Math 533 Week 8 Final Exam Set 1, Exams of Health sciences

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Week 8 : Final

Exam - Final Exam

Page 1

1. (TCO A) Seventeen salespeople reported the

following number of sales calls completed last

month.

a.

Compute the mean, median, mode, &

standard deviation, Q 1

, Q

, Min, & Max

for the above sample data on number of

sales calls per month.

b.

In the context of this situation,

interpret the Median, Q 1

, & Q

. (Points

: 33)

a.

Ans: Mean, median, mode, & standard deviation,

Q1, Q3, Min, & Max for the above sample data on

number of sales calls per month

Mean 91.

Median 91

Mode 93

Standard

Deviation

1st

quartile

3rd

quartile

00

Minimum 72

Maximum 119

b. Median of the above sales calls means that if

all the sales calls data points are arranged in

an ascending order, then 91 Nos. of calls

made would fall in the middle. So, there are

as 8 sales calls data point above this median

& 8 sales calls data point below this median

point.

Q1 is the first quartile points which is 82 nos.

of calls made. It means that there are 25 %

of sales calls data point which lie below this

point.

Q3 is the third quartile points which is 100

nos. of calls made. It means that there are 75

% of sales calls data point which lie below this

point.

2. (TCO B) Cedar Home Furnishings has

collected data on their customers in

terms of whether they reside in an

urban location or a suburban location,

as well as rating the customers as

either “good,” “borderline,” or “poor.”

The data is below.

Urba

n

Subur

ban

Total

Good 60 168 228

Borderl

ine

Poor 24 40 64

Tot

al

If you choose a customer at random, then find the

probability that the customer

a. is considered “borderline.”

Ans: P(Customer is considered “borderline) =

108/400 = 27/

b. is considered “good” & resides in an

urban location.

Ans: P(is considered “good” & resides in an

urban location) = 60/400 = 3/

c. is suburban, given that customer is

considered “poor.” (Points : 18) Ans:

P(is suburban, given that customer is

considered “poor”) = 40 / 64 = 5 / 8

3. (TCO B) Historically, 70% of your customers

at Rodale Emporium pay for their purchases

using credit cards. In a sample of 20

customers, find the probability that

a. exactly 14 customers will pay for

their purchases using credit cards.

Ans:

This is a binomial distribution with

p = 0.70, so q = 1-p = 0.30 n =

Probability distribution:

P(exactly 14 customers will pay for their

purchases using credit cards) = 20 C 14

*(0.70)^14 *(0.30)^6 = 0.

b. at least 10 customers will pay for

their purchases using credit cards.

Ans :

0.0308+0.0653+0.1144+0.1643+0.1916+0.

788+0.1304+0.0716+0.0278+0.0068+0.

008 = 0.

c. at most 12 customers will pay for

their purchases using credit cards.

(Points : 18)

Ans:

0.00000000003+0.000000002+0.00000004+

0.0000005+0.000005+0.00003+0.0002+

.0010+0.0038+0.0120+0.0308+0.0653+0.

44 = 0.

4. (TCO B) The demand for gasoline at a local

service station is normally distributed with a

mean of 27,009 gallons per day & a standard

deviation of 4,530 gallons per day.

a. Find the probability that the demand

for gasoline exceeds 22,000 gallons for a

given day.

Ans:

z- score = (22000-27009)/4530 = -1.

From the Standard Normal

cumulative proportions p-

value = 0.

probability that the demand for gasoline

exceeds 22,000 gallons for a given day = 1-

0.1346 = 0.

b. Find the probability that the demand

for gasoline falls between 20,000 &

23,000 gallons for a given day.

Ans:

z- score for 20000 gallons =

(20000-27009)/4530 = -1.

From the Standard Normal

cumulative proportions p-

value = 0.

z- score for 23000 gallons =

(23000-27009)/4530 = - 0.

From the Standard Normal

cumulative proportions

p- value = 0.

probability that the demand for gasoline falls

between 20,000 & 23,000 gallons for a given day

= (0.1880-0.0606) = 0.

c. How many gallons of gasoline should

be on hand at the beginning of each

day so that we can meet the demand

90% of the time (i.e., the station stands

a 10% chance of running out of

gasoline for that day)? (Points : 18)

Ans: For the demand to be met

90 % of the time, it means that p-

value = 0.

Z-score for (p-value = 0.9) = 1.

gallons of gasoline should be on hand =

1.28*4530 + 27009 = 32807.4 gallons ~

approx 32808 gallons

5. (TCO C) An operations analyst from an airline

company has been asked to develop a fairly

accurate estimate of the mean refueling &

baggage handling time at a foreign airport. A

random sample of 36 refueling & baggage

handling times yields the following results.

