 Problem #351 Get a video solution for your math problem!Problem text: Problem #2
What is the critical F value for a sample of four observations in the numerator and seven in the denominator? Use a onetailed test and the .01 significance level.
Answer price: Free! df (num) = 41=3
df (den) = 71=6
alpha = 0.01, look these numbers up in the F distribution table in the balck of the book to get a critical value of 9.78 
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Problem #352 Get a video solution for your math problem!Problem text: Problem #47
The null hypothesis and the alternate are:
Ho: The cell categories are equal.
H1: The cell categories are not equal.
Category fo
A 10
B 20
C 30
a. State the decision rule, using the .05 significance level.
b. Compute the value of chisquare.
c. What is your decision regarding Ho?
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Problem #353 Get a video solution for your math problem!Problem text: Problem # 53
The following hypothesis are given:
Ho: Forty percent of the observations are in category A, 40 percent are in B, and 20 percent are in C.
H1: The observations are not as described in Ho.
We took a sample of 60, with the following results.
Category fo
A 30
B 20
C 10
a. State the decision rule using the .01 significance level.
b. Compute the value of chisquare.
c. What is your decision regarding Ho?
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Problem #354 Get a video solution for your math problem!Problem text: Problem #59
The director of advertising for the Carolina Sun Times, the largest newspaper in the Carolinas, is studying the relationship between the type of community in which a subscriber resides and the portion of the newspaper he or she reads first. For a sample of readers, she collected the following sample information.
National
News Sports Comics
City 170 124 90
Suburb 120 112 100
Rural 130 90 88
At the .05 significance level, can we conclude there is a relationship between the type of community where the person resides and the portion of the paper read first?
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Problem #355 Get a video solution for your math problem!Problem text: 1. A new process has been implemented and it is stated to have an average cycle time of 120 seconds. A sample of 50 cycles yielded an average cycle time of 115 seconds, with a standard deviation of 2 seconds.
A. Set up the null and alternative hypothesis
B. Test your hypothesis using a level of significance of
.05
2. A. Define the Z distribution
B. Find Z if X = 980, u = 1000, The standard deviation
is 50, and the number of samples is 49
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Problem #356 Get a video solution for your math problem!Problem text: (http://www.webmath.com/Answers/Files/Problem_5511_45,doc)
I need the solution to the following questions from your website. I would like to get the Answers ASAP.
HERE IS THE CASE STUDY FOR THE FOLLOWING 3 QUESTIONS
I picked up the newspaper the next week, and saw a story about a local baker, Dusty, who claimed he could make a cheese cake (from scratch to finished product) in 60 minutes flat. I doubted anyone could make a cheese cake in that amount of time, even though the reporter swore he saw and timed the entire process.
Thinking this might be great opportunity selfpromotion, I devised a plan to uncover Dusty and the reporter as frauds. I called him the next day, told him I was also a baker, and asked him if he would be kind enough to let me come watch him make one of his famous cheese cakes. As he was still basking in the sun of his recent publicity, he readily agreed and invited me over that same afternoon.
I arrived with a stopwatch in my pocket, and a notepad in my hand. Dusty was quite a showman, and I encouraged him to try to beat his personal best record, which was 58:42. He was good, but not good enough to beat his record, which prompted him to invite me back for another try (all part of my plan). Afterwards, I went back to the office, and filed all the information I had gathered.
Two days later I went back and watched again. After eighteen visits and observations (he was obsessed), he threw in the towel and said the record couldn't be beaten.
Back at the office I sifted through the data and found the production time for each time Dusty made the cheese cake. From the 18 readings I calculated that the mean time was 65.37 minutes and the standard deviation was 8.2, which was pretty close to the claim in the newspaper. But was it enough difference to call him a fraud?
What I needed was an hypothesis test!
I used the "Student’s t Distribution" because I have a small sample.
I let my null hypothesis be: Ho: m = 60, and, since I want to prove that his mean time is greater than 60 minutes, I’ll set up my alternate hypothesis like this:
H1: µ > 60
If I find my sample "test statistic" and then test it against the critical value, I can determine, once and for all, whether I can safely reject Dusty's claim. I think that an alpha of .05 should be adequate.
Assume Ho: m = 60, H1: µ > 60, and alpha of .05. Input the key figures, then indicate whether or not to reject Ho.
1. tc =
a. 1.74
b. 1.95
c. 2.00
2. t =
a. 2.330
b. 2.778
c. 1.876
3. Decision on Ho:
a. Fail to reject Ho
b. Reject Ho
c. Retain Ho
HERE IS THE CASE STUDY FOR THE FOLLOWING 3 QUESTIONS
After completing my work on the hypothesis test of Dusty's cheesecake production, I felt it only fair to share my results with him. He was impressed by my ability to use statistics, and suggested that I should contact the reporter to write a follow up story on my findings. This totally caught me off guard, and I felt embarassed about my original intentions. I then realized that Dusty would be a great asset to my business, so I told him about my most recent dilemma.
