Questions
Shipments of TV sets that arrive at a factory have a varying levels of quality. In...

Shipments of TV sets that arrive at a factory have a varying levels of quality. In order to decide whether to accept a particular shipment, inspectors randomly select a sample of 15 TVs and test them; if no more than one TV in the sample is defective, the shipment is accepted. Let X be a random variable representing the number of defective staples in the random sample 15.

a. Explain why X may be treated as a binomial random variable:

•Identify n (the number of trails):
•Specify in words which event would be defined as a “success”
•Explain why the trails may be considered independent:
•Give the value of p (probability of a success)
b. What is the probability that shipment is accepted? ( Use a table or the formula)
c. What is the expected value of the number of defective TV set in the sample?
d. Fill this sentence: According to the Law of Large Numbers, if we have obtained many different simple random samples of size___ from this shipment, the average number of defective TV sets per sample would be approximately ___.

In: Math

For this problem, carry at least four digits after the decimal in your calculations. Answers may...

For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.

In a random sample of 63 professional actors, it was found that 39 were extroverts.

(a) Let p represent the proportion of all actors who are extroverts. Find a point estimate for p. (Round your answer to four decimal places.)


(b) Find a 95% confidence interval for p. (Round your answers to two decimal places.)

lower limit    
upper limit    


Give a brief interpretation of the meaning of the confidence interval you have found.

We are 5% confident that the true proportion of actors who are extroverts falls within this interval.We are 95% confident that the true proportion of actors who are extroverts falls outside this interval.    We are 5% confident that the true proportion of actors who are extroverts falls above this interval.We are 95% confident that the true proportion of actors who are extroverts falls within this interval.


(c) Do you think the conditions np > 5 and nq > 5 are satisfied in this problem? Explain why this would be an important consideration.

Yes, the conditions are satisfied. This is important because it allows us to say that is approximately binomial.No, the conditions are not satisfied. This is important because it allows us to say that is approximately normal.    Yes, the conditions are satisfied. This is important because it allows us to say that is approximately normal.No, the conditions are not satisfied. This is important because it allows us to say that is approximately binomial.

In: Math

A study  examined whether or not noses continue to grow throughout a person’s lifetime.  The study included many...

A study  examined whether or not noses continue to grow throughout a person’s lifetime.  The study included many measurements (including size of the nose as measured by total volume) and included multiple tests.  For each of the tests described below:

            a)  State the null and alternative hypotheses.

            b) Give a formal decision using a 5% significance level, and interpret the conclusion in context.

  • In a test to see whether males, on average, have bigger noses than females, the study indicates that “p < 0.01.”
  • In a test to see whether there is a positive linear relationship between age and nose size, the study indicates that “p < 0.001.”

In: Math

The failure rate in a statistics class is 20%. In a class of 30 students, find...

The failure rate in a statistics class is 20%. In a class of 30 students, find the probability that exactly five students will fail. Use the normal distribution to approximate the binomial distribution.

In: Math

The mean number of pets per household is 2.96 with standard deviation 1.4. A sample of...

The mean number of pets per household is 2.96 with standard deviation 1.4. A sample of 52 households is drawn. Find the probability that the sample mean is less than 3.11.

a.

0.2245

b.

0.5676

c. 0.3254

d.

0.7726

In: Math

The pH of 20 randomly selected lakes is measured. Their average pH is 5.7. Part A....

The pH of 20 randomly selected lakes is measured. Their average pH is 5.7.

Part A. Historically the standard deviation in the pH values is 0.9. Use this standard deviation for the following questions.
Part Ai. Build a 95% confidence interval for the population mean lake pH.
Part Aii. Build a 90% confidence interval for the population mean lake pH.
Part Aiii. Build an 80% confidence interval for the population mean lake pH.
Part Aiv. Compare the intervals you created in Ai, Aii and Aiii. What effect does changing the level of confidence have on the interval?