Sample Size = 36

Sample Mean = 24.2 minutes

Sample Standard Deviation = 4.2 minutes

a. Compute the 90% confidence

interval for the population mean

refueling & baggage time.

Ans:

90 % confidence

interval means

Z(Upper – 95%)

= 1.

Z ( Lower – 5%) = -1.

90 % Confidence interval lower limit

= 24. 2 - 1. 645 * 4. 2 / 6 = 23. 0 49 90 %

Confidence interval Upper limit =

24. 2 + 1. 645 * 4. 2 / 6 = 25. 351

b. Interpret this interval.

Ans: Mean refueling & baggage handling time at

a foreign airport location would lie between

23.049 minutes & 25.351 minutes for 90% of the

times.

c. How many refueling & baggage

handling times should be sampled so

that we may construct a 90%

confidence interval with a sampling

error of .5 minutes for the population

mean refueling & baggage time?

(Points : 18)

Ans:

Nos. of refueling & baggage handling times that

should be sampled so that we may construct a

90% confidence interval = (4.2/0.5) ^ 2 =

70.56 ~approx. 71

6. (TCO C) The manufacturer of a certain brand of

toothpaste claims that a high

percentage of dentists recommend the use of

their toothpaste. A random sample of 400

dentists results in 310 recommending their

toothpaste.

a. Compute the 99% confidence

interval for the population proportion

of dentists who recommend the use

of this toothpaste.

Ans: p value =

310 /400 =

0.775 Sample

Size = 40072.1%

of

99% confidence interval for the

population proportion of dentists =

99% lower population proportion of

dentists = 0.721 = 72.1%

99% Upper population proportion of dentists =

0.829 = 82.9 %

b. Interpret this

confidence

interval. Ans:

So, for 99 % of the cases/times, More than

72.1% of population proportion of dentists &

less than 82.9 % of population proportion of

dentists would recommend the use of their

toothpaste.

c. How large a sample size will need to

be selected if we wish to have a 99%

confidence interval that is accurate to

within 3%? (Points : 18)

Ans: Sample size = 1286

7. (TCO D) A Ford Motor Company quality

improvement team believes that its recently

implemented defect reduction program has

reduced the proportion of paint defects. Prior to

the implementation of the program, the

proportion of paint defects was .03 & had been

stationary for the past 6 months. Ford selects a

random sample of 2,000 cars built after the

implementation of the defect reduction program.

There were 45 cars with paint defects in that

sample. Does the sample data provide evidence

to conclude that the proportion of paint defects is

now less than .03 (with a = .01)? Use the

hypothesis testing procedure outlined below.

a. Formulate the null &

alternative hypotheses.

Ans:

Null Hypotheses: H0: proportion of paint defects

after the implementation =.

Alternative hypotheses: H1: proportion of paint

defects after the implementation is now more

than.

b. State the level

of significance.

Ans:

Level of significance is = 1- 0.01 = 0.99 = 99

%

c. Find the critical value (or

values), & clearly show the

rejection & nonrejection regions.

Ans:

New defect % = 45 /2000 = 0.

At 99 % level of significance (one-sided), the

lower limit = 0.

So, more than 2.1 % of car(= more than 42 cars)

with paint defects would clearly fall into rejection

region.

Anything less than 2.1 % of “car ( less than 42

cars) with paint defects” would mean Non-

rejection region.

d. Compute the

test statistic.

Ans:

Z ( Lower Value) = 0.

e. Decide whether you can

reject Ho & accept Ha or not.

Ans:

New defect % = 45 /2000 = 0.

Since the new defect % 2.25 % is higher than the

lower limit of 2.1%, we would reject the null

Hypotheses (H0). We would accept the Alternative

hypotheses: H1: proportion of paint defects after

the implementation is now more than.

f. Explain & interpret your conclusion

in part e. What does this mean?

Ans:

It means that at 99 % level of significance,

proportion of paint defects after the

implementation of the defect reduction program in

cars is now more than .03. So, the defect

reduction program has not really worked for Ford

Motor Company quality improvement team.

g. Determine the observed p-value for

the hypothesis test & interpret this

value. What does this mean?

Ans:

New defect % = 45 /2000 = 0.

Observed p-value is 2.25 % which means that

sample mean observed value recorded

2.25 % of the cars having paint defects.

h. Does the sample data provide

evidence to conclude that the

proportion of paint defects is now less

than .03 (with a = .01)? (Points : 24)

Ans:

Sample data doesn’t provide evidence to conclude

that the proportion of paint defects is now less

than .03 (with a = .01).

