BA 578 Live Fin Set2

Submitted By: Humayra SalauddinCWID: 50068944

BA 578-Live: Spring 2011: Final Exam: Total 400 points

Starts at 3 pm on Tuesday, May 10th: to be emailed back by 10 pm the same day.

Name:-

True/False (6 points each)

1. The sampling distribution of the sample mean is always normally distributed according to the Central Limit Theorem.(Ch7)

True False

2. In performing a chi-square test of independence, as the difference between the respective observed and expected frequencies decrease, the probability of concluding that the row variable is independent of the column variable increases. (Ch12)

True False

3. In a regression model, a value of the error term depends upon other values of the error term. (Ch13)

True False

4. In a regression model, at any given combination of values of the independent variables, the population of potential error terms is assumed to have an F-distribution. (Chs. 13 and 14)

True False

5. In testing the difference between two means from two independent populations, the sample sizes do not have to be equal to be able to use the Z statistic. (Ch10)

True False

1


6. An estimator is called consistent if its variance and standard deviations consistently remain the same regardless of changes in the sample size.(Ch7)

True False

7. For a binomial probability experiment, with n=150 and p = .1, we can use the normal approximation to the binomial distribution even without continuity correction.(Ch6)

True False

8. The error term in the regression model describes the effects of all factors other than the independent variables on y (response variable). (Ch13)

True False

9. The correlation coefficient is the ratio of explained variation to total variation. (Ch13)

True False

10. For a continuous distribution, Probability of (X less than or equal to 100) is greater than the probability of (X less than 100)(Ch6)

True False

11. We do not need to perform the continuity correction if the population is 20 times or more than the sample size.(Ch6)

True False

12. For a hypothesis test about a population proportion or mean, if the level of significance is less than the p-value, the null hypothesis is rejected.(Ch9)

2


True False

13. When determining the sample size n, if the value found for n is 79.2, we would choose to sample 79 observations. (Ch8)

True False

14. The level of significance indicates the probability of rejecting a false null hypothesis. (Ch9)

True False

15. When the level of confidence and sample standard deviation remain the same, a confidence interval for a population mean based on a sample of n=100 will be narrower than a confidence interval for a population mean based on a sample of n=50. (Ch8)

True False

Multiple Choices(10 points each)

16. To investigate the rate at which employees with cancer are fired or laid off, a telephone survey was taken of 100 cancer survivors who worked while undergoing treatment. Seven (7) were either fired or laid off due to their illness. Construct a 90% confidence interval for the true percentage of all cancer patients who are fired or laid off due to their illness. (Ch 8)

A. [0.0000 0.2034] B. [0.0371 0.1029] C. [0.0039 0.1361] D. [0.0078 0.1400] E. [0.0278 0.1122]

3


17. A manufacturer of a chemical used in glue, attempting to control the amount of a hazardous chemical its workers are exposed to, has given instructions to halt production if the mean amount in the air exceeds 3.0ppm. A random sample of 50 air specimens produced the following statistics: sample mean = 3.1 ppm , sample standard deviation = 0.5 ppm. Calculate the appropriate test statistic to test the hypotheses. (Ch 9)

A. 0.20 B. -0.20 C. 10.00 D. 1.41 E. -1.41

18. When we carry out a chi-square test of independence, as the difference between the respective observed and expected frequencies decrease, the probability of concluding that the row variable is independent of the column variable: (Ch12)

A. Decreases B. Increases C. May increase or decrease depending on the number of rows and columns D. Will be unaffected

19. A manufacturing operation consists of a unique system that produces an average of 15.5 jet engine propulsion parts every hour. After undergoing a complete overhaul, the system was monitored by observing the number of parts produced in each of sixteen randomly selected one-hour periods. The mean is 15.42 with a standard deviation of 0.16. Calculate the appropriate test statistic to test the hypotheses. (Ch 9)

A. -8.00 B. -2.00 C. 8.00 D. 2.00 E. -0.50

4


20. The MPG (Miles per Gallon) for a mid-size car is normally distributed with a mean of 32 and a standard deviation of .8. What is the probability that the MPG for a selected mid-size car would be: More than 33.2? (Ch 6)

A. 43.32% B. 6.68% C. 93.32% D. 86.64% E. 13.36%

21. A multiple regression analysis with 20 observations on each of three independent variables and the dependent variable would yield ______ and ______ degrees of freedom respectively for regression (explained) and error. (Ch 14)

A. 3, 17 B. 3, 16 C. 4, 16 D. 3, 19 E. 3, 20

22. A major airline company is concerned that its proportion of late arrivals has substantially increased in the past month. Historical data shows that on the average 18% of the company airplanes have arrived late. In a random sample of 1,240 airplanes, 310 airplanes have arrived late. If we are conducting a hypothesis test of a single proportion to determine if the proportion of late arrivals has increased: What is the value of the calculated test statistic? (Cch 9)

