Questions
6.31. The exponential distribution can be used to solve Poisson-type problems in which the intervals are...

6.31. The exponential distribution can be used to solve Poisson-type problems in which the intervals are not time. The Air Travel Consumer Report published by the U.S. Department of Transportation reported that in a recent year, Virgin America led the nation in fewest occurrences of mishandled baggage, with a mean rate of 0.95 per 1,000 passengers. Assume that mishandled baggage occurrences are Poisson distributed. Using the exponential distribution to analyze this problem, determine the average number of passengers between occurrences. Suppose baggage has just been mishandled.

  1. What is the probability that at least 500 passengers will have their baggage handled properly before the next mishandling occurs?
  2. What is the probability that the number will be fewer than 200 passengers

Book says answers are

.μ = 1052.6

a..5908

b..1899 but how?

In: Statistics and Probability

2.32 The following are the grades that 50 students obtained on an accounting test: 73 65...

2.32 The following are the grades that 50 students obtained on an accounting test:

73

65

82

70

45

50

70

54

32

75

75

67

65

60

75

87

83

40

72

64

58

75

89

70

73

55

61

78

89

93

43

51

59

38

65

71

75

85

65

85

49

97

55

60

76

75

69

35

45

63

Convert the distribution from classes 30-39, 40-49, 50-59, …, and 90-99 into cumulative “less than” distribution, beginning with “less than 30”.

2.34The following are the miles per gallon obtained with 40 tankful of gas:

24.1

25.0

24.8

24.3

24.2

25.3

24.2

23.6

24.5

24.4

24.5

23.2

24.0

23.8

23.8

25.3

24.5

24.6

24.0

25.2

25.2

24.4

24.7

24.1

24.6

24.9

24.1

25.8

24.2

24.2

24.8

24.1

25.6

24.5

25.1

24.6

24.3

25.2

24.7

23.3

Group these figures into a distribution having the classes 23.0-23.4, 23.5-23.9, 24.0-24.4, 24.5-24.9, 25.0-25.4, and 25.5-25.9.

b) Convert the distribution obtained in part (a) into a cumulative "or more" distribution, beginning with "23.0 or more" and ending with "26.0 or more."

In: Statistics and Probability

A very large study showed that aspirin reduced the rate of first heart attacks by 45%....

A very large study showed that aspirin reduced the rate of first heart attacks by 45%. A pharmaceutical company thinks it has a drug that will be more effective than aspirin, and plans to do a randomized clinical trial to test the new drug. Complete parts a through d below.

a) What is the null hypothesis the company will use?

A. The new drug is not more effective than aspirin.

B. The new drug is more effective than aspirin.

C. The new drug is less effective than aspirin.

D. The new drug is as effective as aspirin.

b) What is their alternative hypothesis?

A. The new drug is as effective as aspirin.

B. The new drug is not more effective than aspirin.

C. The new drug is less effective than aspirin.

D. The new drug is more effective than aspirin.

c) The company conducted the study and found that the group using the new drug had somewhat fewer heart attacks than those in the aspirin group. The P-value from the hypothesis test was 0.002. What do you conclude?

A. There is not sufficient evidence to conclude that the alternative hypothesis is true because the​ P-value was so large.

B. There is not sufficient evidence to conclude that the alternative hypothesis is true because the​ P-value was so small.

C.There is evidence to conclude that the alternative hypothesis is true because the​ P-value was so large.

D. There is evidence to conclude that the alternative hypothesis is true because the​ P-value was so small.

d) What would you have concluded if the P-value had been 0.42?

A.There is evidence to conclude that the alternative hypothesis is true because the​ P-value was so small.

B. There is not sufficient evidence to conclude that the alternative hypothesis is true because the​ P-value was so large.

C. There is evidence to conclude that the alternative hypothesis is true because the​ P-value was so large.

D.There is not sufficient evidence to conclude that the alternative hypothesis is true because the​ P-value was so

small.

In: Statistics and Probability

The box plot shows the undergraduate in-state tuition per credit hour at four-year public colleges. Estimate...

The box plot shows the undergraduate in-state tuition per credit hour at four-year public colleges.