8. (TCO D) A new car dealer calculates that the

dealership must average more than 4.5% profit

on sales of new cars. A random sample of 81 cars

gives the following result.

Sample Size

= 81 Sample

Mean =

4.97%

Sample Standard Deviation = 1.8%

Does the sample data provide evidence to

conclude that the dealership averages more

than 4.5% profit on sales of new cars (using a =

.10)? Use the hypothesis testing procedure

outlined below.

a. Formulate the null &

alternative hypotheses.

Ans:

Null Hypotheses: H0: dealership average profit

on sales of new cars = 4.5%

Alternative hypotheses: H1: dealership average

profit on sales of new cars is more than

4.5 %

b. State the

level of

significance.

Ans:

Level of significance = 1-0.10 = 0.9 = 90 %

c. Find the critical value (or

values), & clearly show the

rejection & nonrejection regions.

Ans:

Critical Value = 4.5 + 1.28 * 0.2 = 4.

So, less than 4.756% of dealership average profit

on sales of new cars would be rejection region.

More than 4.756% of dealership average profit on

sales of new cars would be non- rejection region

d. Compute

the test

statistic. Ans:

Z – higher value = 4.756 %

e. Decide whether you can

reject Ho & accept Ha or not.

Ans:

Since, the Sample Mean = 4.97% lie above the

critical value of 4.756 %, we would reject H0 &

accept Ha.

f. Explain & interpret your

conclusion in part e. What does this

mean? Ans:

It means that at 90 % level of significance, the

dealership averages more than 4.5% profit on

sales of new cars.

g. Determine the observed p-value

for the hypothesis test & interpret

this value. What does this mean?

Ans:

P – value = Sample Mean = 4.97% so, it

means that random sample of 81 cars gave

profit of 4.97 % on sales of new car.

h. Does the sample data provide evidence to

conclude that the dealership averages more

than 4.5% profit on sales of new cars (using a =

.10)? (Points : 24)

Ans:

Yes, the sample data does provide evidence to

conclude that the dealership averages more than

4.5% profit on sales of new cars (using a = .10).

Week 8 : Final Exam -

Final Exam Page 2

1. (TCO E) Bill McFarland is a real estate broker

who specializes in selling farmland in a large

western state. Because Bill advises many of his

clients about pricing their land, he

is interested in developing a pricing formula of

some type. He feels he could increase his

business significantly if he could accurately

determine the value of a farmer’s land. A

geologist tells Bill that the soil & rock

characteristics in most of the area that Bill sells

do not vary much. Thus the price of land should

depend greatly on acreage. Bill selects a sample

of 30 plots recently sold. The data is found below

(in Minitab), where X=Acreage & Y=Price

($1,000s).

PRIC

E

ACR

EAG

E

PRE

DICT

0

• 145 45.
  • 61 19.
• 235 80.
• 250 90.
• 278 95.
• 118 41.
  • 46 14.
  • 69 22.

0

Correlations: PRICE, ACREAGE

Pearson correlation of PRICE

& ACREAGE = 0.997 P-Value =

0.

Regression Analysis: PRICE versus ACREAGE

The regression

equation is PRICE =

2.26 + 2.89 ACREAGE

Predictor Coef SE

Coef T P

Constant 2.257 2.231 1.

0.

ACREAGE 2.89202 0.04353 66.44 0.

S = 7.21461 R-Sq = 99.4% R-Sq(adj) = 99.3%

Analysis of Variance

Source DF SS MS F

P Regression

4414.11 0.

Residual Error 281457 52

Total 29231215

Predicted Values for New Observations

New Obs Fit SE Fit 95% CI 95% PI

1 146.86 1.37 (144.05, 149.66) (131.82,

161.90)

2 725.26 9.18 (706.46, 744.06) (701.35,

749.17)XX

XX denotes a point that is an extreme outlier

in the predictors.

Values of Predictors for

New Observations New Obs

ACREAGE

a. Analyze the above output to

determine the regression

equation. Ans:

The regression equation is

PRICE ($1,000s) = 2.26 + 2.89 ACREAGE

b. Find & interpret in the

context of this problem.

Ans:

Price of land depends on acreage & varies with

the acreage. Price of the land can be predicted

by multiplying the acreage of the land with 2.

& then adding a constant value of 2.26. The land

price arrived would show up in $ 1,000s

(thousands of Dollars).

c. Find & interpret the

coefficient of determination (r-

squared). Ans:

The coefficient of determination, r 2, is useful

because it gives the proportion of the variance

(fluctuation) of price of the land ( in ,000 $) that

can be predicted from the acreage of the land.