Please solve only part B

Part B. For the 20 measured lakes the standard deviation in the pH values is 0.9. Use this standard deviation for the following questions.
Part Bi. Build a 95% confidence interval for the population mean lake pH.
Part Bii. Compare the confidence intervals in Bi and Ai. What effect does not knowing the value of ? have on the interval?
Part Biii. Test whether the population mean pH differs from 6.

In: Math

For 300 trading​ days, the daily closing price of a stock​ (in $) is well modeled...

For 300 trading​ days, the daily closing price of a stock​ (in $) is well modeled by a Normal model with a mean of

​$197.12197.12 and a standard deviation of

​$7.187.18. According to this​ model, what is the probability that on a randomly selected day in this​ period, the stock price closed as follows.

​a) above ​$204.30204.30​?

​b) below ​$211.48211.48​?

​c) between ​$182.76182.76 and ​$211.48211.48​?

In: Math

Hale's TV Productions is considering producing a pilot for a comedy series in the hope of...

Hale's TV Productions is considering producing a pilot for a comedy series in the hope of selling it to a major television network. The network may decide to reject the series, but it may also decide to purchase the rights to the series for either one or two years. At this point in time, Hale may either produce the pilot and wait for the network's decision or transfer the rights for the pilot and series to a competitor for $250,000. Hale's decision alternatives and profits (in thousands of dollars) are as follows:

State of Nature
Decision Alternative Reject, S1 1 Year, S2 2 Years, S3
Produce pilot, d1 -150 50 450
Sell to competitor, d2 250 250 250

The probabilities for the states of nature are P(S1) = 0.20, P(S2) = 0.30, and P(S3) = 0.50. For a consulting fee of $45,000, an agency will review the plans for the comedy series and indicate the overall chances of a favorable network reaction to the series. Assume that the agency review will result in a favorable (F) or an unfavorable (U) review and that the following probabilities are relevant:

P(F) = 0.63 P(S1|F) = 0.07 P(S1|U) = 0.41
P(U) = 0.37 P(S2|F) = 0.27 P(S2|U) = 0.38
P(S3|F) = 0.66 P(S3|U) = 0.21

a. What is the expected value?

b. What is the expected value of perfect information?

c. What is the expected value of the agency's information? Round your answer to two decimal places.

d. What is the maximum that Hale should be willing to pay for the information? Round your answer to two decimal places.

In: Math

In each of parts​ (a)-(c), we have given a likely range for the observed value of...

In each of parts​ (a)-(c), we have given a likely range for the observed value of a sample proportion p. Based on the given​ range, identify the educated guess that should be used for the observed value of p to calculate the required sample size for a prescribed confidence level and margin of error.

a. 0.2 to 0.3

b. 0.1 or less

c. 0.3 or greater

In: Math

Physical activity of obese young adults. In a study on the physical activity of young adults,...

Physical activity of obese young adults. In a study on the physical activity of young adults, pediatric researchers measured overall physical activity as the total number of registered movements (counts) over a period of time and then computed the number of counts per minute (cpm) for each subject (International Journal of Obesity, Jan. 2007). The study revealed that the overall physical activity of obese young adults has a mean of μ = 320 cpm μ = 320   cpm and a standard deviation of σ = 100 c p m . σ = 100 c p m . (In comparison, the mean for young adults of normal weight is 540 cpm.) In a random sample of n = 100 n = 100 obese young adults, consider the sample mean counts per minute, ¯ x x ‾ . Describe the sampling distribution of ¯ x x ‾ . What is the probability that the mean overall physical activity level of the sample is between 300 and 310 cpm? What is the probability that the mean overall physical activity level of the sample is greater than 360 cpm?

In: Math

Please Double Check answers I've recived 3 wrong answers on three diffrent questions today thank you...

Please Double Check answers I've recived 3 wrong answers on three diffrent questions today thank you

CNNBC recently reported that the mean annual cost of auto insurance is 1006 dollars. Assume the standard deviation is 245 dollars. You take a simple random sample of 73 auto insurance policies.