  1. Estimate the median.
  1. Estimate the first and third quartiles.
  1. Determine the interquartile range.
  1. Beyond what point is a value considered an outlier?
  1. Identify any outliers and estimate their value. If there are no outliers select “None”.
  • 1,500

  • 1,300

  • 800

  • None

  1. Is the distribution symmetrical or positively or negatively skewed?
  • symmetric skewed

  • positively skewed

  • negatively skewed

Next Visit question map

Question1of10Total1 of 10

Prev

In: Statistics and Probability

Q: Answer yes or no to each of the items from a to c. If yes,...

Q: Answer yes or no to each of the items from a to c. If yes, briefly justify the statement, if no, provide a counter-example or disprove the statement.

a) We have a sample of an individuals, indexed by i = 1, · · · , n. For the i-th individual, we can observe Xi andYi. If we assume our sample is i.i.d. cross-sectionally, it implies Xi and Yi are independent with each other.

b) In multivariate OLS, we have three regressors X1, X2, X3. If (X1, X2), (X1, X3), and (X3, X2) are not perfectly correlated, then we do not have the perfect multicollinearity.

c) T-test is still valid when OLS assumption 1 correct specification is violated.

In: Statistics and Probability

Table 1: Weight Suspended in kg Luke Rey Anikin week 1 0 10 35 week 2...

Table 1: Weight Suspended in kg
Luke Rey Anikin
week 1 0 10 35
week 2 10 10 41
week

3

15 20 45
week 4 20 20 48
week 5 25 40 50
week 6 30 40 53
week 7 35 80 55
week 8 40 80 61
week 9 45 160 70
week 10 50 160 74
week 11 55 320 77
week 12 60 320 80

In an alternate reality, a temporal anomoly transports Luke Skywalker, Rey and Anikin Skywalker to Yoda's home planet as 8 year olds. Yoda's people decide to train them all in the ways of the Force. As they begin their Jedi training, their progress is tracked weekly. One of the items tracked is weight of object they are able to make hover. The results of the first 12 weeks of training are shown in Table 1.

1. Examine the data in Table 1. Without doing any calculations, of the three Jedi trainees, for which kid(s) would be Simple Linear Regression be a good tool to model their progress? Why? Explain in 50 words or less.

2. Perform a simple linear regression on all three kid's progress. Hint: time is the x variable and weight of stone suspended is the y variable. Perform a simple linear regression on all three kid's progress. Show your work in the analysis tab, putting the output in the designated spaces.

What is the regression equation for Luke? For Rey? For Anikin? Put your results in the indicated space.

What is the r2 for each kid? According to the p-value, which of the regression lines is statistically significant at an alpha of 0.05?

3. Using Luke and Anikin's regression equations, make a prediction during which week Luke will catch up with Anikin.

Table 2: Number of items Suspended
rocks kittens daggers wookies
Luke 25 15 30 3
Rey 30 5 45 2
Anikin 25 25 40 5

4.  Jedi trainee can suspend in the air after 12 weeks of training. Do an ANOVA analysis - Randomized Block Design at alpha = 0.05. Put the output in the indicated area. Is there a difference in the 3 kids overall ability to suspend objects in the air (rows)? Is there a difference in the type of object that is suspended objects in the air (columns)? Put the p-factor and conclusion in the indicated space.

5. Based upon the data in Table 2, perform a Chi-square Test Of Independence to determine if the type of suspended objects is dependent upon the Jedi trainee doing the suspending.

In: Statistics and Probability

The following data is representative of that reported in an article on nitrogen emissions, with x...

The following data is representative of that reported in an article on nitrogen emissions, with x = burner area liberation rate (MBtu/hr-ft2) and y = NOx emission rate (ppm):

x 100 125 125 150 150 200 200 250 250 300 300 350 400 400
y 150 130 190 220 190 330 280 400 420 430 400 590 610 680

(a) Assuming that the simple linear regression model is valid, obtain the least squares estimate of the true regression line. (Round all numerical values to four decimal places.)
y =____



(b) What is the estimate of expected NOx emission rate when burner area liberation rate equals 225? (Round your answer to two decimal places.)
ppm _______

(c) Estimate the amount by which you expect NOx emission rate to change when burner area liberation rate is decreased by 50. (Round your answer to two decimal places.)
ppm_______


(d) Would you use the estimated regression line to predict emission rate for a liberation rate of 500? Why or why not?

