88 - JustAnswer



88. Refer to the Baseball 2005 data, which reports information on the 30 major league teams for the 2005 baseball season.

a. Select the variable team salary and find the mean, median, and the standard deviation.

b. Select the variable that refers to the age the stadium was built. (Hint: Subtract the

year in which the stadium was built from the current year to find the stadium age and

work with that variable.) Find the mean, median, and the standard deviation.

c. Select the variable that refers to the seating capacity of the stadium. Find the mean,

median, and the standard deviation.

Solution:

a. Select the variable team salary and find the mean, median, and the standard deviation.

|Team |Salary( in Million Dollars) |

|Boston |123.5 |

|New York Yankees |208.3 |

|Oakland |55.4 |

|Baltimore |73.9 |

|Los Angeles Angels |97.7 |

|Cleveland |41.5 |

|Chicago White Sox |75.2 |

|Toronto |45.7 |

|Minnesota |56.2 |

|Tampa Bay |29.7 |

|Texas |55.8 |

|Detroit |69.1 |

|Seattle |87.8 |

|Kansas City |36.9 |

|Atlanta |86.5 |

|Arizona |62.3 |

|Houston |76.8 |

|Cincinnati |61.9 |

|New York Mets |101.3 |

|Pittsburgh |38.1 |

|Los Angeles Dodgers |83 |

|San Diego |63.3 |

|Washington |48.6 |

|San Francisco |90.2 |

|St. Louis |92.1 |

|Florida |60.4 |

|Philadelphia |95.5 |

|Milwaukee |39.9 |

|Chicago Cubs |87 |

|Colorado |48.2 |

|Mean |$73.0600 |

|Median |66.2 |

|Standard Deviation |34.23285 |

b. Select the variable that refers to the age the stadium was built. (Hint: Subtract the

year in which the stadium was built from the current year to find the stadium age and

work with that variable.) Find the mean, median, and the standard deviation.

|Team |Built Age |Age (2009-Built Age) |

|Boston |1912 |97 |

|New York Yankees |1923 |86 |

|Oakland |1966 |43 |

|Baltimore |1992 |17 |

|Los Angeles Angels |1966 |43 |

|Cleveland |1994 |15 |

|Chicago White Sox |1991 |18 |

|Toronto |1989 |20 |

|Minnesota |1982 |27 |

|Tampa Bay |1990 |19 |

|Texas |1994 |15 |

|Detroit |2000 |9 |

|Seattle |1999 |10 |

|Kansas City |1973 |36 |

|Atlanta |1993 |16 |

|Arizona |1998 |11 |

|Houston |2000 |9 |

|Cincinnati |2003 |6 |

|New York Mets |1964 |45 |

|Pittsburgh |2001 |8 |

|Los Angeles Dodgers |1962 |47 |

|San Diego |2004 |5 |

|Washington |1961 |48 |

|San Francisco |2000 |9 |

|St. Louis |1966 |43 |

|Florida |1987 |22 |

|Philadelphia |2004 |5 |

|Milwaukee |2001 |8 |

|Chicago Cubs |1914 |95 |

|Colorado |1995 |14 |

|Mean |28.2 |

|Median |17.5 |

|Standard Deviation |25.94477 |

c. Select the variable that refers to the seating capacity of the stadium. Find the mean,

median, and the standard deviation.

|Team |Seating Capacity of the Stadium |

|Boston |33871 |

|New York Yankees |57746 |

|Oakland |43662 |

|Baltimore |48262 |

|Los Angeles Angels |45050 |

|Cleveland |43368 |

|Chicago White Sox |44321 |

|Toronto |50516 |

|Minnesota |48678 |

|Tampa Bay |44027 |

|Texas |52000 |

|Detroit |40000 |

|Seattle |45611 |

|Kansas City |40529 |

|Atlanta |50062 |

|Arizona |49075 |

|Houston |42000 |

|Cincinnati |42059 |

|New York Mets |55775 |

|Pittsburgh |38127 |

|Los Angeles Dodgers |56000 |

|San Diego |42445 |

|Washington |56000 |

|San Francisco |40800 |

|St. Louis |49625 |

|Florida |42531 |

|Philadelphia |43500 |

|Milwaukee |42400 |

|Chicago Cubs |38957 |

|Colorado |50381 |

|Mean |45912.60 |

|Median |44174.00 |

|Standard Deviation |5894.20 |

56. Assume the likelihood that any flight on Northwest Airlines arrives within 15 minutes of the scheduled time is .90. We select four flights from yesterday for study.

a. What is the likelihood all four of the selected flights arrived within 15 minutes of the

scheduled time?

b. What is the likelihood that none of the selected flights arrived within 15 minutes of

the scheduled time?

c. What is the likelihood at least one of the selected flights did not arrive within 15 minutes

of the scheduled time?