I had been looking at http://www.bakerynet.com/, and found a two year old survey about people's preferences for different types of pastries. I told him 50% favored doughnuts; 22% favored apple turnovers; 13% favored doughnut holes; 9% favored Bismarcks; and 6% favored pomegranate cookies.
I asked Dusty if this sort of information would be helpful in his business (I knew it would be for mine), and then wondered out loud if these statistics would hold true locally.
"There's only one way to find out", said Dusty, "and I think you're the right person for the task."
So the next day I made up a form with the appropriate questions, went down to the corner, and started my survey. It wasn’t long before I had 40 people give me their opinions. Their responses seemed to be fair and truthful. And I think it was as random a sample as I could have obtained.
while thumbing through the pages of my statistics book, which I thought might help. It was the symbol X2 and something called "GoodnessofFit Tests."
Here were the results from my survey:
Item Number of Votes
Doughnuts 20
Apple Turnovers 6
Doughnut Holes 5
Bismarcks 4
Pomegranate Cookies 5
Based upon all of this information, do a "goodnessoffit" test to see if the distributions are the same as they were two years ago. Use a level of confidence of .01. Should we accept Ho or reject Ho?
1. Chi Square is equal to:
2. The critical value is:
a. 13.28
b. 11.34
c. 15.09
3. Decision:
a. Retain Ho
b. Reject Ho
HERE IS THE CASE STUDY FOR THE FOLLOWING TWO QUESTIONS
Over the past weeks, with the help of my husband and Norman, we have solved the problem of employees, supplies, types of pastries to sell, how much money we should expect to make (at least initially).
I've also become good friends with Dusty, and we both agree that advertising is essential to building up a clientele, and keeping it.
Dusty keeps meticulous records of how much he spends on each type of advertising and how much he has made with each type. In this way, he is able to budget his expenses so that he got the most out of each.
I took this data and constructed a graph for him to see if the ordered pairs looked like they might be linearly related. I took the type of advertising he did and paired it up with the amount of income it generated. He had tried advertising in the local newspaper for a month, then a month using the Thrifty Dime, a month using local Church bulletins, and then a month using flyers around the neighborhood. He even hired someone for a month to dress up as "Cheese Cake Man".
Here's what I came up with:
TYPE OF ADVERTISING (MY COST,MY INCOME)
Local Newspaper ( $120 , $1100 )
"Thrifty Dime" ( $25 , $600 )
Church Bulletins ( $40 , $500 )
Flyers ( $60 , $800 )
"Cheese Cake Man" ( $80 , $1000 )
Using the process of linear regression, find the equation of the line and calculate the Linear Correlation Coefficient.
1. Equation of the Line:
a. y = 5.5x + 465.76
b. y = 6.3x + 386.37
c. y = 2x + 155.87
2. Correlation Coefficient:
a. 0.875
b. 0.596
c. 0.926
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Problem #359 Get a video solution for your math problem!Problem text: Please see attached, please supply the solutions to the following questions, that comes from your website (www.webmath.com/Answers/Files/problem_5621_47.doc)
1. An independent measures experiment uses two samples with n = 6 in each to compare two experimental treatments. The t statistic from this experiment will have degrees of freedom equal to:
a. 7
b. 10
c. 5
d. 14
2. The type of t statistic that can be used with either a beforeafter design or a matchedsubjects design is called a __________ t test.
3. The Chisquare distribution is ...
a. symmetrical with a mean of zero.
b. positively skewed with all values equal to or greater than zero.
c. negatively skewed with all values equal to or less than zero.
d. symmetrical with a mean equal to n  1.
4. As the differences between fe and fo decreases.
a. the likelihood of rejecting Ho also increases.
b. the likelihood of rejecting Ho decreases.
c. the critical value for Chisquare increases.
d. the critical value for Chisquare decreases.
5. The Chisquare test of independence is used to test for
a. a mean difference between two populations.
b. a difference between a sample distribution and a population distribution.
c. a difference in variance between two populations.
d. a relationship between two variables.
6. The Chisquare Test of Independence has degrees of freedom given by
a. R x C
b. n  R x C
c. R x C (n  1)
d. (R  1)(C1)
7. With df=12 and a critical value of 21.03, the area under the curve
to the right of the critical value is ________.