Yes, the data is perfectly linear, thus lending to accurate predictions.

Yes, this value is between two existing values.    

No, this value is too far away from the known values for useful extrapolation.

No, the data near this point deviates from the overall regression model.

In: Statistics and Probability

1.a We have 12 dice: 8 are regular and 4 are irregular. The probability of getting...

1.a We have 12 dice: 8 are regular and 4 are irregular. The probability of getting a 3 with an irregular dice is twice the probability of anyone of the rest of the numbers. 1) Find the probability of getting a 3 2) If we have got a 3, find the probability of being tossed with a regular dice 3) Find the probability of getting a 3 with an irregular dice (0.8 points)

1.b Which one is true and why? P(A|B) + P(B|A) = 1 or P(A|B) + P(Ac|B) = 1 (0.2 points)

In: Statistics and Probability

A research laboratory was developing a new compound for the relief of severe cases of hay...

A research laboratory was developing a new compound for the relief of severe cases of hay fever. In an experiment with 36 volunteers, the amounts of the two active ingredients (A & B) in the compound were varied at three levels each. Randomization was used in assigning four volunteers to each of the nine treatments. The data on hours of relief can be found in the following .csv file: Fever.csv

State the Null and Alternate Hypothesis for conducting one-way ANOVA for both the variables ‘A’ and ‘B’ individually.

1.2) Perform one-way ANOVA for variable ‘A’ with respect to the variable ‘Relief’. State whether the Null Hypothesis is accepted or rejected based on the ANOVA results.

1.3) Perform one-way ANOVA for variable ‘B’ with respect to the variable ‘Relief’. State whether the Null Hypothesis is accepted or rejected based on the ANOVA results.

1.4) Analyse the effects of one variable on another with the help of an interaction plot.
What is an interaction between two treatments?
[hint: use the ‘pointplot’ function from the ‘seaborn’ function]

1.5) Perform a two-way ANOVA based on the different ingredients (variable ‘A’ & ‘B’) with the variable 'Relief' and state your results.

1.6) Mention the business implications of performing ANOVA for this particular case study.

A B Volunteer Relief
1 1 1 2.4
1 1 2 2.7
1 1 3 2.3
1 1 4 2.5
1 2 1 4.6
1 2 2 4.2
1 2 3 4.9
1 2 4 4.7
1 3 1 4.8
1 3 2 4.5
1 3 3 4.4
1 3 4 4.6
2 1 1 5.8
2 1 2 5.2
2 1 3 5.5
2 1 4 5.3
2 2 1 8.9
2 2 2 9.1
2 2 3 8.7
2 2 4 9
2 3 1 9.1
2 3 2 9.3
2 3 3 8.7
2 3 4 9.4
3 1 1 6.1
3 1 2 5.7
3 1 3 5.9
3 1 4 6.2
3 2 1 9.9
3 2 2 10.5
3 2 3 10.6
3 2 4 10.1
3 3 1 13.5
3 3 2 13
3 3 3 13.3
3 3 4 13.2

In: Statistics and Probability

A manufacturing company produces electrical insulators. If the insulators break when in use, a short circuit...

A manufacturing company produces electrical insulators. If the insulators break when in use, a short circuit is likely to occur. To test the strength of the insulators, destructive testing is carried out to determine how much force is required to break the insulators. Force is measured by observing the number of pounds of force applied to the insulator before it breaks. The following data (stored in Force) are from 30 insulators subjected to this testing:

  1. At the 0.05 level of significance, is there evidence that the population mean force required to break the insulator is greater than 1,500 pounds? Be sure to write out the hypotheses clearly in your output and to clearly state your conclusion.
  2. What assumption about the population distribution is needed in order to conduct the t test in (a)?
  3. Construct a histogram, boxplot, or normal probability plot to evaluate the assumption made in (b). Based on the evidence obtained from the plot, is the assumption needed for the t-test valid? Explain.
Force
1870
1728
1656
1610
1634
1784
1522
1696
1592
1662
1866
1764
1734
1662
1734
1774
1550
1756
1762
1866
1820
1744
1788
1688
1810
1752
1680
1810
1652
1736

Provide steps for both answer

In: Statistics and Probability

Brand Display_size Price Storage Screen_type Purchase Likelihood Combination1 Samsung 10" display $499 32 GB LCD 10...

