How can Chi Square test for independence be used to evaluate test re-test reliability?

The American Bar Association reports that the mean length of time for a hearing in juvenile court is 25 minutes. Assume that this this your population mean. As a lawyer who practices in the juvenile court, you think that the average hearing much shorter than this. You take a sample of 20 other lawyers who do juvenile work and ask them how long their last case in juvenile court was. The mean hearing length for this sample of 20 was 23 minutes., with a standard deviation of 6. Test the null hypotheses that the population mean is 25 minutes against the alternative that is less than 25. Set your alpha at .05.

A copier cost $200 for each repair call, after 3 calls they are free (For example 1 call= $200, 3 or more calls = $600). The following chart shows the probability of how many repair calls you will have in 1 year. Questions 36-37

pls show work

36. What is the expected repair call costs for 1 year?

1. $150.00

2. $200.00

3. $206.00

4. $220.00

5. $300.00

37. What is the probability that one repair call cost will be more than $200 for 1 year?

1. .70

2. .40

3. .30

4. .20

5. .10

38. The daily high temperature of Los Angeles in the month of December is normally distributed with a mean of 58 degrees Fahrenheit and a standard deviation of 2 degrees fahrenheit. What percent of high temperatures are between 60 and 64?

1. 68%

2. 84%

3. 15.85%

4. 16%

5. None of the above

39. If P(A) = 0.38, P(B) = 0.83, and P (A ∩ B) = 0.57; then P(A U B) =

1. 1.21

2. 0.64

3. 0.78

4. 1.78

5. None of the above

Consider the probability that at least 36 out of 298 cell phone calls will be disconnected. Choose the best description of the area under the normal curve that would be used to approximate binomial probability.

The option are

Area to the right of 35.5

Area to the right of 36.5

Area to the left of 35.5

Area to the left of 36.5

Area between 35.5 and 36.5

Must solve in R code

Question 2

The random variable X is a payout table of a casino slot machine. The probability mass function is given:

X   -5   0   2   10   20   40   60   1000
Probability   0.65   0.159   0.10   0.05   0.02   0.01   0.01   0.001
Please simulate the results for playing the slot machine 10,000,000 times. Find the mean of these 10,000,000 simulated payout outcomes .

When one uses an ANOVA, explain the difference between “main effects” and “interaction effects”. Give examples of each.

According to a survey in a​ country, 18​% of adults do not own a credit card. Suppose a simple random sample of 900 adults is obtained. ​(b) What is the probability that in a random sample of 900 ​adults, more than 22​% do not own a credit​ card? The probability is what.? ​(Round to four decimal places as​ needed.) Interpret this probability. If 100 different random samples of 900 adults were​ obtained, one would expect nothing to result in more than 22​% not owning a credit card. ​(Round to the nearest integer as​ needed.).  

(c) What is the probability that in a random sample of 900 ​adults, between 17​% and 22​% do not own a credit​ card?

Interpret this probability. If 100 different random samples of 900 adults were​ obtained, one would expect ? to result in between 17​% and 22​% not owning a credit card.

​(Round to the nearest integer as​ needed.)

Would it be unusual for a random sample of 900 adults to result in 153 or fewer who do not own a credit​ card? Why? Select the correct choice below and fill in the answer box to complete your choice

The result

is notis not

unusual because the probability that

ModifyingAbove p with caretp

is less than or equal to the sample proportion is

nothing​,

which is

greatergreater

than​ 5%.

B.The result

is notis not

unusual because the probability that

ModifyingAbove p with caretp

is less than or equal to the sample proportion is

nothing​,

which is

lessless

than​ 5%.

C.The result

isis

unusual because the probability that

ModifyingAbove p with caretp

is less than or equal to the sample proportion is

nothing​,

which is

lessless

than​ 5%.

D.The result

isis

unusual because the probability that

ModifyingAbove p with caretp

is less than or equal to the sample proportion is

nothing​,

which is

greatergreater

than​ 5%.

Statistics:

Construct a stem and leaf diagram, a frequency distribution table, plus develop a bar chart, pie chart, histogram, polygon and an ogive from the following data set.

