Bachelor’s Degree Recipients: How to Find the Right Program Free Essay Example

The Wall Street Journal reported that bachelor’s degree recipients with majors in business average starting salaries of $53,900 in 2012 ( The Wall Street Journal , March 17, 2014). The results for a sample of 100 business majors receiving a bachelor’s degree in 2013 showed a mean starting salary of $55,144 with a sample standard deviation of $5,200. Conduct a hypothesis test to determine whether the mean starting salary for business majors in 2013 is greater than the mean starting salary in 2012. Use a = .01 as the level of significance.

Solution

We are given the following:

Hypothesis:

Null Hypothesis: H o : There is no difference between the means

Alternative Hypothesis: H a : The mean starting salary for business majors in 2013 is greater than the mean starting salary in 2012.

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Mathematically;

Test-Statistic:

The formula for calculating t-statistic is:

By using the p-value calculator from t-statistic located at Socscistatistics, we can determine the P-value for the calculated t-statistic. Alternatively, we can use the t-statistic table to determine this value. The p-value is given below:

Decision Rule:

Here,

Since the p value is less than level of significance, we reject the null hypothesis (H o ).

Conclusion:

The mean starting salary for business majors in 2013 is greater than the mean starting salary in 2012.

In Born Together – Reared Apart: The Landmark Minnesota Twin Study (2012), Nancy Segal discusses the efforts of research psychologists at the University of Minnesota to understand similarities and differences between twins by studying sets of twins who were raised separately. Below are critical reading SAT scores for several pairs of identical twins (twins who share all of their genes), one of whom was raised in a family with no other children (no siblings) and one of whom was raised in a family with other children (with siblings).

What is the mean difference between the critical reading SAT scores for the twins raised with no siblings and the twins raised with siblings?

No Siblings		With Siblings
Name	SAT Score	Name	SAT Score	di	di^2
Bob	440	Donald	420	20	400
Matthew	610	Ronald	540	70	4900
Shannon	590	Kedriana	630	-40	1600
Tyler	390	Kevin	430	-40	1600
Michelle	410	Erin	460	-50	2500
Darius	430	Michael	490	-60	3600
Wilhelmina	510	Josephine	460	50	2500
Donna	620	Jasmine	540	80	6400
Drew	510	Kraig	460	50	2500
Lucinda	680	Bernadette	650	30	900
Barry	580	Larry	450	130	16900
Julie	610	Jennifer	640	-30	900
Hannah	510	Diedra	460	50	2500
Roger	630	Latishia	580	50	2500
Garrett	570	Bart	490	80	6400
Roger	630	Kara	640	-10	100
Nancy	530	Rachel	560	-30	900
Sam	590	Joey	610	-20	400
Simon	500	Drew	520	-20	400
Megan	610	Annie	640	-30	900
Total	10950		10670	280	58800
Mean	547.5		533.5		3094.737
Standard Deviation	80.37		78.06		55.63036

Alternatively,

The mean difference can be calculated by dividing the sum of differences between the sample by n, i.e.

Provide a 90% confidence interval estimate of the mean difference between the critical reading SAT scores for the twins raised with no siblings and the twins raised with siblings

Using the t-distribution table,

The formula for calculating the confidence interval (CI) is:

Conduct a hypothesis test of equality of the critical reading SAT scores for the twins raised with no siblings and the twins raised with siblings at a = .01. What is your conclusion?

Using the t table,

Thus, the result is not significant.

Conclusion:

There is no evidence that the SAT scores for the twins raised with no siblings is different from twins raised with siblings.

No Siblings		With Siblings
Name	SAT Score	Name	SAT Score
Bob	440	Donald	420
Matthew	610	Ronald	540
Shannon	590	Kedriana	630
Tyler	390	Kevin	430
Michelle	410	Erin	460
Darius	430	Michael	490
Wilhelmina	510	Josephine	460
Donna	620	Jasmine	540
Drew	510	Kraig	460
Lucinda	680	Bernadette	650
Barry	580	Larry	450
Julie	610	Jennifer	640
Hannah	510	Diedra	460
Roger	630	Latishia	580
Garrett	570	Bart	490
Roger	630	Kara	640
Nancy	530	Rachel	560
Sam	590	Joey	610
Simon	500	Drew	520
Megan	610	Annie	640

Part variability is critical in the manufacturing of ball bearings. Large variances in the size of the ball bearings cause bearing failure and rapid wearout. Production standards call for a maximum variance of .0001 when the bearing sizes are measured in inches. A sample of 15 bearings shows a sample standard deviation of .014 inches.

