As per the author, Hirschey, how are the following methods used to effectively answer the question, What do customers want?: consumer Interviews, surveys, market experiments and regression analysis.

A fast food company uses two management-training methods. Method 1 is a traditional method of training and Method 2 is a new and innovative method. The company has just hired 40 new management trainees. Nineteen of the trainees are randomly selected and assigned to the first method, and the remaining twenty-one trainees are assigned to the second training method. After three months of training, the management trainees took a standardized test. The test was designed to evaluate their performance and learning from training. The sample mean test score and sample standard deviation of the two methods are given below. The management wants to determine if the company should implement the new training method. Using α = .025, is the average test score for the traditional method significantly lower than the average test score for the innovative method? Do a complete and appropriate hypothesis test.

	Sample Mean	Sample standard deviation
Population 1: Method 1 (traditional)	83	2.7
Population 2: Method 2 (innovative)	81	3.3

H₀: (Click to select)p1π1σ1n1s1μ1x-bar1 (Click to select)=≠≤>≥< (Click to select)s2x-bar2π2n2p2σ2μ2

H_A: (Click to select)μ1x-bar1π1p1σ1s1n1 (Click to select)=≠≤>≥< (Click to select)μ2x-bar2π2p2σ2s2n2

Using only the appropriate statistical table in your textbook, the critical value for rejecting H₀ is (Click to select)+-± . (report your answer to 3 decimal places, using conventional rounding rules)

Using the sample data, the calculated value of the test statistic is (Click to select)+-± . (report your answer to 2 decimal places, using conventional rounding rules)

Should the null hypothesis be rejected? (Click to select)yes or no

Should the management conclude that the average test score for the traditional method is significantly lower than the average test score for the innovative method? yes or no

Using only the appropriate statistical table in your textbook, what is the most accurate statement you can make about the numerical value of the p-value of this hypothesis test?      

Answer: ____________________________________________________________________________ (provide a one-sentence statement about the p-value)

A treadmill manufacturer has developed a new machine with softer tread and better fans than its current model. The manufacturer believes these new features will enable runners to run for longer times than they can on its current machines. To determine whether the desired result is achieved, the manufacturer randomly sampled 10 runners. Each runner was measured for one week on the current treadmill and for one week on the new treadmill. The weekly total number of minutes for each runner on the two types of treadmills was collected, and is provided in the table below.

Runner	New treadmill (minutes run)	Current treadmill (minutes run)
1	269	270
2	280	268
3	260	254
4	271	256
5	273	258
6	264	264
7	263	262
8	260	251
9	263	264
10	274	258

Construct a 98% confidence interval estimate of the mean difference in running time (in minutes) on the new and current treadmills.

ANSWER:  (Click to select)-±+  ≤ (Click to select)(μ1 - μ2)(n1 - n2)(σ1 - σ2)μd(p1 - p2)(s1 - s2)(π1 - π2)(x-bar1 - x-bar2) ≤ (Click to select)±+-  (report your answers to 2 decimal places, using conventional rounding rules.)

Consider the following hypothesis test.

H₀: μ ≤ 12

H_a: μ > 12

A sample of 25 provided a sample mean

x = 14

and a sample standard deviation

s = 4.37.

(a)

Compute the value of the test statistic. (Round your answer to three decimal places.)

(b)

Use the t distribution table to compute a range for the p-value.

p-value > 0.200

0.100 < p-value < 0.200    

0.050 < p-value < 0.100

0.025 < p-value < 0.050

0.010 < p-value < 0.025

p-value < 0.010

(c)

At

α = 0.05,

what is your conclusion?

Reject H₀. There is sufficient evidence to conclude that μ > 12.

Do not reject H₀. There is sufficient evidence to conclude that μ > 12.    

Do not reject H₀. There is insufficient evidence to conclude that μ > 12.

Reject H₀. There is insufficient evidence to conclude that μ > 12.

(d)

What is the rejection rule using the critical value? (If the test is one-tailed, enter NONE for the unused tail. Round your answer to three decimal places.)

test statistic≤

test statistic≥

What is your conclusion?

Reject H₀. There is sufficient evidence to conclude that μ > 12.

Do not reject H₀. There is sufficient evidence to conclude that μ > 12.    

Do not reject H₀. There is insufficient evidence to conclude that μ > 12.

Reject H₀. There is insufficient evidence to conclude that μ > 12.