A. Z = 3.208 B. Z = 6.416 C. Z = -3.208 D. Z = -6.416 E. Z = 1.833

23. A property of continuous distributions is that: (Ch6)

A. As with discrete random variables, the probability distribution can be approximated by a smooth curve B. Probabilities for continuous variables can be approximated using discrete

5


random variables C. Unlike discrete random variables, probabilities can be found using tables D. Unlike discrete random variables, the probability that a continuous random variable equals a specific value is zero [P(X=x)=0] E. The distribution is symmetric

24. In a manufacturing process, we are interested in measuring the average length of a certain type of bolt. Based on a preliminary sample of 100 bolts, the sample standard deviation is .3 inches. How many bolts should be sampled in order to make us 90% confident that the sample mean bolt length is within .02 inches of the true mean bolt length? (Ch8)

A. 865 B. 609 C. 1493 D. 100 E. 1000

25. The changing ecology of the swamps in Louisiana has been the subject of much environmental research. One water-quality parameter of concern is the total phosphorous level. Suppose that the EPA makes 15 measurements in one area of the swamp, yielding a mean level of total phosphorus of 12.3 parts per billion (ppb) and a standard deviation of 5.4 ppb. The EPA wants to test whether the data support the conclusion that the mean level is less than 15 ppb. Calculate the appropriate test statistic to test the hypotheses. (Ch 9)

A. 7.50 B. 1.94 C. 3.88 D. -1.94 E. -7.50

26. A state education agency designs and administers high school proficiency exams. Historically, time to complete the exam was an average of 120 minutes. Recently the format of the exam changed and the claim has been made that the time to complete the exam has changed. A sample of 50 new exam times yielded

6


an average time of 118 minutes. The standard deviation is assumed to be 5 minutes. Calculate a 99% confidence interval. (Ch 8)

A. [117.61 120.09] B. [117.36 119.39] C. [116.18 119.82] D. [115.67 120.33] E. [115.82 120.18]

27. In a manufacturing process a random sample of 36 bolts manufactured has a mean length of 3 inches with a standard deviation of .3 inches. What is the 99% confidence interval for the true mean length of the bolt? (Ch 8)

A. 2.902 to 3.098 B. 2.884 to 3.117 C. 2.871 to 3.129 D. 2.228 to 3.772 E. 2.902 to 3.098

28. A new company is in the process of evaluating its customer service. The company offers two types of sales: 1. Internet sales; 2. Store sales. The marketing research manager believes that the Internet sales are more than 10% higher than store sales. The null hypothesis would be: (Ch 10)

A. Pinternet–Pstore > .10 B. Pinternet–Pstore < .10 C. Pinternet–Pstore ≥ .10 D. Pinternet–Pstore ≤ .10 E. Pinternet–Pstore = .10

29. The range of feasible values for the multiple coefficient of determination is from: (Ch 14)

A. 0 to infinity B. -1 to 0 C. -1 to 1 D. 0 to 1 E. Any real value

7


30. If we are testing the significance of the independent variable X1 and we reject the null hypothesis H0: b1 = 0, we conclude that:(Chs. 13 & 14)

A. X1 is significantly related to Y B. X1 is not significantly related to Y C. X1 is an unimportant independent variable D. b1 is significantly related to the dependent variable Y

Essay Type (16 points each)

31. An apple juice producer buys all his apples from a conglomerate of apple growers in one northwest state. The amount of juice squeezed from each of these apples is approximately normally distributed with a mean of 2.25 ounces and a standard deviation of 0.15 ounce.

Between what two values (in ounces) symmetrically distributed around the population mean will 80% of the apples fall? (Ch8)

Solution:

The Z value for 80% is 1.28; The margin of error is Z*STD = 1.28 * 0.15 = 0.192

The confidence interval is: 2.25 ± 0.192 = [2.058 , 2.442]

32. A human resource manager is interested in whether absences occur during the week with equal frequency. The manager took a random sample of 100 absences and created the following table:

Monday 28Tuesday 20Wednesday 12Thursday 18Friday 22

At a significance level of α = .05 test the Null that the probabilities of absences are the same for all five days.(Ch 12)

Solution:

8


Expected frequency = 100/5 = 20

df = number of variables -1 = 5-1 = 4

OBS EXP Fo-Fe (Fo-Fe)^2/Fe % of chi square

28 20 8 3.2 47.06

20 20 0 0 0

12 20 -8 3.2 47.06

18 20 -2 0.2 2.94

22 20 2 0.2 2.94

Total 6.8 100

The calculated chi-square = 6.8

tcritical chi-square at df = 4 & 5% level = 9.48

as calculated Chi-square < the critical Chi-square , the Null

Hypothesis of Independence is rejected at 5% level

33. Consider the following partial computer output for a multiple regression model.

Predictor Coefficient Standard DeviationConstant 41.225 6.380X1 1.081 1.353X2 -18.404 4.547

