Buec Assignment 2 - Andrew Flostrand PDF

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Andrew Flostrand...

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1. The question we are investigating is what is the most popular colour of outdoor jacket that male college students wear. We have defined “outdoor jackets” as one that is has insulating material and water resistant. Included in our definition of outdoor jackets in our definition are winter jackets, windbreakers, rain coats, trench coats and puffer jackets. Some clothes that did not fit our definition of outdoor jacket were thin jackets such as shells, bomber jackets, denim jackets and hoodies. Our population of interest is male SFU students and the population it applies to is male college/university students in Canada. While someone often owns several hoodies or shirts, most people only own a few jackets that are worn regularly on a day-to-day basis. When given a limited number of choices, color can be one of the major deciding factors during the decision making process when purchasing an item. In a business context, when a firm is given a limited amount of product color variations they can produce, this statistical information may motivate them to decide which color their product should be. For example, if a clothing brand wanted to release a luxury modeled jacket with a single color, the firm may be motivated to choose the most popular color if there is a significant proportion of the population wearing a certain color compared to any other color. To avoid coverage error, we have collected half of our data from the lower bus loop and half at the upper bus loop of SFU because there may be a disproportionate amount of students from each faculty if it was only recorded in one area. To keep our sample more representative, we have observed - How it avoids Coverage Error, Nonresponse Error and Measurement Error 2. black

109

55.61%

white

4

2.04%

gray

30

15.31%

green

15

7.65%

red

3

1.53%

blue

26

13.27%

Other

9

4.59%

Total

196

100.00%

We chose to use a 95% confidence level because it is conservative and the most commonly used for confidence intervals in statistical estimates of a population’s proportion. We chose to use a margin of error of ± 7% because we expected a large proportion of the jackets to be

black. Thus, we decided a larger margin of error would make the sample reasonable to conduct while still keeping the results statistically meaningful for decision making. 1-α = 0.95 (1-α)/2 = 0.475 Z = 1.96 e = 0.07

π = 0.5333 (according to our pilot survey, π = 16 black jackets/30 = 0.5333) Z 2 2 [¿ ¿ 2 π (1−π )]/ e =[ 1.9 6 (0.5333)(1−0.5333 )]/ 0.0 72=¿ 195.12 → n = 196 required n=¿ Pilot survey results: black

16

white

1

gray

5

green

3

red

0

blue

4

Other

1

Total

30

Readjusted values black

119

55.09%

white

5

2.31%

gray

33

15.28%

green

17

7.87%

red

3

1.39%

blue

29

13.43%

Other

10

4.63%

Total

216

100.00%

Mean = 0.5509 Standard Deviation = √ (P(1−P))/n ¿

= √ (0.5509 (1−0.5509))/216 = 0.0338 The sample mean or point estimate of the population proportion of males that wear black jackets is 0.5509 with a standard deviation of 0.0338 from the sample mean. 5. Use the expanded n sample size of your survey from part 3, to calculate confidence intervals for 90%, 95% and 99%. Show all calculations, neatly sketch the sampling distribution for each confidence interval by hand and show the appropriate Z or t values corresponding to your data (similar to sketch on p.285 of text but include values for x) and state and interpretation for each case. 90% confidence interval 1-α = 0.90 (1-α)/2 = 0.45 Z = 1.645 n = 216 P = 0.5509

P± Z (√ (P(1−P))/n) =

0.5509 ±1.645 (√ (0.5509 (1− 0.5509))/ 216 ) ¿ 0.5509 ± 0.0557 We are 90% confident that the population proportion that wears black jackets is between 0.4952 to 0.6066.

95% confidence interval 1-α = 0.95 (1-α)/2 = 0.475 Z = 1.96 e = 0.07 n = 216

P± (√ (P (1− P ))/n) = 0.5509 ± 1.96 (√ (0.5509 (1−0.5509))/ 216 ) = 0.5509 ± 0.0663 We are 95% confident that the population proportion that wears black jackets is between 0.4846 to 0.6172.

99% confidence interval 1-α = 0.99

P± (√ (P (1− P ))/n)

(1-α)/2 = 0.495 Z = 2.576 e = 0.07 n = 216

¿ 0.5509 ± 2.576 (√ (0.5509 (1−0.5509))/ 216 ) ¿ 0.5509 ± 0.0871 We are 99% confident that the population proportion that wears black jackets is between 0.4638 to 0.6380....