Brand Display_size Price Storage Screen_type Purchase Likelihood
Combination1 Samsung 10" display $499 32 GB LCD 10
Combination2 Apple 10" display $499 16 GB LCD 7
Combination3 Apple 8" display $399 32 GB LCD 2
Combination4 Samsung 8" display $399 16 GB LCD 8
Combination5 Samsung 8" display $499 32 GB LED 9
Combination6 Apple 10" display $399 32 GB LED 9
Combination7 Samsung 10" display $399 16 GB LED 10
Combination8 Apple 8" display $499 16 GB LED 6

QUESTION 1

  1. Graph includes responses of a consumer. The consumer was given 8 different combinations of tablets and indicated how likely he/she would be to purchase each combination.


    Run regression analysis for the consumer (Code Apple as 1, 10'' display as 10, $499 as 499, 32GB as 32, LED as 1).

    What is the coefficient of Brand?

In: Statistics and Probability

The table shows the number of confirmed COVID-19 infected cases, number of death and number of...

The table shows the number of confirmed COVID-19 infected cases, number of death and number of recovered cases in four countries according to 5/4/2020 global record. The total confirmed cases around the world is 1249636, the total number of death around the world is 68069, and the total number of recovered cases around the world is 256860.

Calculate the following:

1-     The % of death of the number of the infected cases in each of the four countries.

2-     The % of death of the total recorded death cases around the world in each of the four countries.

3-     The % of the recovered cases of the infected cases in each of the four countries.

4-     The % of the recovered cases of the total recovered cases around the world in each of the four countries.

5-     The % of COVID-19 infected cases of the total confirmed cases around the world for each of the four countries.

Country

China

Italy

Spain

USA

Infected cases

81669

128948

130759

327848

Death

3329

15887

12418

9325

Recovered cases

76964

21815

38080

16700

In: Statistics and Probability

According to U.S. Department of Transportation report, 8% of all the registered cars fail the emissions...

According to U.S. Department of Transportation report, 8% of all the registered cars fail the emissions test.

a)For a random sample of 10 cars, what is the chance that 4 of them will fail the test?

b) how many of them are expected to fail the test?

c)For a random sample of 10 cars, find the probability that more than 8 of them will fail the test?

In: Statistics and Probability

What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=0.27?...

What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=0.27? (Do not round your answer; compute your answer using a value from the table below.)

z0.10.20.30.40.50.000.5400.5790.6180.6550.6910.010.5440.5830.6220.6590.6950.020.5480.5870.6260.6630.6980.030.5520.5910.6290.6660.7020.040.5560.5950.6330.6700.7050.050.5600.5990.6370.6740.7090.060.5640.6030.6410.6770.7120.070.5670.6060.6440.6810.7160.080.5710.6100.6480.6840.7190.090.5750.6140.6520.6880.722

In: Statistics and Probability

Suppose a grocery store is considering the purchase of a new self-checkout machine that will get...

Suppose a grocery store is considering the purchase of a new self-checkout machine that will get customers through the checkout line faster than their current machine. Before he spends the money on the equipment, he wants to know how much faster the customers will check out compared to the current machine. The store manager recorded the checkout times, in seconds, for a randomly selected sample of checkouts from each machine. The summary statistics are provided in the table.

Group Description Sample
size
Sample
mean (min)
Sample standard
deviation (min)
Standard error
estimate (min)
1 old machine n1=51n1=51 ¯¯¯x1=128.4x¯1=128.4 s1=29.8s1=29.8 SE1=4.17283SE1=4.17283
2 new machine n2=48n2=48 ¯¯¯x2=113.0x¯2=113.0 s2=24.2s2=24.2 SE2=3.49297SE2=3.49297

df=94.99897df=94.99897

Compute the lower and upper limits of a 95% confidence interval to estimate the difference of the mean checkout times for all customers. Estimate the difference for the old machine minus the new machine, so that a positive result reflects faster checkout times with the new machine. Use the Satterthwaite approximate degrees of freedom, 94.99897. Give your answers precise to at least three decimal places.

upper limit:

lower limit:

In: Statistics and Probability