MATH 533 Final Exam Questions – Answered
(TCO A) A random sample of 20 cars driving down I-294 is selected and their speed is monitored. The results are as follows (in mph).
68 65 50 79 77 60 55
61 78 75
75 67 72 58 70 62 67
72 70 74
- Compute the mean, median, mode, and standard deviation, Q1, Q3, Min, and Max for the above sample data on speed per car.
- In the context of this situation, interpret the Median, Q1, and Q3. (Points: 33)
Answers:
Mean StDev Minimum Q1 Median Q3 Maximum Mode
67.75 8.04 50.00 61.25 69.00 74.75 79.00 67, 70, 72, 75
- The Median value is 69 implying that 50% of the values are less than or equal to 69 and other 50% is more than 69. The Q1 value is 61.25 implying that 25% of the values are less than or equal to 61.25 and other 75% is more than 61.25. And the Q3 value is 74.75 implying that 75% of the values are less than or equal to 74.75 and other 25% is more than 74.75.
(TCO B) Consider the following data on newly hired employees in relation to which part of the country they were born and their highest degree attained.
HS | BS | MS | PHD | Total | ||
East | 3 | 5 | 2 | 1 | 11 | |
Midwest | 7 | 9 | 2 | 0 | 18 | |
South | 5 | 8 | 6 | 2 | 21 | |
West | 1 | 7 | 8 | 6 | 22 | |
Total | 16 | 29 | 18 | 9 | 72 |
Of you choose one person at random, then find the probability that the person
- Has a PHD
- is from the East and has a BS as the highest degree arraigned
- has only a HS degree, given that person is from the West. (Points: 18)
Answers:
- P(Has a PHD) = 9/ 72 = 0.125
- P(East and BS) = 5/72 = 0.0694
- P(HS|West) = P(HS and West)/P(West) = (1/72)/(22/72) = 1/22 = 0.0455
(TCO B) Midwest Airlines has had an 80% on time departure rate. A random sample of 20 flights is selected. Find the probability that
- exactly 15 flights depart on time in the sample
- at least 17 flights depart on time in the sample
- less than 11 flights depart on time in the sample. (Points: 18)
Answers:
Here,
X = Number of flights depart on time out of 20 ~Binomial(n=20,p=0.8)
- The obtained Minitab output is as follows,
Probability Density Function
Binomial with n = 20 and p = 0.8
x P( X = x )
15 0.174560
Thus required probability is 0.174560.
- The obtained Minitab output is as follows,
Cumulative Distribution Function
Binomial with n = 20 and p = 0.8
x P( X ≤ x )
16 0.588551
As P(X≥17) = 1-P(X≤16) thus required probability is (1-0.588551)= 0.411449.
- The obtained Minitab output is as follows,
Cumulative Distribution Function
Binomial with n = 20 and p = 0.8
x P( X ≤ x )
10 0.0025948
As P(X < 11 ) = P(X≤10) thus required probability is 0.0025948.
(TCO B) The Federal Government is stepping up efforts to reduce average response times of fire departments to fire calls. The distribution of mean response times to fire calls follows a normal distribution with a mean of 12.8 minutes and a standard deviation of 3.7 minutes.
- Find the probability that a randomly selected response time is less than 15 minutes.
- Find the probability that a randomly selected response time is less than 13 minutes.
- The fastest 20% of fire departments will be singled out for a special safety award. How fast must a fire department be I order to qualify for the special safety award? (Points: 18)
Answers:
- The obtained Minitab output is given below,
Cumulative Distribution Function
Normal with mean = 12.8 and standard deviation = 3.7
x P( X ≤ x )
15 0.723943
So the required probability is 0.723943.
- The obtained Minitab output is given below,
Cumulative Distribution Function
Normal with mean = 12.8 and standard deviation = 3.7
x P( X ≤ x )
13 0.521554
15 0.723943
P(13 < X < 15) = P(X<15) – P(X<13) = 0.723943- 0.521554 = 0.202389
So the required probability is 0.202389.
