perform Levene’s test for equal variance. Note, this is a one‐way ANOVA testing for the equality of 16 variances (each combination of promotion/discount). 0.1 signigicance

2. perform 2-way anova with replication

To answer these questions, an experiment was designed using laundry detergent pods. For ten weeks, 160 subjects received information about the products. The factors under consideration were the number of promotions (1, 3, 5, or 7) that were described during this ten‐ week period and the percent that the product was discounted (10%, 20%, 30%, or 40%) off the average non‐promotional price. Ten individuals were randomly assigned to each of the sixteen combinations. The data reflecting what the sub‐ jects would expect to pay for the product (i.e., their reference price) at the end of the 10‐week period. 0.1 significance

Stop 'N Shop Reference Pricing Data					Note: the headings reflect the number of promotions and the percent discount										Stop 'N Shop Case.xlsx
					(for example: N5D30 represents 5 promotions with a 30 percent discount).
N1D10	N3D10	N5D10	N7D10	N1D20	N3D20	N5D20	N7D20	N1D30	N3D30	N5D30	N7D30	N1D40	N3D40	N5D40	N7D40
11.36	11.33	11.15	10.82	10.83	11.46	11.16	10.71	12.20	12.14	11.37	11.15	12.45	12.16	11.57	11.30
11.76	11.39	11.44	11.17	11.03	11.20	11.03	11.32	11.85	12.06	11.61	11.71	12.14	12.41	11.62	11.48
11.73	11.51	11.08	11.31	11.16	11.46	11.12	10.61	11.84	11.72	11.43	11.06	12.04	11.94	12.01	11.65
11.68	11.49	11.35	11.17	11.75	11.14	11.36	10.93	11.74	11.99	11.37	11.41	12.15	12.24	11.88	11.15
11.82	11.83	11.20	11.37	11.26	11.61	11.36	11.00	11.81	11.22	11.28	11.67	11.95	11.92	11.00	11.52
11.95	11.59	11.67	10.87	11.92	11.25	11.07	11.06	11.79	11.68	11.67	11.01	12.22	11.72	11.60	11.67
11.68	11.43	11.40	10.98	11.74	11.27	11.23	11.16	11.85	11.56	11.74	11.24	12.26	11.96	11.78	11.65
11.43	11.73	11.41	10.95	11.90	11.48	10.93	11.34	11.92	11.94	11.02	11.33	12.19	11.63	11.63	11.78
11.57	11.86	11.32	11.05	11.57	10.96	11.31	10.78	11.99	11.71	11.92	11.47	12.36	11.95	11.66	11.13
11.85	11.28	11.16	10.71	11.69	11.74	10.93	11.29	12.50	11.82	11.70	11.49	12.04	12.23	11.78	11.96

To characterize random uncertainty of a pressure measuring technique, twelve pressure measurements were made of a certain constant pressure source, giving the following results in kPa: 125, 128, 129, 122, 126, 125, 125, 130, 126, 127, 124, and 123. (a) Estimate the 95% confidence interval of the next measurement obtained with this technique; (b) Estimate the 95% confidence interval of the average of next 5 measurements obtained with this technique; (c) Using only the 12 measurements available, how would you report the value of measured pressure, including the 95% confidence interval?

Consider a Bernoulli random variable X such that P(X=1) = p. Calculate the following and show steps of your work:

a) E[X]

b) E[X²]

c) Var[X]

d) E[(1 – X)¹⁰]

e) E[(X – p)⁴]

f) E[3^x4^1-x]

g) var[3^x4^1-x]

1-what is the dummy variable and what is the purpose of included in regression model ?

2- explaining the meaning of adjust r square?

3-if adjust r square computed. would it be higher , equal of lower than value of r square?

The U.S. Bureau of Economic Statistics reports that the average annual salary in the metropolitan Boston area is $50,542. Suppose annual salaries in the metropolitan Boston area are normally distributed with a standard deviation of $4,246. A Boston worker is randomly selected. (Round the values of z to 2 decimal places. Round your answers to 4 decimal places.)

(a) What is the probability that the worker’s annual salary is more than $60,000?