Question 7

Acid rain is an environmental challenge in many places around the world. It refers to rain or any other form of precipitation that is unusually acidic, i.e. rainwater having elevated levels of hydrogen ions (low pH). The measure of pH is a measure of the acidity or basicity of a solution and has a scale ranging from 0 to 14. Distilled water, with carbon dioxide removed, has a neutral pH level of 7. Liquids with a pH less than 7 are acidic. However, even unpolluted rainwater is slightly acidic with pH varying between 5.2 to 6.0 due to the fact that carbon dioxide and water in the air react together to form carbonic acid. Thus, rainwater is only considered acidic if the pH level is less than 5.2..

In a remote region of Algonquin Park, a biologist measured the pH levels of rainwater and obtained the following data for 16 rainwater samples on 16 different lakes:

4.73	4.79	4.87	4.88
5.04	5.06	5.07	5.09
5.11	5.16	5.18	5.21
5.23	5.24	5.25	5.25

Is there reason to believe that the is considered acidic (less than 5.20)?

Use the Sign Test to test the claim at 95% confidence that the rainwater from this region has a median pH level less than 5.20.

a) State the Hypotheses (1)

b) State the decision rule (1)

c) Determine the test statistic (1)

d) State your decision and interpretation. (1,1)

A company has developed a forecast using the Delphi method for January through June

1. Using the data provided, what is the Mean Absolute Deviation for the forecast method?

2. Using the data provided, what is the MAPD?

Month	Demand	Forecast
January	232	234
February	225	222
March	239	240
April	260	261
May	234	231
June	260	262

Assume Monthly Advertising Expenditure is the independent variable.

Store	Monthly Profit	Monthly Advertising expenditure
A1	$13,593.02	$1,710.00
A2	$23,896.73	$1,983.00
A3	$9,932.64	$952.00
A4	$9,100.41	$1,009.00
A5	$15,220.08	$2,315.00
A6	$33,900.67	$2,197.00
A7	$6,935.36	$934.00
A8	$10,112.62	$2,375.00
A9	$9,155.14	$1,065.00
A10	$8,513.94	$812.00
A11	$7,933.25	$1,052.00
A12	$11,388.13	$2,234.00
A13	$26,299.72	$2,962.00
A14	$21,423.87	$1,699.00
A15	$21,430.21	$1,991.00
A16	$19,984.96	$2,181.00
A17	$11,575.09	$1,831.00
A18	$18,900.44	$1,819.00
A19	$21,815.24	$2,394.00
A20	$35,642.73	$2,296.00

What is the sum of squares total value?

	Friendly	Cooperative	Impatient	Total
Male	674	166	40	800
Female	886	161	38	1085
Total	1560	327	78	1965

The table shows a cross tabulation of two variables, gender and the attitudes of the survey respondents towards the interviewers who collected data from them.

Can you conclude with at least 95% confidence that gender and respondent attitudes towards the interviewers are related? Explain why.

The mortgage foreclosure crisis that preceded the Great Recession impacted the U.S. economy in many ways, but it also impacted the foreclosure process itself as community activists better learned how to delay foreclosure, and lenders became more wary of filing faulty documentation. Suppose the duration of the eight most recent foreclosures filed in the city of Boston (from the beginning of foreclosure proceedings to the filing of the foreclosure deed, transferring the property) has been 230 days, 420 days, 340 days, 367 days, 295 days, 314 days, 385 days, and 311 days. Assume the duration is normally distributed.

Refer to Exhibit 8-2. Construct a 90% confidence interval for the mean duration of the foreclosure process in Boston.

[259.7400, 405.7600]

[293.2137, 372.2863]

[298.4296, 367.0704]

[303.2282, 362.2718]

Text book: Essentials of Biostatistics in Public Health 3 ed. by Lisa Sullivan.

Instructor provided problem: An observational study is conducted to compare experiences of men and women between the ages of 50-59 years following coronary artery bypass surgery. Participants undergo the surgery and are followed until the time of death, until they are lost to follow-up or up to 30 years, whichever comes first. The following table details the experiences of participating men and women. The data below are years of death or years of last contact for men and women. Estimate the Estimate the survival functions for each treatment group using the Kaplan-Meier approach. Test if there is a significant difference in survival between treatment groups using the log rank test and a 5% level of significance.