- The obtained Minitab output is given below,
Inverse Cumulative Distribution Function
Normal with mean = 12.8 and standard deviation = 3.7
P( X ≤ x ) x
0.2 9.68600
So the required value is 9.68600.
(TCO C) A transportation company wants to estimate the average length of time goods are in transit across country. A random sample of 20 shipments yields the following results.
Sample Size = 20
Sample Mean = 4.6 days
Sample Standard Deviation = 1.50 days
- Compute the 90% confidence interval for the population mean transit time.
- Interpret this interval.
- How many shipments should be sampled if we wish to generate a 99% confidence interval for the population mean transit time that is accurate to within .25 days? (Points: 18)
Answers:
The obtained output is given below,
One-Sample T
- N Mean StDev SE Mean 90% CI
20 4.600 1.500 0.335 (4.020, 5.180)
- We are 90% confident that the average length of time goods are in transit across the country is between (4.020, 5.180).
- The obtained output is given below,
Sample Size for Estimation
Method
Parameter Mean
Distribution Normal
Standard deviation 1.5 (estimate)
Confidence level 99%
Confidence interval Two-sided
Results
Margin Sample
of Error Size
0.25 243
Required sample size is 243.
(TCO C) United Express Delivery is interested in estimating the percentage of packages delivered damaged. A simple random sample of 500 packages yields 12 delivered damages and 488 delivered undamaged.
- Compute the 99% confidence interval for the population proportion of packages that are delivered damaged.
- Interpret this confidence interval
- How many packages should be sampled in to order to be 99% confident of being within .5% of the actual population proportion of packages delivered damaged? (Points: 18)
Answers
- The obtained output is given below,
Test and CI for One Proportion
Sample X N Sample p 99% CI
1 12 500 0.024000 (0.006370, 0.041630)
Using the normal approximation.
- We are 99% confident that the proportion of packages that are delivered damaged is between (0.006370, 0.041630).
- The obtained output is given below,
Sample Size for Estimation
Method
Parameter Proportion
Distribution Binomial
Proportion 0.024
Confidence level 99%
Confidence interval Two-sided
Results
Margin Sample
of Error Size
0.005 7310
Required sample size is 7310.
(TCO D) A manager at Travis Savings and loan believes that less than 52% of their depositors won their homes. A random sample of 100 depositors is selected with the results that 46 depositors own their homes and the other 54 do not own their homes. Does the sample data provide evidence to conclude that less than 52% of all depositors at Travis Savings and Loan own their homes (with α = .05)? Use the hypothesis testing procedure outlined below.
- Formulate the null and alternative hypotheses.
- State the level of significance.
- Find the critical value (or values), and clearly show the rejection and nonrejection regions.
- Compute the test statistic.
- Decide whether you can reject Ho and accept Ha or not.
- Explain and interpret your conclusion in part e. What does this mean?
- Determine the observed p-value for the hypothesis test and interpret this value. What does this mean?
- Does the sample data provide evidence to conclude that less than 52% of all depositors at Travis Savings and Loan own their homes (with α = .05)? )Points: 24)
Answers
- The null and alternative hypotheses are,
Ho: p ≥ 0.52 and Ha:p<0.52
- The considered level of significance is 0.05.
- The critical value is –Z(0.05) = -1.645 and the rejection region is Test Statistic < -1.645.
- The obtained output is given below,
Test and CI for One Proportion
Test of p = 0.52 vs p < 0.52
Sample X N Sample p 95% Upper Bound Z-Value P-Value
1 54 100 0.540000 0.621979 0.40 0.656
Using the normal approximation.
From the above output we can see that the test statistic is 0.40.
- As the test statistic is not falling in the rejection region so do not reject the null hypothesis.
- We conclude the data was not sufficient enough to provide an evidence to reject null hypothesis at α=0.05
- The observed p-value is 0.656. Which has the interpretation that if the null hypothesis is true then getting the test statistic which we obtained (or extreme) has a probability 0.656. As the probability is large so we can’t reject the null hypothesis.