Analysis of VarianceSource DF SSRegression 2 2270.11Error 26 3585.75

9


Find Total Sum of Squares, Explained Variation, SSE, MSE, R-Squared, and Test the overall usefulness of the model at 1% level of significance calculating the F-Statistic. Ch 14)

Solution:

Y bar = 41.225 + 1.081 X1 - 18.404 X2

n = 2 + 26 + 1 = 29

SST = SSR + SSE = 2270.11 + 3585.75 = 5855.86

explained variation = SSR = 2270.11

SSE = 5855.86 - 2270.11 = 3585.75

MSE = SSE/(n-k-1) = 3585.75 / (29 - 2 - 1) = 3585.75 / 26 = 137.91

R2 = SSR/SST = 2270.11/5855.86 = 0.3877

R2 adjusted = [R2 – k/(n-1)][(n-1)/(n-k-1)] =

[0.3877 – 2/(29-1)][(29-1)/(29-2-1)]

= (0.3877 - 0.0714)(28/26) = 0.3163 * 1.0769

= 0.3406

MSR = SSR/k = 2270.11/2 = 1,135.055

MSE = SSE/(n-k-1) = 3585.75 / 26 = 137.913

F = MSR / MSE = 1135.055/137.913 = 8.23

F critical (0.01,2,26) = 5.53

since F (8.23) > F Critical (5.53), H0 is rejected,

The coefficient of determination is highly significant and the model is useful.

10


34. A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 8.

Only 5% of the students taking the test scored higher than what grade? (Ch 6)

Solution:

µ=69 ;σ=8

X=M+6z

X=69+(8*1.96)

X=69+15.68=84.68

35. A data set with 7 observations yielded the following. Use the simple linear regression model.

∑X =21.57∑X2 =68.31∑Y = 188.9 ∑Y2 =5,140.23∑XY =590.83

Calculate the Correlation coefficient, Coefficient of determination and SSE, and the Standard Error of Estimate.(Ch 13)

Solution:

X average = ΣX/n =21.57/7 = 3.08

Y average = ΣY/n =188.9/7 = 26.99

SSxx = ΣX2 – ((ΣX)^2/n) = 68.31 – (21.57^2/7) = 1.84

SSyy = ΣY2 – ((ΣY)^2/n) = 5140.23 – (188.9^2/7) = 42.63

SSxy = ΣXY – ((ΣX ΣY/n)= 590.83 - (188.9 * 21.57 /7) = 8.75

B1 = SSxy/ SSxx = 8.75/1.84 = 4.75

B0 = 26.99-(4.75*3.08) = 12.35

11


Correlation Coefficient r

= SSxy / (sqrt(SSxxSSyy))

= 8.75 / sqrt(1.84*42.63)

= 8.75/ 8.8565 = 0.9879

Coefficient of Determination R squared

= SSR / SST

= b1*SSxy /SSyy

=4.75 * 8.75/42.63 = 0.9749

SSE = SST – SSR

= SSyy – b1*SSxy

= 42.63 – (4.75*8.75) = 1.0675

MSE = SSE/df

= 1.0675/(7-2) = 0.2135

Standard Error of Estimate = sqrt(MSE) = 0.462

36. A data set with 7 observations yielded the following. Use the simple linear regression model.

∑X =21.57∑X2 =68.31∑Y = 188.9 ∑Y2 =5,140.23∑XY =590.83

Write the Regression Equation showing Intercept and slope and test the significance of slope at 1% significance level.(Ch13)

12


Solution:

X average = ΣX/n =21.57/7 = 3.08

Y average = ΣY/n =188.9/7 = 26.99

SSxx = ΣX2 – ((ΣX)^2/n) = 68.31 – (21.57^2/7) = 1.84

SSyy = ΣY2 – ((ΣY)^2/n) = 5140.23 – (188.9^2/7) = 42.63

SSxy = ΣXY – ((ΣX ΣY/n)= 590.83 - (188.9 * 21.57 /7) = 8.75

B1 = SSxy/ SSxx = 8.75/1.84 = 4.75

B0 = 26.99-(4.75*3.08) = 12.35

Y = B0 + B1*X

Y = 12.35 + 4.75 * X

SSE = SST – SSR = SSyy – b1*SSxy = 42.63 – (4.75*8.75) = 1.0675

MSE = SSE/df = 1.0675/(7-2) = 0.2135

Standard Error of Estimate Se = sqrt(MSE) = 0.462

Sb1 = Se/sqrt(SSxx) = 0.462/sqrt(1.84) = 0.340

t b1 = b1 / Sb1 = 4.75/0.340 = 13.95

the critical t value at df = 5 & 1% test = 4.032

as t b1 > Critical t , the slope coefficient appears significant at

1% level

37.Test H0: π1– π2 ≤ .01, HA: π1– π2 > .01 at α = .05 where p1 = .08, p2 = .035, n1 = 200, n2 = 400. Indicate which test you are performing; show the test statistic and the critical values and mention whether one-tailed or two-tailed.