Solution:

Let X be the sample parameter for any flight on Northwest Airlines arrives within 15 minutes of

the scheduled time.

Sample of size, n = 4

p = 0.9.

q= 0.1

a. P(X=4) = 0.94 = 0.6561

b. P(X=0) = (1-0.9)4 = 0.0001

c. P (X≤3) = 1 - 0.6561 = 0.3439 since P (X≤3) = 1- P(X=4)

64. An internal study by the Technology Services department at Lahey Electronics revealed

company employees receive an average of two emails per hour. Assume the arrival of

these emails is approximated by the Poisson distribution.

a. What is the probability Linda Lahey, company president, received exactly 1 email

between 4 P.M. and 5 P.M. yesterday?

b. What is the probability she received 5 or more email during the same period?

c. What is the probability she did not receive any email during the period?

Solution:

[pic]

where

• e is the base of the natural logarithm (e = 2.71828...)

• k is the number of occurrences of an event - the probability of which is given by the function

• k! is the factorial of k

• λ is a positive real number, equal to the expected number of occurrences that occur during the given interval. For instance, if the events occur on average 4 times per minute, and you are interested in the number of events occurring in a 10 minute interval, you would use as your model a Poisson distribution with λ = 10×4 = 40.

(Source: )

(a) P(received exactly 1 email between 4 P.M. and 5 P.M. yesterday)=f(1,2) = 2e-2 = 0.2707

(b)P(received 5 or more email during the same period) =1-k=0∑k=4f(k;λ)

= 1-[(e-2+2e-2+2e-2+(4/3)e-2+(2/3)e-2] = 0.05265

(c) P(she did not receive anyemail during period) = f(0,2)= e-2 = 0.13534

50. Fast Service Truck Lines uses the Ford Super Duty F-750 exclusively. Management made

a study of the maintenance costs and determined the number of miles traveled during

the year followed the normal distribution. The mean of the distribution was 60,000 miles

and the standard deviation 2,000 miles.

a. What percent of the Ford Super Duty F-750s logged 65,200 miles or more?

b. What percent of the trucks logged more than 57,060 but less than 58,280 miles?

c. What percent of the Fords traveled 62,000 miles or less during the year?

d. Is it reasonable to conclude that any of the trucks were driven more than 70,000 miles?

Explain.

Solution:

a. What percent of the Ford Super Duty F-750s logged 65,200 miles or more?

Here, sample mean is 65,200 miles and population mean is 60,000 miles.

the standard deviation is 2,000 miles.

(a) Calculation of test-statistics,

z = (X - μ) / σx

Where X is a normal random variable, μ is the mean, and σx is the standard deviation.

Hence, z= 65,200-60,000/2,000 = 5,200/2000 = 2.6

P (the Ford Super Duty F-750s logged 65,200 miles or more) = P(Z>2.6)=1-P(Z 106.33 < n

This gives the sample size of 107.

42. During recent seasons, Major League Baseball has been criticized for the length of the

games. A report indicated that the average game lasts 3 hours and 30 minutes. A sample

of 17 games revealed the following times to completion. (Note that the minutes have

been changed to fractions of hours, so that a game that lasted 2 hours and 24 minutes

is reported at 2.40 hours.)

