A component is purchased from 3 suppliers, A, B, and C, where the suppliers have

respective defective rates of 2%, 6%, and 4%. Of all the components purchased, 20%

comes from supplier A, 50% from supplier B, and 30% from supplier C, that is, each

shipment comes from each of these suppliers with these probabilities. The company uses

the following quality control policy. A sample of 15 units is randomly selected from each

shipment of components. If at most 2 defective units are found in the sample, then the

entire shipment is accepted; otherwise, the entire shipment is rejected.

Determine the following:

(a-1) probability that a shipment will be accepted given that it came from Supplier A.

(a-2) probability that a shipment will be accepted given that it came from Supplier B.

(a-3) probability that a shipment will be accepted given that it came from Supplier C.

(a-4) Using the law of total probability and your responses to parts (a-1)-(a-3) above,

determine the probability that a shipment will be accepted.

(b-1) expected number of defectives in a random sample of 15 units, if the shipment came

from Supplier A.

(b-2) expected number of defectives in a random sample of 15 units, if the shipment came

from Supplier B.

(b-3) expected number of defectives in a random sample of 15 units, if the shipment came

from Supplier C.

(b-4) Using the law of total expectation and your responses to parts (b-1)-(b-3) above,

determine the expected number of defectives in a random sample of 15 units.

A restaurant manager asks all of the customers this month to take an online survey (and get a free appetizer their next visit). It turns out that 241 customers actually take the survey, and of these 209 were ''extremely satisfied'' with their visit. Set up an approximate 95% confidence interval for the proportion of all the restaurant's customers that say they were ''extremely satisfied'' with their visit..

Select only one of the boxes below.

A. The Normal curve cannot be used to make the requested confidence interval.

B. Making the requested confidence interval does not make any sense.

C. It is appropriate to compute a confidence interval for this problem using the Normal curve.

If it is appropriate to compute a confidence interval for this problem using the Normal curve, then enter the confidence interval below. Otherwise, enter [0,0] for the confidence interval.

15.3-11. Management of the Telemore Company is considering developing
and marketing a new product. It is estimated to be twice
as likely that the product would prove to be successful as unsuccessful.
It it were successful, the expected profit would be
$1,500,000. If unsuccessful, the expected loss would be
$1,800,000. A marketing survey can be conducted at a cost of
$300,000 to predict whether the product would be successful. Past
experience with such surveys indicates that successful products
have been predicted to be successful 80 percent of the time, whereas
unsuccessful products have been predicted to be unsuccessful 70
percent of the time.
(a) Develop a decision analysis formulation of this problem by
identifying the alternative actions, the states of nature, and the
payoff table when the market survey is not conducted.
T (b) Assuming the market survey is not conducted, use Bayes’
decision rule to determine which decision alternative should
be chosen.
T (c) Find EVPI. Does this answer indicate that consideration
should be given to conducting the market survey?
T (d) Assume now that the market survey is conducted. Find the
posterior probabilities of the respective states of nature for
each of the two possible predictions from the market survey.
(e) Find the optimal policy regarding whether to conduct the market
survey and whether to develop and market the new product.

Variable Mean Median Q1 Q3 Two-Star 65.08 69.50 46.50 79.25 Three-Star 89.74 87.00 70.75 107.00 Four-Star 127.74 125.50 90.75 150.00 B. Variable Range Interquartile Range Variance Standard deviation Coefficient of variation Two - star 94.00 32.75 473.80 21.77 33.45 Three-star 126.00 36.25 761.66 27.60 30.75 Four-star 134.00 59.25 1583.60 39.79 31.15 Based on the results, what conclusions can you reach concerning these variables?

6. List below are the numbers of words spoken in a day by each member of six different couples.

Use a 0.05 significance level to test the claim that among couples, males speak more words in a

day than females. Assume that the population differences is normally distributed.

Male

5638

21,319

17,572

26,429

46,978

25,835

Female

5198

11,661

19,624

13,397

31,553

18,667

7. A sociologist wanted to estimate the difference in the amount of daily leisure time (in hours) of adults who do not have children under the age of 18 years and adults who have children under the age of 18 years. A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.62 hours, with a standard deviation of 2.43 hours. A random sample of 40 adults with children under the age of 18 years results in a mean daily leisure time of 4.10 hours, with a standard deviation of 1.82 hours. Construct a 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children.

Question1

A market researcher wants to study TV viewing habits of residents in a particular area. A random sample 0f 70 respondents is selected. The results are as follows

Viewing time per week Mean=14 hours, S=3.8 hours

50 respondents watch the evening news on at least three week nights

(a)      Construct a 90% confidence interval estimate for the mean amount of TV watched per week.

(b)      Construct a 95% CI estimate of population proportion who watches evening news on at least three week nights.

Question 2

Following data gives the amount that a sample of 10 customers spent for lunch ($) at a fast food restaurant:

$8.50, $7, $5.80,  $9, $5.60,  $7.80,  $6.50, $5.40,  $7.50,  $5.50

Construct a 95% confidence interval estimate for the population mean amount spent on lunch at the fast food restaurant.

Question 3

A cell phone provider has the business objective of wanting to estimate the proportion of subscribers who would upgrade to a new cell phone with improved features if it were made available at a substantially reduced cost. Data are collected from a random sample of 500 subscribers. The results indicated that 135 of the subscribers would upgrade to a new cell phone at a reduced cost. Construct a 95% confidence interval estimate for the population proportion of subscribers that would upgrade to a new cell phone at a reduced cost.

