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Quantitive Methods - questions and answers

Extracts from this document...

Introduction

Quantitative Methods Part I

Course Assignment 1

Question One

A pharmaceuticals company has consistently observed a strong correlation between the number of doctors their sales reps visit to inform them about new products and the unit volume of sales achieved.  It samples 15 sales reps and asks them to provide a count of the number of doctors they visited in the previous week

Sales Rep

Number of Doctor

Visits

1

28

2

32

3

41

4

51

5

33

6

19

7

46

8

22

9

37

10

24

11

30

12

18

13

27

14

19

15

36

Total

463

  1. What is the mean number of doctor visits for the previous week?

Mean, image00.png        =        image01.png

=         image35.png

=        30.87

  1. What is the median number of doctor visits for the previous week?

Number of Doctor

Visits

Sales Rep

18

12

19

6

19

14

22

8

24

10

27

13

28

1

30

11

32

2

33

5

36

15

37

9

41

3

46

7

51

4

Median                =        value at the observation image46.png

                =        value at the observation image51.png

=        value at the 8th observation

                =        30

  1. What is the variance for the number of doctor visits for the previous week?

Sales Rep

image56.png

Number of Doctor Visits

image60.png

image70.png

image81.png

1

28

-2.87

8.24

2

32

1.13

1.28

3

41

10.13

102.62

4

51

20.13

405.22

5

33

2.13

4.54

6

19

-11.87

140.90

7

46

15.13

228.92

8

22

-8.87

78.68

9

37

6.13

37.58

10

24

-6.87

47.20

11

30

-0.87

0.76

12

18

-12.87

165.64

13

27

-3.87

14.98

14

19

-11.87

140.90

15

36

5.13

26.32

Total

463

Total

1403.73

Variance, s2 (for a sample)        =        image91.png

=        image02.png

                                        =        100.27

  1. What is the standard deviation for the number of doctor visits for the previous week?

Standard deviation, s        =        image13.png

                        =        image22.pngimage28.png

                        =        10.013

  1. What is the 25th percentile for doctor visits for the previous week?

25th percentile                =        value at observation 0.25 × (n + 1)

                                =        value at observation 0.25 × 16

                                =         4th observation

=        22

This means 25% of the distribution lies below 22 and 75% of the distribution lies above 22.

  1. What is the 75th percentile for doctor visits for the previous week?

75th percentile                =        value at observation 0.75 × (n + 1)

                        =        value at observation 0.75 × 16

                        =         12th observation

                        =        37

This means 25% of the distribution lies above 37 and 75% of the distribution lies above 37.

  1. Using an appropriate technique, present this data graphically

...read more.

Middle

                                =        value at observationimage49.png

=        value at observation 48th

=        within image38.pnggroup

Median for Region A                =         image50.png

=         image52.png

=        40.56

Median Group Region B        =        value at observation image53.png

                                =        value at observation image49.png

=        value at observation 48th

=        within image38.pnggroup

Median for Region B                =         image54.png

=         image55.png

=        41.50

  1. Calculate the variance and standard deviation of the clients’ age in each region

Age

Mid-point (mp)

Freq

(Region A)

mp x Freq

(Region A)

mp -mean

(mp – mean)2

*Freq

image36.png

12.495

10

124.950

-29.075

845.356

8453.560

image37.png

29.995

25

749.875

-11.579

134.073

3351.825

image38.png

42.495

35

1487.325

0.921

0.848

29.680

image39.png

57.495

15

862.425

15.921

253.478

3802.170

image40.png

72.495

10

724.950

30.921

956.108

9561.08

Total

95

3949.525

25,198.315

Mean

41.574

Variance

265.245

Standard Deviation

16.286

Variance, σ2                =        265.245

Standard Deviation, σ        =        image57.png

                        =        image58.png

=        16.286

Age

Mid-point (mp)

Freq

(Region B)

mp x freq

(Region B)

mp -mean

(mp – mean)2

*Freq

image36.png

12.495

15

187.425

-28.947

837.929

12568.932

image37.png

29.995

20

599.900

-11.447

131.034

2620.676

image38.png

42.495

30

1274.850

1.053

1.109

33.264

image39.png

57.495

20

1149.900

16.053

257.699

5153.976

image40.png

72.495

10

724.950

31.053

964.289

9642.888

Total

95

3937.025

30019.737

Mean

41.442

Variance

315.997

Standard Deviation

17.776

Variance, σ2                =        315.997

Standard Deviation, σ        =        image57.png

                        =        image59.png

=        17.776

  1. Calculate the 25th and 75th percentile of the clients’ age for each region.

25th percentile for Region A        =        value at observation 0.25 × (nA + 1)

                                        =        value at observation 0.25 × 96

                                        =        value at position 24

(within the 25 – 34.99 class interval)

                                        =        image61.png

                                        =        30.594

75th percentile for Region A        =        value at observation 0.75 × (nA + 1)

                                        =        value at observation 72

(within the 50 – 64.99 class interval)

                                        =        image62.png

                                        =        51.999

25th percentile for Region B        =        value at observation 0.25 × (nB + 1)

                                        =        value at observation 24

(within the 25 – 34.99 class interval)

                                        =        image63.png

                                        =        29.496

75th percentile for Region B        =        value at observation 0.75 × (nB + 1)

                                        =        value at observation 72

(within the 50 – 64.99 class interval)

                                        =        image64.png

                                        =        55.247

  1. The firm is considering a special marketing campaign for younger clients if the 25th percentile value is less than 30.  Based on the calculations above, should they undertake this campaign in either region?

The firm should only undertake the marketing campaign in Region B because the 25th Percentile of this region is 29.50, which is less than 30. However, the 25th Percentile for Region A is 30.59 which is higher than 30.

...read more.

Conclusion

  1. Conduct a statistical test to determine (at a 5% level of significance) if males and females play different prices for haircuts.

H0: μf – μm = 0

H1: μf – μm ≠ 0

Degree of freedom, df                =        nf + nm – 2

                                =        25 + 20 – 2

                                =        43

Testing at a significance level of 5% and df = 43, the critical value of t = 2.018

Pooled standard error, sp                =        image19.png

                                =        image20.png

                                        =        9.2082

t                =        image21.png

                                        =        image23.png

                                        =        2.6208

2.6208 > 2.018. The test statistic is greater than the critical value. Therefore, we can reject the Null Hypothesis in favour of the alternative. The sample data does not indicate if the price for male hair cuts is increasing. This indicates that statistically, males and females pay different prices for haircuts.

Question Nine

Part A

A health authority is concerned about the waiting times at emergency rooms in hospitals.  The authority intends to conduct a statistical test on this.  What sample size would be required to estimate the true (population) within a bound of 10 minutes, at a 95% confidence level?  The population standard deviation is known to be 20 minutes.

Taking a confidence level of 95%,

z = 1.96

σ = 20

e = 10

n        =        image24.png

        =        image25.png

        =        15.3664

A sample of 16 would need to be drawn to ensure the variation of no greater than ±10 minutes from the mean.

Part B

A polling firm has been hired to find out whether citizens support a particular policy initiative.  If they want to be accurate within a range of +/- 3.5% at a 95% level of confidence, how many individuals would they need to survey?

Taking a confidence level of 95%,

z = 1.96

e = 0.035

Since we have not given any further information on p, it is assumed to be 0.5

n        =        image26.png

        =        image27.png

        =        784

A sample of 784 individuals would need to be drawn to ensure the accuracy is within a range of ±3.5% from the population mean.

...read more.

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