Find the probability that a single randomly selected value is less than 973 dollars. P(X < 973) =

Find the probability that a sample of size n = 73 is randomly selected with a mean less than 973 dollars. P(M < 973) =

In: Math

A sample containing years to maturity and yield for 40 corporate bonds are contained in the...

A sample containing years to maturity and yield for 40 corporate bonds are contained in the data given below.

Years to Maturity Yield Years to Maturity Yield
23.50 4.757 3.75 2.769
21.75 2.473 12.00 6.293
21.50 4.464 17.50 7.411
23.50 4.684 18.00 3.558
27.00 4.799 8.25 0.945
18.25 3.755 23.25 2.966
15.75 7.068 14.75 1.476
2.00 7.043 10.00 1.382
8.75 6.540 23.00 6.334
5.25 7.000 15.25 0.887
11.25 4.823 4.75 4.810
25.75 1.874 18.00 1.238
14.25 5.654 3.00 6.767
19.25 1.745 9.50 3.745
25.00 8.153 17.50 4.186
6.75 6.571 17.00 5.991
23.00 7.506 9.50 7.322
19.00 2.857 5.50 4.871
10.75 8.010 27.50 2.403
21.25 4.214 26.00 4.500

a. What is the sample mean years to maturity for corporate bonds and what is the sample standard deviation?

Mean ? (to 4 decimals)
Standard deviation ? (to 4 decimals)

b. Develop a 95% confidence interval for the population mean years to maturity. Round the answer to four decimal places.

( , ) years

c. What is the sample mean yield on corporate bonds and what is the sample standard deviation?

Mean ?(to 4 decimals)
Standard deviation ?(to 4 decimals)

d. Develop a 95% confidence interval for the population mean yield on corporate bonds. Round the answer to four decimal places.

( , )percent

In: Math

As a data scientist of a company, you want to analyze the following data collected by...

As a data scientist of a company, you want to analyze the following data collected by your company which relates the advertising expenditure A in thousands of dollars to total sales S in thousands of dollars. The following table shows this relationship

Advertising Expenditure (A) Total Sales (S)
18.6 312
18.8 322
18.8 333
18.8 317
19 301
19 320
19.2 305

Using Advertising expenditure (A) as the domain and Total Sales (S) as the range, the data is not a function because the value 18.8 and 19 appear in the domain more than once with a different corresponding value of the range each time.

--Interpret the slope and y-intercept of this equation.

--Express this equation as a function S of A and find its domain.

--Predict the sales if the advertising expenditure is $25000.

In: Math

A prominent university conducted a survey on the effect of part-time work on student grade point...

A prominent university conducted a survey on the effect of part-time work on student grade point average (GPA). Let x be the hours worked per week and y the GPA for the year. A summary of the results is below. What can the university conclude with an α of 0.05?

n = 21

sigmay = 55

,sigma x = 520

sigmay2 = 171

, aigmax2 = 15288

sigmayx = 1275

, sigma ( yŷ2)  = 24

a) Compute the quantities below.

Bhat0 =  , Bhat =

What GPA is predicted when a students works 9 hours a week?

b) Compute the appropriate test statistic(s) for H1: β < 0.

Critical value =  ; Test statistic =

Decision:  

---Select---

Reject H0

Fail to reject H0

c) Compute the corresponding effect size(s) and indicate magnitude(s).

If not appropriate, input and/or select "na" below.

Effect size =  ;  

---Select---

na

trivial effect

small effect

medium effect

large effect

d) Make an interpretation based on the results.

More hours of part-time work significantly predicts a higher GPA.

More hours of part-time work significantly predicts a lower GPA.    

Part-time work does not significantly predict GPA.

In: Math

Financial analysts know that January credit card charges will generally be much lower than those of...