When you construct the pie charts, bar graphs, etc. Use the calculation of 2^K to find the number of classes you will need to construct the frequency distribution table. You will need to use lower class limits, lower-class boundaries, midpoints upper-class boundaries, upper-class limits, frequency, relative frequency, and cumulative frequency.

DATA:

90, 92, 94, 100, 100, 91, 95, 96, 93, 23, 31, 57, 42, 59, 48, 58, 49, 37, 62, 75, 83, 88, 89, 70, 71, 79, 85, 77, 69, 82, 85.

The Pew Research Center took a random sample of 2928 adults in the United States in September 2008. In this sample, 53% of 2928 people believed that reducing the spread of acquired immune sample deficiency disease (AIDS) and other infectious diseases was an important policy goal for the U.S. government.

1. If the significance level α=0.05, should we reject or fail to reject the null hypothesis? How would you interpret your results of hypothesis testing?        REJECT H_O      DO NOT REJECT H_O

INTERPRET:

2. Based on the above questions, what is the difference between confidence interval and hypothesis testing? When should we use confidence interval and when should we use hypothesis testing?

Directions: Answer the following questions. Round probabilities to four digits after the decimal. For full credit, you will need to show your work justifying how you determined numerical values. You may consider adding additional blank space to this document, printing, filling out by hand, and then uploading a scan or pictures of your answers. You may also re-write these questions on a separate sheet of paper and use as much space as you need to answer the questions.

Charles is a notoriously bad student who never studies for his quizzes. One day, his teacher gives a four question, multiple-choice quiz. Each question has five answer options (a, b, c, d, and e) and each question has only one correct answer option. Since Charles doesn’t study, we might assume that he is going to guess on each of the four questions.

Let X be a discrete random variable representing the number of questions Charles may correctly answer.

Part 1. (5 points) Complete the following probability distribution table:

Part 2. (4 points) Using the table you constructed, find the expected value, μ, and the standard deviation, σ. Please show your work.

Part 3. (2 points) A statistician might consider an outcome with probability less than 0.05 to be “unusual”. Based on this criterion, should we be surprised if Charles guesses three or more questions correctly? Please explain briefly.

Part 4. (2 points) Alternatively, a statistician might consider an outcome to be “unusual” if it is more than two standard deviations away from the mean. Based on this criterion, should we be surprised if Charles guesses three or more questions correctly? Please explain briefly.

Part  5. (2 points) Suppose Charles happens to get three out of four questions correct. When computing the probabilities in problem 1, we assumed that Charles was guessing on every question; based on your answers to problems 3 and 4, do you think our assumption is plausible? What is an alternative explanation for Charles’s performance?

The following eight observations were drawn from a normal population whose variance is 100:

12

8

22

15

30

6

29

58

Part A

What is the standard error of the sample mean, based on the known population variance? Give your answer to two decimal places in the form x.xx

Standard error:

Part B

Find the lower and upper limits of a 90% confidence interval for the population mean. Give your answer to two decimal places in the form xx.xx

Lower limit:

Upper limit:

Part C

True or false:

If the population was not normally distributed the confidence interval calculation above would not be valid. Answer by writing T or F in the space provided.

Answer:

A major department store chain is interested in estimating the mean amount its credit card customers spent on their first visit to the chain’s new store in the mall. Fifteen credit card accounts were randomly sampled and analyzed with the following results: X = $50.50 and S = 20. Assuming that the amount spent follows a normal distribution, construct a 95% confidence interval for the mean amount its credit card customers spent on their first visit to the chain’s new store in the mall. assuming that the amount spent follows a normal distribution.

A sales and marketing management magazine conducted a survey on salespeople cheating on their expense reports and other unethical conduct. In the survey on 200 managers, 58% of the managers have caught salespeople cheating on an expense report, 50% have caught salespeople working a second job on company time, 22% have caught salespeople listing a “bar” as a restaurant on an expense report, and 19% have caught salespeople giving a kickback to a customer. To convert the data into useful information, construct a 95% confidence interval estimate of the population proportion of managers who have caught salespeople cheating on an expense report, determine the critical value, and sampling error. Explain what management can learn from the 95% confidence interval estimate of the population proportion of managers who have caught salespeople cheating on an expense report.