Use a = .10 to determine whether the sample indicates that the maximum acceptable variance is being exceeded.

Solution:

We are given the following:

Hypothesis Testing:

The result is not significant. Thus, we reject the null hypothesis.

Conclusion:

The sample indicates that the maximum acceptable variance is being exceeded.

Compute the 90% confidence interval estimate of the variance of the ball bearings in the population.

Solution:

We are given the following:

The CI can be calculated using the following formula:

Using the z table,

Thus,

The Transactional Records Access Clearinghouse at Syracuse University reported data showing the odds of an Internal Revenue Service audit. The following table shows the average adjusted gross income reported and the percent of the returns that were audited for 20 selected IRS districts.

Develop the estimated regression equation that could be used to predict the percent audited given the average adjusted gross income reported.

Solution:

Microsoft Excel was used to carry out the regression analysis. The table below shows the regression output.

SUMMARY OUTPUT

Regression Statistics
Multiple R	0.465892
R Square	0.217056
Adjusted R Square	0.173559
Standard Error	0.208768
Observations	20

ANOVA
	df	SS	MS	F	Significance F
Regression	1	0.21749	0.21749	4.990136	0.038419
Residual	18	0.78451	0.043584
Total	19	1.002

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	-0.47095	0.584247	-0.80609	0.430714	-1.69841	0.756503	-1.69841	0.7565034
X Variable 1	3.87E-05	1.73E-05	2.233861	0.038419	2.3E-06	7.51E-05	2.3E-06	7.50539E-05

From the regression output,

Thus, the regression equation is:

Here,

At the .05 level of significance, determine whether the adjusted gross income and the percent audited are related.

From the regression output,

From the F-distribution table,

The result is significant.

Thus, the adjusted gross income and the percent audited are related.

Did the estimated regression equation provide a good fit? Explain.

No. The estimated regression equation did not provide a good fit. This is because the r-square value for the regression model is not close to 1.

Use the estimated regression equation developed in part (a) to calculate a 95% confidence interval for the expected percent audited for districts with an average adjusted gross income of $35,000.

From the regression output,

From the t-distribution table,

Thus,

District	Adjusted Gross Income ($)	Percent Audited
Los Angeles	36,664	1.3
Sacramento	38,845	1.1
Atlanta	34,886	1.1
Boise	32,512	1.1
Dallas	34,531	1.0
Providence	35,995	1.0
San Jose	37,799	0.9
Cheyenne	33,876	0.9
Fargo	30,513	0.9
New Orleans	30,174	0.9
Oklahoma City	30,060	0.8
Houston	37,153	0.8
Portland	34,918	0.7
Phoenix	33,291	0.7
Augusta	31,504	0.7
Albuquerque	29,199	0.6
Greensboro	33,072	0.6
Columbia	30,859	0.5
Nashville	32,566	0.5
Buffalo	34,296	0.5

The Tire Rack, America’s leading online distributor of tires and wheels, conducts extensive testing to provide customers with products that are right for their vehicle, driving style, and driving conditions. In addition, the Tire Rack maintains and independent consumer survey to help drivers help each other by sharing their long-term driving experiences. The following data show survey ratings (1 to 10 scale with 10 the highest rating) for 18 maximum performance summer tires. The variable Steering rates the tire’s steering responsiveness, Tread Wear rates quickness of wear based on the driver’s expectations, and Buy Again rates the driver’s overall tire satisfaction and desire to purchase the same tire again.

Develop the estimated regression equation that can be used to predict the Buy Again rating given based on the Steering rating. At the .05 level of significance, test for a significant relationship.

SUMMARY OUTPUT

Regression Statistics
Multiple R	0.918166
R Square	0.843029
Adjusted R Square	0.833218
Standard Error	0.841071
Observations	18

ANOVA
	df	SS	MS	F	Significance F
Regression	1	60.78659	60.78659	85.9295	7.8E-08
Residual	16	11.31841	0.707401
Total	17	72.105

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	-7.52183	1.466759	-5.1282	0.000101	-10.6312	-4.41244	-10.6312	-4.41244
X Variable 1	1.815067	0.195804	9.269817	7.8E-08	1.399981	2.230153	1.399981	2.230153

From the regression output,

Thus,

Did the estimated regression equation developed in part (a) provide a good fit to the data? Explain.