[pic]

Solution:

Size of sample, n = 17

Degree of freedom = n-1 = 17-1 = 16

Sum of sample = i=1∑n=17xi = (2.98 + 2.40 + 2.70 + 2.25 + 3.23 + 3.17 + 2.93 + 3.18 +2.80

+2.38+ 3.75+ 3.20+ 3.27+ 2.52+ 2.58+ 4.45+ 2.45) = 50.24

| |Time (in Hours) |Sum |Mean(xm) |(xi- xm)2 |Standard Deviation |

|1 |2.98 |50.24 |Mean = Sum/n |0.000605 |σ = √( i=1∑n=25(xi-xm)2/(n-1)) |

| | | |= 50.24/17 | |= √(5.009623/16) |

| | | |= 2.9552 | |= √0.3131 |

| | | | | |=0.5595 |

|2 |2.4 | | |0.308462 | |

|3 |2.7 | | |0.065226 | |

|4 |2.25 | | |0.497581 | |

|5 |3.23 | | |0.075408 | |

|6 |3.17 | | |0.046056 | |

|7 |2.93 | | |0.000645 | |

|8 |3.18 | | |0.050448 | |

|9 |2.8 | | |0.024147 | |

|10 |2.38 | | |0.331078 | |

|11 |3.75 | | |0.631399 | |

|12 |3.2 | | |0.059832 | |

|13 |3.27 | | |0.098977 | |

|14 |2.52 | | |0.189568 | |

|15 |2.58 | | |0.140921 | |

|16 |4.45 | | |2.233847 | |

|17 |2.45 | | |0.255423 | |

Here X = 2.9552

σ = 0.5595

We can define one-tailed statistics for above observations as follows:

Null Hypothesis: H0: ( =3.5 vs. Ha: ( < 3.5

Rejection region:

t < -t(

Here significance level is 0.05,

So, t0.05 = - 2.120 (Using Statistical Ratio Calculator

From for calculating t with 0.05 significance and DF =16)

I am assuming the sample is selected independently and randomly from population. Population size is sufficiently large in sample.

Now,

t = (X - μ) / (s/√n)

Where X is a normal random variable, μ is the mean, and s is the standard deviation.

and n is the sample size.

Calculating t-test statistics:

t = 2.9552-3.5/(0.5595/√17) = -0.5448* √17/0.5595= -4.0147

Calculating p-value:

Degree of freedom = DF = 17-1 = 16

P(t16 < -4.0147) = 0.000501 (Using with t=-4.0147and DF = 16)

The p-value of 0.0501% is less than significant level of 5%. Hence, we can reject the null hypothesis and conclude that the mean time for a game is less than 3.50 hours.

58. The amount of income spent on housing is an important component of the cost of living.

The total costs of housing for homeowners might include mortgage payments, property

taxes, and utility costs (water, heat, electricity). An economist selected a sample of

20 homeowners in New England and then calculated these total housing costs as a percent

of monthly income, five years ago and now. The information is reported below. Is it

reasonable to conclude the percent is less now than five years ago?

[pic]

Solution:

Let us define,

Total housing costs as a percent of monthly income five years ago = C-5

Current total housing costs as a percent of monthly income five years ago = C0

So, differenec between housing costs as a percent of monthly income five years ago and current value = ∆C = C-5 - C0

Null Hypothesis: H0: ∆C = 0 vs H1: ∆C > 0

Rejection region:

t > t(

Here significance level is 0.05,

So, t0.05 = 1.729 (Using Statistical Ratio Calculator

From for calculating t with 0.05 significance and DF =19)

Table for Paired t-Test Calculation:

|Homeowner |5 Years Ago |Now |∆C |

|1 |17 |10 |7 |

|2 |20 |39 |-19 |

|3 |29 |37 |-8 |

|4 |43 |27 |16 |

|5 |36 |12 |24 |

|6 |43 |41 |3 |

|7 |45 |24 |21 |

|8 |19 |26 |-7 |

|9 |49 |28 |21 |

|10 |49 |26 |23 |

|11 |35 |32 |3 |

|12 |16 |32 |-16 |

|13 |23 |21 |2 |

|14 |33 |12 |21 |

|15 |44 |40 |4 |

|16 |44 |42 |2 |

|17 |28 |22 |6 |

|18 |29 |19 |10 |

|19 |39 |35 |4 |

|20 |22 |12 |10 |

Mean X =6.35

Standard Deviation (xs) = 12.4778

I am assuming the sample is selected independently and randomly from population. Population size is sufficiently large in sample.

Now,

t = (X - μ) / (s/√n)

Where X is a normal random variable, μ is the mean, and s is the standard deviation.

n is the sample size.

Calculating t-test statistics:

t = (6.35 -0) / (12.4778 / √20) = 2.275

Conclusion: Since calculated value of t is greater than critical value of t i.e.2.275 > 1.729. We reject the null hypothesis and conclude that the percent is less now than 5 years ago.