(b) What is the probability that the worker’s annual salary is less than $42,000?

(c) What is the probability that the worker’s annual salary is more than $39,000?

(d) What is the probability that the worker’s annual salary is between $43,000 and $51,000?

A credit reporting agency claims that the mean credit card debt in a town is greater than $3500. A random sample of the credit card debt of 20 residents in that town has a mean credit card debt of $3619 and a standard deviation of $391. At α=0.10, can the credit agency’s claim be supported?

The following table shows ceremonial ranking and type of pottery sherd for a random sample of 434 sherds at an archaeological location.

Ceremonial Ranking	Cooking Jar Sherds	Decorated Jar Sherds (Noncooking)	Row Total
A	91	44	135
B	96	49	145
C	81	73	154
Column Total	268	166	434

Use a chi-square test to determine if ceremonial ranking and pottery type are independent at the 0.05 level of significance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: Ceremonial ranking and pottery type are not independent.
H₁: Ceremonial ranking and pottery type are not independent.

H₀: Ceremonial ranking and pottery type are independent.
H₁: Ceremonial ranking and pottery type are not independent.    

H₀: Ceremonial ranking and pottery type are not independent.
H₁: Ceremonial ranking and pottery type are independent.

H₀: Ceremonial ranking and pottery type are independent.
H₁: Ceremonial ranking and pottery type are independent.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least three decimal places. Round the test statistic to three decimal places.)


Are all the expected frequencies greater than 5?

Yes

No    

What sampling distribution will you use?

Student's t

uniform    

normal

chi-square

binomial

What are the degrees of freedom?


(c) Find or estimate the P-value of the sample test statistic. (Round your answer to three decimal places.)

p-value > 0.100

0.050 < p-value < 0.100    

0.025 < p-value < 0.050

0.010 < p-value < 0.025

0.005 < p-value < 0.010

p-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

Since the P-value > α, we fail to reject the null hypothesis.

Since the P-value > α, we reject the null hypothesis.   

Since the P-value ≤ α, we reject the null hypothesis.

Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is sufficient evidence to conclude that ceremonial ranking and pottery type are not independent.

At the 5% level of significance, there is insufficient evidence to conclude that ceremonial ranking and pottery type are not independent.    

Health self-evaluation as an indicator of general health was investigated in two groups of older adults. Older adults in one group were dog owners while in the second group older adults did not own dogs. Follow the 8-step hypothesis testing, using the information given below to determine if the two groups of older adults had different health self-evaluation scores. Use alpha level of .05. For all calculations, please keep two decimals.

Group	Sample Size (n)	Mean Health Evaluation Score	Std. Deviation
Dog Owners	18	62.5	6.5
Not a dog owner	17	58	5.7

It is known that the length of a certain product X is normally distributed with μ = 37 inches. How is the probability P(X > 33) related to P(X < 33)?

Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an advance indication of illness such as gout, leukemia, or lymphoma.† Over a period of months, an adult male patient has taken fifteen blood tests for uric acid. The mean concentration was x = 5.35 mg/dl. The distribution of uric acid in healthy adult males can be assumed to be normal, with σ = 1.91 mg/dl. (a) Find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. What is the margin of error? (Round your answers to two decimal places.) lower limit upper limit margin of error

Problem 1: Oil Production Data: The Data in the following are the annual world crude oil production in millions of barrels for the period 1880-1988. The data are taken from Moore and McCabe( 1993, p. 147).

Here is the code help you to paste the data into your R.

data5<-'year barrels
1880 30
1890 77
1900 149
1905 215
1910 328
1915 432
1920 689
1925 1069
1930 1412
1935 1655
1940 2150
1945 2595
1950 3803
1955 5626
1960 7674
1962 8882
1964 10310
1966 12016
1968 14104
1970 16690
1972 18584
1974 20389
1976 20188
1978 21922
1980 21722
1982 19411
1984 19837
1986 20246
1988 21388
'
data5n<-read.table(textConnection(object=data5),
header=TRUE,
sep="",
stringsAsFactors = FALSE)

1. Construct a scatter plot of the oil production variable (OIL) versus Year and observe that the scatter of points on the graph is not linear. In order to fit a linear model to these data, OIL must be transformed.
2. Construct a scatter plot of log(OIL) versus Year. The scatter of points now follows a straight line from 1880 to 1973. Political turmoil in the oil-producing regions of the Middle East affected patterns of oil production after 1973.