Solution:

13


it is a one tailed (right) test

P0 =(( 0.08*200) + (0.035*400 ))/ (200+400) = 16+14 / 600 = 30/600 =

0.05

the standard error S p1-p2 = square root (p0(1- p0)[(1/ n1) + (1/ n2)] =

the standard error S p1-p2 = sqrt(0.05*0.95*(1/200+1/400))

the standard error S p1-p2 = sqrt(0.05*0.95*0.0075)=sqrt(0.00035625)

the standard error S p1-p2 = 0.0189

Z = [(p1-p2) - D1] / [standard error S p1-p2] = (0.08-0.035) - 0.01 /

0.0189 = 0.035/0.0189 = 1.85

Z= 1.85

One-tailed critical value for the 5% is 1.645, as the calculated

Z > Critical Z

1.85 > 1.645

the null hypothesis is rejected at 5% significance test level

38. A recent study conducted by the state government attempts to determine whether the voting public supports further increase in cigarette taxes. The opinion poll recently sampled 1500 voting age citizens. 1020 of the sampled citizens were in favor of an increase in cigarette taxes. The state government would like to decide if there is enough evidence to establish whether the proportion of citizens supporting an increase in cigarette taxes is significantly greater than .66 at 5% and 10% significance levels. Indicate which test you are performing; show the hypotheses, the test statistic and the critical values and mention whether one-tailed or two-tailed. (Ch9)

14


Solution:

H0 : p <= 0.66

H1 : p > 0.66

Here p = 1020/1500 = 0.68;

Pi(population proportion) = 0.66 and n = 1500;

n*pi= 1500*(0.66) = 990;

and n(1- pi) = 1500*(1-0.66) = 495. So,normal approximation is

appropriate

The std. error σp = square root (pi(1-pi)/n)= square root

(0.66(1-0.66)/1500)= 0.0122

z = 0.68-0.66/.0122 = 0.02/0.0122 = 1.635

it is a one-tailed test.

For 5% significance level Zc =1.645

for 10% significance level Zc = 1.28

as Z > Zc in the 10% the null hypothesis is rejected at 10%

significance level

as Z < Zc in the 5% the null hypothesis is not rejected at 5%

significance level

39.Historically the average age of European soccer players is reported as 26 years with a standard deviation of 4 years. A random sample of 81 European professional soccer players has an average age of 27 years. We would like to decide if there is enough evidence to establish that average age of European soccer

15


players has increased significantly. What is the decision at α=.05 and 0.01? Indicate which test you are performing; show the hypotheses, the test statistic and the critical values and mention whether one-tailed or two-tailed.(Ch9)

Solution:

H0 : mean avg. ≤ 26

H1 : mean avg. > 26

μ0 = 26 ,

σ = 4 ,

n = 81 ,

X average = 27

σ sample = σ / square root (n) = 4 / 9 = 0.44

z = (X average- μ0)/ σ sample = (27 – 26) / 0.44 = 1/0.44 = 2.25

it is a one-tailed (right-tailed) test.

Z critical for one-tailed test are 2.33, 1.645 for 1% and 5% significance level tests.

As z > z critical at 5% , its rejected at 5% significance level test. But as z < z critical at 1% , its not rejected under the 1% significance level test.

40.Test H0: µ = 42 versus HA: µ ≠ 42 when X = 42.8, s =1.2 and n =16 at α = .01 and .05. Assume that the population from which the sample is selected is normally distributed. Indicate which test you are performing; show the test statistic and the critical values and mention whether one-tailed or two-tailed.(Ch 9)

Solution:

σx̄ = s/sq rt(n)

= 1.2/4

= 0.3

16


t̂ = (x̄ -µ)/σx̄

= (42.8-42)/0.3

= 2.667

The df = n-1

= 16-1

= 15

The degree of freedom (df) = 16-1 = 15,

t-values are: 2.131 for 5% significance level and

2.947 for 1% significance level.

So the calculated test statistic is less than the critical values

t̂ > tc at 5% level of significance, So its rejected at 5% level of significance.

t̂ < tc at 1% level of significance, we cannot reject the null hypothesis

17

Documents

BA 578 Live Fin Set2