42. Martin Motors has in stock three cars of the same make and model. The president would

like to compare the gas consumption of the three cars (labeled car A, car B, and car C)

using four different types of gasoline. For each trial, a gallon of gasoline was added to

an empty tank, and the car was driven until it ran out of gas. The following table shows

the number of miles driven in each trial.

[pic]

Solution:

|Results of ANOVA | |

| | | | | | | | |

| | | | | | | | |

| |Summary |Count |Sum |Average |Variance | | |

| |Regular |3 |64.7 |21.57 |0.64 | | |

| |Super Regular |3 |57.1 |19.03 |3.52 | | |

| |Unleaded |3 |60.6 |20.2 |1.00 | | |

| |Premium Unleaded |3 |59.3 |19.77 |1.02 | | |

| | | | | | | | |

| |Car A |4 |78.9 |19.73 |5.06 | | |

| |Car B |4 |79 |19.75 |0.92 | | |

| |Car C |4 |83.8 |20.95 |0.24 | | |

| | | | | | | | |

| | | | | | | | |

| |ANOVA | | | | | | |

| |Source of Variation |SS |DF |MS |F |P-Value |FCrit |

| |Rows |10.21 |3 |3.40 |2.41 |0.16 |4.76 |

| |Columns |3.92 |2 |1.96 |1.39 |0.32 |5.14 |

| |Errors |8.46 |6 |1.41 | | | |

| | | | | | | | |

| | | | | | | | |

| |Hence applying 0.05 level of significance, |

| |a.     F = 2.41, no difference as F < FCrit | | | | | |

| |b.     F = 1.39, no difference as F < FCrit | | | | | |

37. A regional commuter airline selected a random sample of 25 flights and found that the

correlation between the number of passengers and the total weight, in pounds, of luggage

stored in the luggage compartment is 0.94. Using the .05 significance level, can we

conclude that there is a positive association between the two variables?

Solution:

Null Hypothesis: H0: ρ = 0 ( There is no positive association the number of passengers and the total weight) vs.

Ha: ρ > 0 ( There is positive association between the number of passengers and the total weight)

|Degre of Freedom= |n-2= |23 | |

|one-tailed test, critical value of t = |1.713872 | |

|Correlation Coefficient |r= |0.94 | | | | |

| |13.21 | | | | | |

| | | | | | | |

| | | | | | | |

| | | | | | | |

|Since 13.21 > 1.31, we can conclude that there is a positive association between no of | | | | |

|passengers and total weight of luggages | | | | |

| | | | | |

| | | | | |

| | | | | |

40. A suburban hotel derives its gross income from its hotel and restaurant operations. The

owners are interested in the relationship between the number of rooms occupied on a

nightly basis and the revenue per day in the restaurant. Below is a sample of 25 days

(Monday through Thursday) from last year showing the restaurant income and number of

rooms occupied.

[pic]

Solution :

Attached excel sheet has solution.

[pic]

17. The district manager of Jasons, a large discount electronics chain, is investigating why

certain stores in her region are performing better than others. She believes that three factors

are related to total sales: the number of competitors in the region, the population in

the surrounding area, and the amount spent on advertising. From her district, consisting

of several hundred stores, she selects a random sample of 30 stores. For each store she

gathered the following information.

Y _ total sales last year (in $ thousands).

X1 _ number of competitors in the region.

X2 _ population of the region (in millions).

X3 _ advertising expense (in $ thousands).

[pic]

Multiple Regression and Correlation Analysis 549

a. What are the estimated sales for the Bryne store, which has four competitors, a

regional population of 0.4 (400,000), and advertising expense of 30 ($30,000)?

b. Compute the R2 value.

c. Compute the multiple standard error of estimate.

d. Conduct a global test of hypothesis to determine whether any of the regression coefficients

are not equal to zero. Use the .05 level of significance.

e. Conduct tests of hypotheses to determine which of the independent variables have

significant regression coefficients. Which variables would you consider eliminating?

Use the .05 significance level.