- No the sample data does not provide evidence to conclude that less than 52% of all depositors at Travis Savings and Loan own their homes.
(TCO D) Bill Smith is the Worthington Township manager. When citizens request a traffic light, the staff assesses the traffic flow at the requested intersection. Township policy requires the installation of a traffic light when an intersection averages more than 150 vehicles per hour. A random sample of 48 vehicles counts is done. The results are as follows:
Sample Size = 48
Sample Mean = 158.3 vehicles/hr.
Sample Standard Deviation = 27.6 vehicles/hr.
Does the sample data provide evidence to conclude that the installation of the traffic light is warranted (using α = .10)? Use the hypothesis testing procedure outlined below.
- Formulate the null and alternative hypotheses
H_{0}: µ≤150,
H_{a}: µ>150.
- State the level of significance.
α = 0.10
- Find the critical value (or values), and clearly show the rejection and norejection regions.
The critical value T_{.10, 47}=1.302, we reject H_{0} if t computed is greater than 1.302(t computed), we fail to reject H_{0} for values of T statistics that are less than/equal to 1.302.
- Compute the test statistic
One-Sample T
Test of μ = 150 vs > 150
N Mean StDev SE Mean 90% Lower Bound T P
48 158.30 27.60 3.98 153.12 2.08 0.021
T statistics=2.08
- Decide whether you can reject Ho and accept Ha or not.
We reject H_{0} in farvour of H_{a}: µ>150.
- Explain and interpret your conclusion in part e. What does this mean?
The sample data provide evidence to conclude that the installation of the traffic light is warranted (using α = .10). This means that the intersection averages is more than required 150 vehicles per hour
- Find the observed p-value for the hypothesis test and interpret this value. What does this mean?
P_value =2(0.021) = 0.042. One side test for p value < α thus we reject Null in fervor of H_{a}: µ>150.
- Does this sample data provide evidence (with α = 0.10), that the installation of the traffic light is warranted? (Points; 24)
The sample data provide evidence to conclude that the installation of the traffic light is warranted (using α = .10)
(TCO E) The management of JAL Airlines assumes a direct relationship between advertising expenditures and the number of passengers who choose to fly JAL. The following data is collected over the past 15 months of performance by JAL Airlines. Note that X=ADEXP (Advertising Expenditures in $1,000s), and Y=Passengers (number of passengers in 1,000s). The MINITAV printout can be found below.
ADEXP | PASSANGERS | PREDICT | ||
100 | 15 | 120 | ||
120 | 17 | 250 | ||
80 | 13 | |||
170 | 23 | |||
100 | 16 | |||
150 | 21 | |||
100 | 14 | |||
140 | 20 | |||
190 | 24 | |||
100 | 17 | |||
110 | 16 | |||
130 | 18 | |||
160 | 23 | |||
100 | 15 | |||
120 | 16 |
- Analyze the above output to determine the regression equation.
PASSANGERS = 4.386 + 0.10813 ADEXP
- Find and interpret in the context of this problem.
Β_{1}=0.10813, increase of advertisement expenditure by 0.10813 would increase of passengers by one unit
- Find and interpret the coefficient of determination (r-squared).
Model Summary
S R-sq R-sq(adj) R-sq(pred)
0.906780 93.78% 93.30% 91.47%
The coefficient of determination =93.78% this means that advertising cost explains 93.78% of changes with the passenger numbers
- Find and interpret coefficient of correlation.
Correlation: PASSANGERS, ADEXP
Pearson correlation of PASSANGERS and ADEXP = 0.968
P-Value = 0.000
R=968, there is strong positive association between advertising expenditure and the increase in passengers, the association is statistically significant at α=0.05.
- Does the data provide significant evidence (α = .05) that advertising expenditures can be used to predict the number of passengers? Test the utility of this model using a two-tailed test. Find the observed p-value and interpret.
T_{0.025,13}=2.1604 T computed from the table=13.99 since T computed > t tabulated, we reject null hypothesis that B=0. Thus we conclude that the advertising expenditures can be used to predict the number of passengers
- Find the 95% confidence interval for the mean number of passengers when advertising expenditures were $120,000. Interpret this interval.