Financial analysts know that January credit card charges will generally be much lower than those of the month before. What about the difference between January and the next​ month? Does the trend​ continue? The accompanying data set contains the monthly credit card charges of a random sample of 99 cardholders. Complete parts​ a) through​ e) below.

January

February

902.74

641.04

7212.18

4565.35

4235.42

2270.56

79.92

300.09

4045.57

1377.72

89.29

−120.74

3289.59

1928.85  

2419.54

2609.97

83.81

144.83

6.42

392.85

0.00

40.46

564.69

295.63

2712.23

848.62

187.12

162.12

3265.86

2412.45

1523.59

956.31

1359.23

38.03

733.33

2656.79

75.09

64.94

70.29

−70.32

634.53

1862.61  

1041.23

478.07

553.08

994.64

1016.27

774.54

1304.94

3368.08

249.39

5.52

48.78

96.93

872.34

890.89

485.94

485.21

616.52

1485.52

1574.18

890.46

422.34

391.43

770.85

323.19

56.53

0.00

1486.78

2253.73

495.28

390.19

1064.88

1065.85  

510.65  

131.33

5637.68

4942.63

5.49

5.51

871.63

591.44

1636.66

3364.19

92.13

85.99

669.34

1367.13

829.32

280.85

69.24

67.99

830.54

1057.56

2301.44

3317.76

270.67

14.13

210.42

160.52

1012.36

519.35

1044.96

2021.35

298.64

635.44

−29.99

0.00

1634.61

393.34

1731.93

1323.33

0.00

65.16

31.43

28.75

4.95

77.15

1088.69

892.78

26.88

29.03

120.31120.31

32.23

2007.48

815.63

291.31

779.47

104.02

0.00

53.01

66.25

2842.52

1530.91

675.47

293.45

221.86

171.92

37.79

4.78

533.25

880.96

1932.71

1063.55

692.17

915.55

6804.35

5941.41

393.36

466.47

1309.18

302.89  

796.21

497.02

0.00

266.64

1040.29

59.45

565.12

206.62

339.14

412.34

5275.34

5324.54

40.09

72.58

43.39

38.45

653.63

480.25

1071.23

416.29

2337.04

1787.19

91.47

175.32

1433.01

1107.78

719.86

307.79

28.61

24.19

980.34

1216.35

1576.18

1810.23

0.00

468.24

161.96

147.68

494.32

1995.28

534.11

935.24

462.45

114.51

1478.23

2093.37

​a) Build a regression model to predict February charges from January charges.

Feb=____+____Jan ​(Round to two decimal places as​ needed.)

Check the conditions for this model. Select all of the true statements related to checking the conditions.

A. All of the conditions are definitely satisfied.

B. The Linearity Condition is not satisfied.

C. The Randomization Condition is not satisfied.

D. The Equal Spread Condition is not satisfied.

E. The Nearly Normal Condition is not satisfied. ​

b) How​ much, on​ average, will cardholders who charged ​$2000 in January charge in​ February? ​$____ ​(Round to the nearest cent as​ needed.)

​c) Give a​ 95% confidence interval for the average February charges of cardholders who charged ​$2000 in January.

($___,$___) ​(Round to the nearest cent as​ needed.) ​

d) From part​ c), give a​ 95% confidence interval for the average decrease in the charges of cardholders who charged ​$2000 in January.

​($___,$___) (Round to the nearest cent as​ needed.)

​e) What​ reservations, if​ any, would a researcher have about the confidence intervals made in parts​ c) and​ d)? Select all that apply.

A. The residuals show increasing​ spread, so the confidence intervals may not be valid.

B. The residuals show a curvilinear​ pattern, so the confidence intervals may not be valid.

C. The data are not​ linear, so the confidence intervals are not valid.

D. The data are not​ independent, so the confidence intervals are not valid.

E. A researcher would not have any reservations. The confidence intervals are valid. Click to select your answer(s).

In: Math