Solution:

|Y = total sales last year (in $ thousands). |

|X1 = number of competitors in the region. |

|X2 =population of the region (in millions). | |

|X3 = advertising expense (in $ thousands). |

|Analysis of variance |  |  |

|SOURCE |DF |SS |MS |

|Regression |3 |3050.00 |1016.67 |

|Error |26 |2200.00 |84.62 |

|Total |29 |5250.00 | |

|  | | | | |

|Predictor |Coef |StDev |t-ratio |

|Constant |14.00 |7.00 |2.00 |

|X1 |-1.00 |0.70 |-1.43 |

|X2 |30.00 |5.20 |5.77 |

|X3 |0.20 |0.08 |2.50 |

a. What are the estimated sales for the Bryne store, which has four competitors, a regional population of 0.4 (400,000), and advertising expense of 30 ($30,000)?

Solution:

The regression equation for sales, Y = -1.00 * X1 + 30 * X2 + 0.20 * X3 + 14

= - X1 + 30 X2 + 0.20 X3 + 14

So, the estimated sales for Bryne store,

Y = -4+30*0.4+0.20*30+14 = -4+12+6+14 = $28 thousands

b. Compute the value of R2

Solution:

R² = SSR/SST = 3050.00/5250.00 = 0.58

c. Compute the multiple standard error of estimate.

Solution:

Multiple Standard error of the estimate = ( MSE =(0.58 = 0.76

d. Conduct a global test of hypothesis to determine whether any of the regression coefficients are not equal to zero. Use the .05 level of significance.

Solution:

Null Hypothesis:

Ho: the regression coefficients are equal to zero.

Ha: the regression coefficients are not equal to zero.

Significant level:

α = 0.05

Critical Value of F:

DFn = 3

DFd = 26

Critical Table value of F at 0.05 significant level,

Probability = 0.05 DFn=3 DFd=26

Fcrit= 2.9752

5.00% of a F distribution has values greater than 2.975.

(Using )

Calculating F-ratio:

F = MSR/MSE = 1016.67/84.62 = 12.01

Conclusion:

Since F > Fcrit , we can reject the null hypothesis and conclude that the regression coefficients are not equal to zero.

e. Conduct tests of hypotheses to determine which of the independent variables have significant regression coefficients. Which variables would you consider eliminating?

Use the .05 significance level.

Solution:

|Predictor |Coef |StDev |t-ratio |

|Constant |14.00 |7.00 |2.00 |

|X1 |-1.00 |0.70 |-1.43 |

|X2 |30.00 |5.20 |5.77 |

|X3 |0.20 |0.08 |2.50 |

Null Hypothesis:

Ho: Independent variables do not affect the sales price

Ha: Independent variables affect the sales price

Significant level:

α = 0.05

Decision Rule:

Reject the null hypothesis if t > tcrit

Critical value of t:

Probability = 0.0499999156603 df=29

Two tailed: tcrit= 2.0452

5.000% of a t distribution has values greater than 2.045 or less than -2.0452.

(Using )

Conclusion:

Since 5.77 > 2.045 for X2, we reject the null hypothesis.

Since calculated value of t is less that tcrit for X1 , X3 , we fail to reject the null hypothesis.

Hence, we can consider eliminating X1 ,and X3 ,

18. Suppose that the sales manager of a large automotive parts distributor wants to estimate

as early as April the total annual sales of a region. On the basis of regional sales,

the total sales for the company can also be estimated. If, based on past experience, it

is found that the April estimates of annual sales are reasonably accurate, then in future

years the April forecast could be used to revise production schedules and maintain the

correct inventory at the retail outlets.

Several factors appear to be related to sales, including the number of retail outlets

in the region stocking the company’s parts, the number of automobiles in the region registered

as of April 1, and the total personal income for the first quarter of the year. Five

independent variables were finally selected as being the most important (according to

the sales manager). Then the data were gathered for a recent year. The total annual sales

for that year for each region were also recorded. Note in the following table that for region

1 there were 1,739 retail outlets stocking the company’s automotive parts, there were

9,270,000 registered automobiles in the region as of April 1 and so on. The sales for that

year were $37,702,000.

[pic]

a. Consider the following correlation matrix. Which single variable has the strongest correlation

with the dependent variable? The correlations between the independent variables

outlets and income and between cars and outlets are fairly strong. Could this

be a problem? What is this condition called?

[pic]

b. The output for all five variables is on the following page. What percent of the variation

is explained by the regression equation?

[pic]

c. Conduct a global test of hypothesis to determine whether any of the regression coefficients

are not zero. Use the .05 significance level.

d. Conduct a test of hypothesis on each of the independent variables. Would you consider

eliminating “outlets” and “bosses”? Use the .05 significance level.

e. The regression has been rerun below with “outlets” and “bosses” eliminated. Compute

the coefficient of determination. How much has R2 changed from the previous analysis?