Prediction for PASSANGERS
Variable Setting
ADEXP 120
Fit SE Fit 95% CI 95% PI
17.3621 0.236890 (16.8503, 17.8738) (15.3373, 19.3868)
The CI gives prediction interval for the mean of passengers give that advertising expenditures were $120,000. We are 95% confident the mean for passengers advertising expenditures were $120,000 is between (16.8503, 17.8738
- Find the 95% prediction interval for the number of passengers when advertising expenditures were $120,000. Interpret this interval.
Prediction for PASSANGERS
Variable Setting
ADEXP 120
Fit SE Fit 95% CI 95% PI
17.3621 0.236890 (16.8503, 17.8738) (15.3373, 19.3868)
We are 95% confident that for every $120,000 spent on advertisement, the number of passengers was 17.3621 with lower and upper limits of (15.3373, 19.3868)
- What can we say about the number of passengers when advertising expenditures were $250,000? (Points : 48)
Prediction for PASSANGERS
Variable Setting
ADEXP 250
Fit SE Fit 95% CI 95% PI
31.4192 0.996288 (29.2668, 33.5715) (28.5088, 34.3295) XX
XX denotes an extremely unusual point relative to predictor levels used to fit the model.
The number of passengers when the advertising expenditures were $250,000 was 31.42
(TCO E) A newly developed low-pressure snow tire has been tested to see how it wears under normal dry weather conditions. Twenty of these tires were tested on standard passenger cars. These cars were driven at high speeds on a dry test track for varying lengths of time. We are interested in finding the relationship between hours driven (HOURS, X1), brand of car driven (BRAND, X2, where 0=Ford and 1 =General Motors), and tread wear (TREAD, Y in inches). The data is found below.
Hours | Brand | Tread |
13 | 0 | 0.1 |
25 | 0 | 0.2 |
27 | 0 | 0.2 |
46 | 0 | 0.3 |
18 | 0 | 0.1 |
31 | 0 | 0.2 |
46 | 0 | 0.3 |
57 | 0 | 0.4 |
75 | 0 | 0.5 |
87 | 0 | 0.6 |
62 | 1 | 0.4 |
105 | 1 | 0.7 |
88 | 1 | 0.6 |
63 | 1 | 0.4 |
77 | 1 | 0.5 |
109 | 1 | 0.7 |
117 | 1 | 0.8 |
35 | 1 | 0.2 |
98 | 1 | 0.6 |
121 | 1 | 0.8 |
- Analyze the above output to determine the multiple regression equation
- Find and interpret the multiple index of determination (R-Sq).
- Perform the multiple regression t-tests on (use two tailed test with (α = .10). Interpret your results.
- Predict the tread wear for tires from General Motors that were driven for 50 hours. Use both a point estimate and the appropriate interval estimate. (Points: 31)
Answers:
- The above output informs us that the multiple linear regression is,
Tread = -0.00146 + 0.00686*Hours – 0.0286*Brand
- The multiple index of determination is 99.35% implying that the predictors explain 99.35% of the total variability of the Thread wears.
- Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant -0.00146 0.00961 -0.15 0.881
Hours 0.006858 0.000174 39.40 0.000 1.82
Brand
1 -0.0286 0.0117 -2.45 0.026 1.82
The computed t statistics are 39.35 and -2.45 for B1 and B2 respectively, the tabulated t value at 0.05 and 18 degrees of freedom gives us t0.05, 18=2.1009, comparing t statistics with t tabulated, we reject Null hypothesis since TC>Tb. That is B1≠B2≠0. We conclude that the effects of hours and brands on thread is statistically significant at α=0.1
- Prediction for Tread
Variable Setting
Hours 50
Brand 1
Fit SE Fit 90% CI 90% PI
0.312828 0.0089550 (0.297250, 0.328406) (0.275676, 0.349980)
The output suggests that the predicted point estimate is 0.312828 and the interval estimate is (0.275676, 0.349980)
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