The coefficient of determination is a measure of

how well the regression line (PRICE ($1,000s) =

2.26 + 2.89 ACREAGE) represents the data. In

this case, R-Sq value is 99.4% which means that

variation in the price of the land can be

explained/predicted to the extent of 99.4 % by

the variation in the acreage of the land.

d. Find & interpret coefficient of

correlation.

Ans:

Coefficient of correlation. R, measures the

strength & the direction of a linear relationship

between two variables. In this case, the r value is

0.997 & so there is a string relationship between

the acreage of the land & price of the land ( in

,000 $).

e. Does the data provide significant

evidence ( 𝛼 = .05) that the acreage can

be used to predict the price? Test the

utility of this model using a two-tailed

test. Find the observed p- value &

interpret.

Ans:

at 95 % level of significance, the observed p –

value is 0.000 which is less than 0.05. So,

reject the null hypothesis. So, the data

provides significant evidence (at 𝛼 = .05) that the

acreage can be used to predict the price.

f. Find the 95% confidence interval

for mean price of plots of farmland

that are 50 acres. Interpret this

interval.

Ans:

95% confidence interval for mean price of plots of

farmland that are 50 acres is (144.05, 149.66).

So, at 95% confidence interval, the mean price

of plots of farmland that are 50 acres would lie in

the interval of (144.05, 149.66).

g. Find the 95% prediction interval for

the price of a single plot of farmland

that is 50 acres. Interpret this interval.

Ans:

95% prediction interval for the price of a single

plot of farmland that are 50 acres (131.82,

161.90.

So, at 95% prediction interval, the price of a

single plot of farmland that is 50 acres would lie

in the interval of (131.82, 161.90.

h. What can we say about the price for a

plot of farmland that is 250 acres?

(Points : 48)

Ans:

Price for a plot of farmland that is 250 acres =

PRICE ($1,000s) = 2.26 + 2.89 ACREAGE =

2.26 + 2.89 * 250 = 724.760 ( In ,000 $ )

Week 8 : Final Exam -

Final Exam 4

1. (TCO E) An insurance firm wishes to

study the relationship between driving

experience (X1, in years), number of

driving violations in the past three years

(X2), & current monthly auto insurance

premium (Y). A sample of 12 insured

drivers is selected at random. The data

is given below (in MINITAB):

Y X1 X2 Predict

X1

Predict

X2

74 5 2 8 1

38 14 0

50 6 1

63 10 3

97 4 6

55 8 2

57 11 3

43 16 1

99 3 5

Regression Analysis: Y versus X1, X2

The regression equation is

Y = 55.1 - 1.37 X1 + 8.05 X2

Predictor Coef

SE Coef T P Constant

55.138 7.309 7.54

0.000

X1 -1.3736 0.4885 -2.81 0.020

X2 8.053 1.307 6.16 0.000

S = 6.07296 R-Sq = 93.1% R-Sq(adj) =

91.6%

Analysis of Variance

Source DF SS MS F

P Regression

2 4490.3 2245.2

60.88 0.000

Residual Error 9 331.9 36.9

Total 11 4822.3

Predicted Values for New Observations

New Obs Fit SE Fit 95% CI 95% PI

1 52.20 2.91 (45.62, 58.79) (36.97,

67.44)

Values of Predictors for

New Observations New Obs

X1 X2

1 8.00 1.00

Correlations: Y, X1, X2

Y X1

X1 -

X2 0.933 -

0.000

Cell Contents: Pearson correlation

P-Value

a. Analyze the above output to determine

the multiple regression equation.

Ans:

The multiple Regression equation is

The regression equation is

Y = 55.1 - 1.37 X1 + 8.05 X2

b. Find & interpret the multiple index of

determination (R-Sq).

Ans:

R-Sq = 93.1%

So, the variation in the current monthly auto

insurance premium (Y) is explained by driving

experience (X1, in years) & number of driving

violations in the past three years (X2) to the

extent of 93.1%.

c. Perform the t-tests on & on (use two

tailed test with ( 𝛼 = .05). Interpret your

results.

Ans:

A t-stat of greater than 1.96 with a significance

less than 0.05 indicates that the independent

variable is a significant predictor of the dependent

variable within & beyond the sample. The greater

the t-stat the greater the relative influence of the

independent variable on the dependent variable.

A t-stat of less than

1.96 with significance greater than 0.05 indicates

that the independent variable is NOT a significant