Yes. The estimated regression equation did provide a good fit. This is because the r-square value for the regression model is very close to 1.

Develop an estimated regression equation that can be used to predict the Buy Again rating given the Steering rating and the Tread Wear rating.

SUMMARY OUTPUT

Regression Statistics

Multiple R	0.965279
R Square	0.931764
Adjusted R Square	0.922665
Standard Error	0.572723
Observations	18

ANOVA
	df	SS	MS	F	Significance F
Regression	2	67.18482	33.59241	102.4121	1.8E-09
Residual	15	4.920181	0.328012
Total	17	72.105

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	-5.38767	1.109533	-4.8558	0.00021	-7.75259	-3.02276	-7.75259	-3.02276
X Variable 1	0.689871	0.287547	2.399155	0.029874	0.076978	1.302764	0.076978	1.302764
X Variable 2	0.91133	0.206343	4.416569	0.0005	0.47152	1.351141	0.47152	1.351141

From the regression output,

Thus,

Is the addition of the Tread Wear independent variable significant? Use a = .05.

From the regression output,

Thus, there is sufficient evidence to support the claim that the addition of the Tread Wear independent variable is significant.

Tire	Steering	Tread Wear	Buy Again
Goodyear Assurance Triple Tred	8.9	8.5	8.1
Michelin HydroEdge	8.9	9.0	8.3
Michelin Harmony	8.3	8.8	8.2
Dunlop SP60	8.2	8.5	7.9
Goodyear Assurance ComforTred	7.9	7.7	7.1
Yokohama Y372	8.4	8.2	8.9
Yokohama Aegis LS4	7.9	7.0	7.1
Kumho Power Star 758	7.9	7.9	8.3
Goodyear Assurance	7.6	5.8	4.5
Hankook H406	7.8	6.8	6.2
Michelin Energy LX4	7.4	5.7	4.8
Michelin MX4	7.0	6.5	5.3
Michelin Symmetry	6.9	5.7	4.2
Kumho 722	7.2	6.6	5.0
Dunlop SP 40 A/S	6.2	4.2	3.4
Bridgestone Insignia SE200	5.7	5.5	3.6
Goodyear Integrity	5.7	5.4	2.9
Dunlop SP20 FE	5.7	5.0	3.3

The following data show the daily closing prices (in dollars per share) for a stock.

Define the independent variable Period, where Period = 1 corresponds to the data for November 3, Period = 2 corresponds to the data for November 4, and so on. Develop the estimated regression equation that can be used to predict the closing price given the value of the Period.

SUMMARY OUTPUT

Regression Statistics
Multiple R	0.970858
R Square	0.942564
Adjusted R Square	0.939373
Standard Error	0.603819
Observations	20

ANOVA
	df	SS	MS	F	Significance F
Regression	1	107.6999	107.6999	295.3942	1.3E-12
Residual	18	6.562754	0.364597
Total	19	114.2627

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	81.98942	0.280493	292.3048	1.51E-34	81.40013	82.57871	81.40013	82.57871
X Variable 1	0.402436	0.023415	17.18703	1.3E-12	0.353243	0.451629	0.353243	0.451629

From the regression output,

Thus,

At the .05 level of significance, test for any positive autocorrelation in the data.

From the regression output,

Thus,

Positive autocorrelation is present.

Date	Price ($)
Nov. 3	82.87
Nov. 4	83.00
Nov. 7	83.61
Nov. 8	83.15
Nov. 9	82.84
Nov. 10	83.99
Nov. 11	84.55
Nov. 14	84.36
Nov. 15	85.53
Nov. 16	86.54
Nov. 17	86.89
Nov. 18	87.77
Nov. 21	87.29
Nov. 22	87.99
Nov. 23	88.80
Nov. 25	88.80
Nov. 28	89.11
Nov. 29	89.10
Nov. 30	88.90
Dec. 1	89.21

Bachelor’s Degree Recipients: How to Find the Right Program

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