Men			Women
Year of Death	Year of Last Contact		Year of Death	Year of Last Contact
5	8		19	4
12	17		20	9
14	24		21	14
23	26		24	15
29	26			17
	27			19
	29			21
	30			22
	30			24
	30			25
				30

Beer bottles are filled so that they contain an average of 385 ml of beer in each bottle. Suppose that the amount of beer in a bottle is normally distributed with a standard deviation of 9 ml. Use Table 1.

a.	What is the probability that a randomly selected bottle will have less than 380 ml of beer?(Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)

Probability

b.	What is the probability that a randomly selected 7-pack of beer will have a mean amount less than 380 ml? (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)

Probability

c.	What is the probability that a randomly selected 24-pack of beer will have a mean amount less than 380 ml? (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)

Text book: Essentials of Biostatistics in Public Health 3rd ed. by Lisa Sullivan

Instructor provided question: An SBB instructor wants to determine if there is a correlation between the students incoming GPA and their overall score on their ASCP exam. Several potential confounders were also considered including years of clinical practice in blood bank and gap between completing the program and taking the exam. Perform the appropriate analysis on the data below and interpret the results. Do SBB ASCP scores correlate with student’s GPAs? Do confounders exist? Justify your response.

SBB ASCP Score	GPA	BB Experience (years)	Gap (months)
698	3.98	3	0.5
540	3.12	8	0.25
572	3.24	2	2
440	3.46	10	3
401	3.71	7	12
380	2.95	13	5
660	3.55	15	0.5
398	2.62	3	4
498	3.16	2	2
557	3.32	20	0.5
613	3.48	2	1
474	3.21	5	3
487	2.89	7	2
522	3.81	2	3
501	3.65	9	2
422	2.86	5	1
475	2.98	8	1
555	3.71	10	1
514	2.98	17	0.5
439	2.62	2	2

A​ government-sponsored organization that was established in 1938 after the Depression to provide local banks with money from the federal government to be used for residential mortgages in an effort to increase homeownership rates. As lending standards have tightened during the most recent housing market​ crisis, credit scores for borrowers of approved mortgages have increased. In​ 2012, the average credit score for loans that were purchased by the company was 782. A random sample of 25 mortgages recently purchased by the company was selected and it was found that the average credit score was 789 with a sample standard deviation of 25. Complete parts​ (a) through​ (c) below. a. Using alphaequals0.05​, is there enough evidence from this sample to conclude that the average credit score increased as the lending standards have tightened​? Determine the null and alternative hypotheses. Choose the correct answer below. A.Upper H 0​: mu not equals 782 and Upper H 1​: mu equals 782 B.Upper H 0​: mu greater than or equals 782 and Upper H 1​: mu less than 782 C.Upper H 0​: mu less than or equals 782 and Upper H 1​: mu greater than 782 D.Upper H 0​: mu equals 782 and Upper H 1​: mu not equals 782 Determine the critical​ value(s). The critical​ value(s) is(are) nothing. ​(Round to three decimal places as needed. Use a comma to separate answers as​ needed.) Determine the test​ statistic, t Subscript x overbar. t Subscript x overbarequals nothing ​(Round to two decimal places as​ needed.) What conclusion should be​ drawn? A.Do not reject Upper H 0. There is not sufficient evidence to conclude that the average credit score increased as the lending standards have tightened. B.Do not reject Upper H 0. There is sufficient evidence to conclude that the average credit score increased as the lending standards have tightened. C.Reject Upper H 0. There is sufficient evidence to conclude that the average credit score increased as the lending standards have tightened. D.Reject Upper H 0. There is not sufficient evidence to conclude that the average credit score increased as the lending standards have tightened. b. Use technology to determine the​ p-value for this test. ​p-valueequals nothing ​(Round to three decimal places as​ needed.) c. What assumptions need to be made to perform this​ analysis? A.That there was less than 30 users sampled. B.Because the sample size is less than​ 30, the population needs to be normally distributed. C.That the median was less than the mean value. D.No assumptions were needed to preform the analysis. Click to select your answer(s).