8. A twotailed test is a common choice when doing a Chisquare analysis.
a. True
b. False
9. The value for degrees of freedom for a Chisquare test does not depend on the sample size (n).
a. True
b. False
10. Using the Chisquare Goodness of Fit, degrees of freedom are based on the number of subjects.
a. True
b. False
11. Expected frequencies are calculated based on
a. the observed sample's distribution.
b. the distribution according to a predetermined model.
c. a comparison of two dependent samples.
d. degrees of freedom.
12. What is the critical value for df=3 and alpha =.05?
13. Which of the following sets of correlations correctly shows the lowest to highest degree of relationship?
a. 0.03, +0.10, 0.91, +0.83
b. +0.83, +0.10, 0.91, 0.03
c. +0.83, +0.10, 0.03, 0.91
d. 0.03, +0.10, +0.64, 0.81
14. Suppose the correlation between height and weight for adults is +0.70. What percent of the variability in weight is predicted by the relationship with height?
a. 70%
b. 30%
c. 35%
d. 49%
15. For the test of significance of a correlation, the null hypothesis states
a. the population correlation is zero.
b. the population correlation is not zero.
c. the sample correlation is zero.
d. the sample correlation is not zero.
16. If the value of the correlation is r = 1.00, then all of the data
points on a scatter plot would fall on a straight line.
a. True
b. False
17. The correlational rule states that if X and Y are correlated, X has caused Y.
a. True
b. False
18. A linear regression
a. is the line that best fits on a scatter plot.
b. is an equation used to predict a y score from an x.
c. cannot be used if r = 0.
d. all of the above.
19. The closer the points on a scatter plot are to the regression line, the larger the Se will be.
a. True
b. False
20. The manager at Air Express claims that the weights of packages shipped recently are less than in the past. Records show that, in past years, packages have had a mean weight of 36.7 lbs. So he took a sample of 64 packages from last month's shipping records and found that their mean weight was 32.1 lbs. with a standard deviation of 14.2. Is this sufficient evidence to reject the null hypothesis in favor of the manager's claim? Use alpha = .01. ANSWER ALL 5 PARTS PLEASE
a. (1 pt.) Is this test lefttailed, righttailed or twotailed?
b. (1 pt.) What is the critical value?
c. (2 pts.) What is the test statistic value?
d. (1 pt.) Is the Ho accepted or rejected?
e. (2 pts.) What is the P value?
21. A new law has been passed giving city police greater power to apprehend suspected criminals. The number of crimes reported for six neighborhoods, 1 year before and 1 year after the new law, are shown. Use the test for "paired differences" to test the claim that 1 year after the new law, the number of crimes has dropped. Use alpha = .05. ANSWER BOTH PARTS PLEASE
Neighborhood Before Law After Law
1 24 21
2 16 23
3 37 30
4 20 19
5 28 24
6 29 29
a. (3 pts.) In the table shown, the obtained t is __________.
b. (1 pt.) Is the Ho accepted or rejected?
22. Professor Fair believes that giving more than one hour's time to take an exam is not related to improved grades on the exams. He randomly divided a group of 280 students into two groups and gave them the same test. One group had exactly one hour in which to finish the test. The other group stayed as long as they wanted behind the one hour. Use the results below and a Chisquare to test that the time to do an exam and the grades are independent.
Use alpha = .01. ANSWER BOTH PARTS PLEASE
Score
Time A B C
1 Hour 23 42 65
Unlimited 17 48 85
a. (3 pts.) In the table shown, what is the obtained Chisquare?
b. (1 pt.) Is the Ho accepted or rejected?
23. A sociologist is studying the age of the population in the Rocky Ridge area. Ten years ago the population was such that 20% were under 20 years old, 15% were in the 20  35 year old bracket, 30% were between 36 and 50, 25% were between 51 and 65, and 10% were over 65. A study done this year used a random sample of 300 residents and yielded the data below. Use the Chisquare Goodness to Fit to see if the population of the Rocky Ridge area has changed. Use alpha = .01. ANSWER BOTH PARTS PLEASE
Under 20 20  35 36  50 51  65 Over 65
15 50 120 95 20
a. (3 pts.) In the table shown, what is the obtained Chisquare?
b. (1 pt.) Is the Ho accepted or rejected?
24. A ticket broker feels that the number of employees a company has and the number of sports tickets a corporation will purchase for "perks" are correlated. If correlated, the marketer may focus on selling to a specific size of company. Last year's sales yielded these data where X = number of employees and Y = number of tickets sold. ANSWER BOTH PARTS PLEASE
X: 15 26 31 280 300 312 390
Y: 0 2 2 6 6 4 16
a. (4 pts.) What is the correlation?
b. (1 pt.) Is the correlation statistically significant with
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