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.
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 |
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 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.
1,500
1,300
800
None
symmetric skewed
positively skewed
negatively skewed
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In: Statistics and Probability
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
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.
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 = 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 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 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 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:
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 |
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
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 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 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? (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 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