligned}\)

\(\begin{aligned}{l}S{2^2} &= \frac{{\sum {{\left( {yi - \bar y} \right)}^2}}}{{n - 1}}\\S{2^2} = \frac{{2,561,574,544}}{{1,399}}\\S{2^2} &= 1,831,003.9628305\end{aligned}\)

The test statistics will therefore be.

\(\begin{aligned}{l}F &= {\textstyle{{Larger\,\,sample\,\,variance} \over {Smaller\,\,sample\,\,variance}}}\\F &= \frac{{{S^2}1}}{{{S^2}2}}\\F &= \frac{{3,348,912.5832737}}{{1,831,003.9628305}}\\F &= 1.8290034599907\end{aligned}\)

Therefore, due to distinct sample variances and dependent variables, it is impossible to draw any conclusions from the data in the table.

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Most popular questions from this chapter

Question: Independent random samples selected from two normal populations produced the sample means and standard deviations shown below.

Sample 1

Sample 2

n1= 17x¯1= 5.4s1= 3.4

role="math" localid="1660287338175" n2= 12x¯2=7.9s2=4.8

a. Conduct the testH0:(μ1-μ2)>10against Ha:(μ1-μ2)≠10. Interpret the results.

b. Estimateμ1-μ2 using a 95% confidence interval

A random sample of n observations is selected from a normal population to test the null hypothesis that σ2=25. Specify the rejection region for each of the following combinations of Ha,αand n.

a.Ha:σ2≠25;α=0.5;n=16

b.Ha:σ2>25;α=.10;n=15

c.Ha:σ2>25;α=.01;n=23

d. Ha:σ2<25;α=.01;n=13

e. Ha:σ2≠25;α=.10;n=7

f. Ha:σ2<25;α=.05;n=25

Find the following probabilities for the standard normal random variable z:

a.P(0<z<2.25)b.P(-2.25<z<0)b.P(-2.25<z<1.25)d.P(-2.50<z<1.50)e.P(z<-2.33orz>2.33)

4.134 Refer to Exercise 4.133. Find the following probabilities:

a.P(20≤x≤30)b.P(20<x≤30)c.P(x≥30)d.P(x≥45)e.(x≤40)f.(x<40)g.P(15≤x≤35)h.P(21.5≤x≤31.5)

The data for a random sample of 10 paired observations is shown below.

PairSample from Population 1

(Observation 1)

Sample from Population 2 (Observation 2)
12345678910
19253152493459471751
24273653553466512055

a. If you wish to test whether these data are sufficient to indicate that the mean for population 2 is larger than that for population 1, what are the appropriate null and alternative hypotheses? Define any symbols you use.

b. Conduct the test, part a, usingα=.10.

c. Find a 90%confidence interval for μd. Interpret this result.

d. What assumptions are necessary to ensure the validity of this analysis?

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