The accompanying data are the percentage of babies born prematurely in a particular year for the 50 U.S. states and the District of Columbia (DC). State Premature Percent State Premature Percent State Premature Percent Alabama 12.3 Kentucky 11.3 North Dakota 9.0 Alaska 9.1 Louisiana 12.9 Ohio 10.9 Arizona 9.6 Maine 9.0 Oklahoma 10.9 Arkansas 10.6 Maryland 10.7 Oregon 8.3 California 8.9 Massachusetts 9.2 Pennsylvania 10.0 Colorado 9.0 Michigan 10.4 Rhode Island 9.2 Connecticut 9.8 Minnesota 9.3 South Carolina 11.4 Delaware 9.9 Mississippi 13.5 South Dakota 9.1 DC 10.2 Missouri 10.4 Tennessee 11.4 Florida 10.5 Montana 9.9 Texas 11.0 Georgia 11.4 Nebraska 9.7 Utah 9.7 Hawaii 10.6 Nevada 10.7 Vermont 8.5 Idaho 8.8 New Hampshire 8.8 Virginia 9.8 Illinois 10.7 New Jersey 10.2 Washington 8.7 Indiana 10.3 New Mexico 9.8 West Virginia 11.4 Iowa 9.9 New York 9.5 Wisconsin 9.8 Kansas 9.3 North Carolina 10.3 Wyoming 11.8 (a) The smallest value in the data set is 8.3 (Oregon), and the largest value is 13.5 (Mississippi). Are these values outliers? Explain. Any observations smaller than 8.3 Incorrect: Your answer is incorrect. % or larger than 13.5 Incorrect: Your answer is incorrect. % are considered outliers. Therefore, Oregon's data value (8.3%) Correct: Your answer is correct. an outlier and Mississippi's data value (13.5%) Changed: Your submitted answer was incorrect. Your current answer has not been submitted. an outlier. (b) Construct a boxplot for this data set. Comment on the interesting features of the plot. The boxplot shows Incorrect: Your answer is incorrect. and the distribution is . The minimum value is %, the lower quartile is %, the median is %, the upper quartile is %, and the maximum value is %.

15. A Pew Poll in 2010, found that 50% of adults aged 25-29 had access to only a cell phone, while 40% had access to both a cellphone and landline and 10% had access to only a landline. We wish to conduct a Goodness of Fit Test to see if the results of a poll of a random sample of 130 adults aged 25-29 are significantly different from the 2010 poll results.

The random sample of 130 adults found that 69 had access to only a cell phone, 39 had access to only a landline, and 22 had access to both types of phones.

Which of the following are the correct null and alternative hypothesis for this test?

	A.	Ho: p = 0.50 Ha: p =/= 0.50
	B.	Ho: The distribution is the same as in 2010. Ha: The distribution is different than in 2010.
	C.	Ho: The distribution is different than in 2010. Ha: The distribution is the same as in 2010.
	D.	Ho: pcell=0.33 , pboth=0.33, pland=0.33 Ha: pcell=0.50 , pboth=0.40, pland=0.10

16. A Pew Poll in 2010, found that 50% of adults aged 25-29 had access to only a cell phone, while 40% had access to both a cellphone and landline and 10% had access to only a landline. We wish to conduct a Goodness of Fit Test to see if the results of a poll of a random sample of 130 adults aged 25-29 are significantly different from the 2010 poll results. The table below presents the observed results.

	Cell only	Landline Only	Both
Observed	69	39	22
Expected

Compute the expected count for the "BOTH" cell.

17. A Pew Poll in 2010, found that 50% of adults aged 25-29 had access to only a cell phone, while 40% had access to both a cellphone and landline and 10% had access to only a landline. We wish to conduct a Goodness of Fit Test to see if the results of a poll of a random sample of 130 adults aged 25-29 are significantly different from the 2010 poll results.

The random sample of 130 adults found that 69 had access to only a cell phone, 39 had access to only a landline, and 22 had access to both types of phones.

The chi-square statistic was calculated as 9.73 and has a p-value of 0.0077.Using α=0.01 what can you conclude?

	A.	There is enough evidence that the distribution is different than in 2010.
	B.	There is enough evidence that the distribution is the same as in 2010.
	C.	There is not enough evidence that the distribution is different than in 2010.
	D.	There is not enough evidence the distribution is the same as in 2010.

18. According to the CDC 2.8% of high school students currently use electronic cigarettes. A high school counselor is concerned that the use of e-cigs at her school is higher.

A test of the following hypothesis Ho: p = 0.028 Ha: p > 0.028 is conducted using a random sample of 50 students at this school and the null hypothesis is rejected. What conclusion can the counselor make?

	A.	There is not enough evidence that the percent of students using e-cigs is higher than2.8%.
	B.	There is not enough evidence that the percent of students using e-cigs is 2.8%.
	C.	There is enough evidence that the percent of students using e-cigs is higher than2.8%.
	D.	There is enough evidence that the percent of students using e-cigs is 2.8%.