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# Quantitive Methods - questions and answers

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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, = = =        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 =        value at the observation =        value at the 8th observation

=        30

1. What is the variance for the number of doctor visits for the previous week?
 Sales Rep Number of Doctor Visits   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)        = = =        100.27

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

Standard deviation, s        = =  =        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

Middle

=        value at observation =        value at observation 48th

=        within group

Median for Region A                = = =        40.56

Median Group Region B        =        value at observation =        value at observation =        value at observation 48th

=        within group

Median for Region B                = = =        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 12.495 10 124.950 -29.075 845.356 8453.560 29.995 25 749.875 -11.579 134.073 3351.825 42.495 35 1487.325 0.921 0.848 29.680 57.495 15 862.425 15.921 253.478 3802.170 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        = = =        16.286

 Age Mid-point (mp) Freq (Region B) mp x freq (Region B) mp -mean (mp – mean)2 *Freq 12.495 15 187.425 -28.947 837.929 12568.932 29.995 20 599.900 -11.447 131.034 2620.676 42.495 30 1274.850 1.053 1.109 33.264 57.495 20 1149.900 16.053 257.699 5153.976 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        = = =        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)

= =        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)

= =        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)

= =        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)

= =        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.

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                = = =        9.2082

t                = = =        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        = = =        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        = = =        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.

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