[pic]

f. Following is a histogram and a stem-and-leaf chart of the residuals. Does the normality

assumption appear reasonable?

[pic]

g. Following is a plot of the fitted values of Y (i.e., Yˆ) and the residuals. Do you see any

violations of the assumptions?

Multiple Regression and Correlation Analysis

[pic]

Solution:

|  |  |Number of |  |Average |  |

|Annual |Number of |Automobiles |Personal |Age of |  |

|Sales |Retail |Registered |Income |Automobiles |Number of |

|($ millions), |Outlets, |(millions), |($ billions), |(years) |Supervisors |

|Y |X1 |X2 |X3 |X4 |X5 |

| 37,702 | 1,739 |9.27 |85.4 |3.5 |9 |

| 24,196 | 1,221 |5.86 |60.7 |5 |5 |

| 32,055 | 1,846 |8.81 |68.1 |4.4 |7 |

| 3,611 |120 |3.81 |20.2 |4 |5 |

| 17,625 | 1,096 |10.31 |33.8 |3.5 |7 |

| 45,919 | 2,290 |11.62 |95.1 |4.1 |13 |

| 29,600 | 1,687 |8.96 |69.3 |4.1 |15 |

| 8,114 |241 |6.28 |16.3 |5.9 |11 |

| 20,116 | | |34.9 |5.5 |16 |

| |649 |7.77 | | | |

| 12,994 | 1,427 | | | | |

| | |10.92 |15.1 |4.1 |10 |

a. Consider the following correlation matrix. Which single variable has the strongest correlation with the dependent variable? The correlations between the independent variables outlets and income and between cars and outlets are fairly strong. Could this be a problem? What is this condition called?

|  | sales |outlets |cars |inome |age |

| | 0.899 | | | |  |

|outlets | | | | | |

| | | | | |  |

| | | | | | |

|cars |0.605 |0.775 | | | |

| | 0.964 | | | |  |

|income | |0.825 |0.409 | | |

| | | | | |  |

|age |-0.323 |-0.489 |-0.447 |-0.349 | |

| | | | 0.395 | | |

|bosses |0.286 |0.183 | |0.155 |0.291 |

Solution:

From above correlation matrix, highest value of correlation 0.964 is between Income and Sales. Other correlations between independent variables like Income and Outlets, Outlets and Sales, Cars and Outlets has also strong positive correlation since correlation co-efficient for all these related variables are greater than 0.7. This condition is called “MultiCollinearity”. Yes, it could be a problem since the Multiple Standard Error of Estimate (MSE) can be distorted by multicollinearity which can affect the result of experiment by changing the degree of statistical significance of independent variables.

b. The output for all five variables is on the following page. What percent of the variation is explained by the regression equation?

|The regression equation is |  |  |  |  |  |

| sales = -19.7 - 0.00063 outlets + 1.74 cars + 0.410 income + 2.04 age - 0.034 bosses |  |

|  | | | | | |

|  |SOURCE |DF |SS |MS | |  |

|  |Regression |5 |1593.81 |318.76 | |  |

|  |Error |4 |9.08 |2.27 | |  |

|  |Total |9 |1602.89 | | |  |

|  |  |  |  |  |  |

| sales = -18.9 + 1.61 cars + 0.400 income + 1.96 age | | |  |

|  | | | | | |

|  |SOURCE |DF |SS |MS | |  |

|  |Regression |3 |1593.66 |531.22 | |  |

|  |Error |6 |9.23 |1.54 | |  |

|  |Total |9 |1602.89 | | |  |

|  |  |  |  |  |

|0 |50 |40.601 |2.176 | |

|1 |77 |81.201 |0.217 | |

|2 |81 |81.201 |0.000 | |

|3 |48 |54.134 |0.695 | |

|4 |31 |27.067 |0.571 | |

|5+ |13 |15.796 |0.495 | |

|Total: |300 |300 |4.155 |0.5272 |

[pic]

(Source: )

Since calculated value of p is 0.5272 which is greater than significance level 0.05. Hence, we do not reject the null hypothesis and we don’t have enough evidence to say that the population distribution is not Poisson with a mean of 2.0.

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