Question 4

You are the manager of a restaurant for a fast food. Last month the mean waiting time at the drive through window for branches in your region was 3.8 minutes. You select a random sample of 49 orders. The sample mean waiting time is 3.57 minutes, with a sample standard deviation of 0.8 minutes. At the .05 level of significance is there evidence that population mean waiting time is different from 3.8 minutes?

Two types of medication for hives are being tested to determine if there is a difference in the proportions of adult patient reactions. Twenty out of a random sample of 200 adults given medication A still had hives 30 minutes after taking the medication. Twelve out of another random sample of 180 adults given medication B still had hives 30 minutes after taking the medication. Test at a 1% level of significance.

1. (Write all answers as a decimal rounded to the 3rd decimal place)

Personnel tests are designed to test a job​ applicant's cognitive​ and/or physical abilities. A particular dexterity test is administered nationwide by a private testing service. It is known that for all tests administered last​ year, the distribution of scores was approximately normal with mean 74 and standard deviation 7.7. a. A particular employer requires job candidates to score at least 79 on the dexterity test. Approximately what percentage of the test scores during the past year exceeded 79​? b. The testing service reported to a particular employer that one of its job​ candidate's scores fell at the 90th percentile of the distribution​ (i.e., approximately 90​% of the scores were lower than the​ candidate's, and only 10​% were​ higher). What was the​ candidate's score?

Patients with chronic kidney failure may be treated by dialysis, in which a machine removes toxic wastes from the blood, a function normally performed by the kidneys. Kidney failure and dialysis can cause other changes, such as retention of phosphorus, that must be corrected by changes in diet. A study of the nutrition of dialysis patients measured the level of phosphorus in the blood of several patients on six occasions. Here are the data for one patient (in milligrams of phosphorus per deciliter of blood).

The measurements are separated in time and can be considered an SRS of the patient's blood phosphorus level. Assume that this level varies Normally with

σ = 0.8 mg/dl.

(a) Give a 95% confidence interval for the mean blood phosphorus level.

Fill in the blanks for "x" and "y". ) A bulk supply of CAT6 cables were purchased to implement a wired office network. The network designer randomly tests 12 of the cables and finds the attenuation is on average 22.7dB with a sample standard deviation of 1.9 dB. She wishes to estimate the 95% confidence interval for the mean attenuation of all of the cables. Find the 95% confidence interval (α=0.05) for the mean attenuation rounded to the

Suppose these data show the number of gallons of gasoline sold by a gasoline distributor in Bennington, Vermont, over the past 12 weeks.

(a)

Compute four-week and five-week moving averages for the time series.

(b)

Compute the MSE for the four-week moving average forecasts. (Round your answer to two decimal places.) __________

Compute the MSE for the five-week moving average forecasts. (Round your answer to two decimal places.) ____________

Basic guide to developing a research design ( Please provide EXAMPLES in terms to Politics !!)
1. What are the treatment and outcome variables? How are they defined and measured?
2. What are the two (or more) groups in your research? That is, what is the treatment
group and what is the control group?
a. How is the treatment assigned to the group? Random vs. selected by something
else? If not randomly assigned, what determines the assignment of the
treatment?
b. Other than the value(s) of the treatment variable, how else might the two
groups differ? What confounding variables might be different between the two?
These variables are potential threats to internal validity

Use R.  Provide Solution and R Code within each problem.

For this section use the dataset “PlantGrowth”, available in base R (you do not need to download any packages).

a.Construct a 95% confidence interval for the true mean weight.

b.Interpret the confidence interval in 1. in the context of the problem.

c.Write down the null and alternative hypothesis to determine if the mean weight of the plants is less than 5.

d.Conduct a statistical test to determine if the mean weight of the plants is less than 5. Use α = 0.05.

i.Pvalue

ii.Conclusion

In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and in the United Kingdom. Suppose that independent random samples of television ads are taken in the two countries. A random sample of 400 television ads in the United Kingdom reveals that 142 use humor, while a random sample of 500 television ads in the United States reveals that 126 use humor. (a) Set up the null and alternative hypotheses needed to determine whether the proportion of ads using humor in the United Kingdom differs from the proportion of ads using humor in the United States. H0: p1 − p2 0 versus Ha: p1 − p2 0. (b) Test the hypotheses you set up in part a by using critical values and by setting α equal to .10, .05, .01, and .001. How much evidence is there that the proportions of U.K. and U.S. ads using humor are different? (Round the proportion values to 3 decimal places. Round your answer to 2 decimal places.) z H0 at each value of α; evidence. (c) Set up the hypotheses needed to attempt to establish that the difference between the proportions of U.K. and U.S. ads using humor is more than .05 (five percentage points). Test these hypotheses by using a p-value and by setting α equal to .10, .05, .01, and .001. How much evidence is there that the difference between the proportions exceeds .05? (Round the proportion values to 3 decimal places. Round your z value to 2 decimal places and p-value to 4 decimal places.) z p-value H0 at each value of α = .10 and α = .05; evidence. (d) Calculate a 95 percent confidence interval for the difference between the proportion of U.K. ads using humor and the proportion of U.S. ads using humor. Interpret this interval. Can we be 95 percent confident that the proportion of U.K. ads using humor is greater than the proportion of U.S. ads using humor? (Round the proportion values to 3 decimal places. Round your answers to 4 decimal places.) 95% of Confidence Interval [ , ] the entire interval is above zero.