You have taken a random sample of 286 children in Virginia’s homeless shelters to establish their needs (using an establish Needs Scale). The mean is 5.32, the mode is 6, and the median is 6.
A. Provide a complete description of this sample (including shape and appropriate measures of center and spread).
B. Why did you choose the measure of center detailed above?
C. Regardless of 10b, why is a confidence interval of the mean a valid estimate of the mean?
D. Calculate and interpret a 95% confidence interval for the mean.
E. What is the population for this analysis?

A researcher wants to determine the impact of soil type on the growth of a certain type of plant. She grows three plants in each of four different types of soil and measures the growth in inches for each plant after one month resulting in the data below.

Inches	Soil
12.2	1
12.8	1
11.9	1
10.8	2
12.2	2
12.3	2
9.3	3
9.9	3
10.8	3
13	4
11.8	4
11.9	4

a) What null hypothesis is the researcher testing if she runs an ANOVA with this data?

The mean growth of the plant is different in each type of soil.

The variability in growth of the plant in each type of soil is the same.    

The mean growth of the plant in each type of soil is the same.

One type of soil has a higher mean growth for the plant than the others.

Soil 3 provides a lower mean growth for the plant than the other types of soil.

b) What is the SS_trt for the ANOVA? Give your answer to at least three decimal places.  

c) What is DF_err for the ANOVA?  

d) What is the value of the F statistic for the ANOVA? Give your answer to at least three decimal places.  
e) Using a 0.1 level of significance, what conclusion should the researcher reach?

There is not enough evidence to reject the claim that the mean growth of the plant is the same in each type of soil.

Soil 3 has a lower mean growth for the plant than the other types of soil.    

Soil 1 has a higher mean growth for the plant than the other types of soil.

The mean growth of the plant is not the same for all soil types.

A marketing research firm wishes to study the relationship between wine consumption and whether a person likes to watch professional tennis on television. One hundred randomly selected people are asked whether they drink wine and whether they watch tennis. The following results are obtained:

	Watch Tennis	Do Not Watch Tennis	Totals
Drink Wine	10	36	46
Do Not Drink Wine	10	44	54
Totals	20	80	100

(a) For each row and column total, calculate the corresponding row or column percentage.

Row 1	%
Row 2	%
Column 1	%
Column 2	%

  
(b) For each cell, calculate the corresponding cell, row, and column percentages. (Round your answers to the nearest whole number.)

	Watch Tennis	Do Not Watch Tennis
Drink Wine	Cell= %	Cell= %
	Row= %	Row= %
	Column= %	Column= %
Do Not Drink Wine	Cell= %	Cell= %
	Row= %	Row= %
	Column= %	Column= %

(c) Test the hypothesis that whether people drink wine is independent of whether people watch tennis. Set α = .05. (Round your answer to 3 decimal places.)

χ2χ2 =          

An advertising executive claims that there is a difference in the mean household income for credit cardholders of Visa Gold and of MasterCard Gold. A random survey of 8 Visa Gold cardholders resulted in a mean household income of $78,320 with a standard deviation of $11,100. A random survey of 12 MasterCard Gold cardholders resulted in a mean household income of $68,070 with a standard deviation of $10,700. Is there enough evidence to support the executive's claim? Let μ1 be the true mean household income for Visa Gold cardholders and μ2 be the true mean household income for MasterCard Gold cardholders. Use a significance level of α=0.05α=0.05 for the test. Assume that the population variances are not equal and that the two populations are normally distributed.

Step 1 of 4: State the null and alternative hypotheses for the test.

Step 2 of 4: Compute the value of the t test statistic. Round your answer to three decimal places.

Step 3 of 4: Determine the decision rule for rejecting the null hypothesis H0H0. Round your answer to three decimal places.

Step 4 of 4: State the test's conclusion.

....reject null hypothesis...fail to reject null hypothesis