BUSN1009 Quantitative Methods

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17-06-2021

Question:

A city’s telephone book lists 100 000 people. If the telephone book is the frame for a study, how large would the sample size be if systematic sampling were done on every 200th person

If a university employed 3500 academic staff, and a random sample of 175 academic staff was selected using systematic sampling, what value of k was used? Between what two values would the sampling selection process have started? Where would a frame for this sampling of academic staff been obtained?

Compute the 20th percentile, the 60th percentile, Q₁, Q₂, Q₃ and IQR for the following data.

470, 320, 146, 204, 800, 280, 957, 298, 445, 356, 450, 786, 964, 918, 849, 820

A data set contains the following eight values.

4 3 0 5 2 9 4 5

Calculate the sample variance.
Calculate the sample standard deviation.

According to a Human Resources report, a worker in the IT industry spends on average 419 minutes (or 6.98 hours) a day on the job. Suppose the standard deviation of time spent on the job is 27 minutes. a. If the distribution of time spent on the job is approximately bell shaped, between what two times would 68% of the data fall? 95%? 99.7%?

Use the values in the following contingency table to solve the equations given.

	A	B	C
X	42	48	36
Y	54	30	18
Z	21	39	12

P(A) = ___ c.P(A Ç X) = ___ e. P(A È C) = ___

P(Z) = ___ d.P(B ÇZ) = ___ f. P(A Ç B)

Convert the following contingency table to a probability matrix to solve the equations given.

	D	E
A	16	8
B	10	6
C	8	2

P(A) = ___ c. P(A ÇE) = ___
P(E) = ___ d. P(A ÷E) = ___

According to the ABS, 73% of women aged 25 to 54 participate in the labour force. Suppose that 78% of the women aged 25 to 54 are married. Suppose also that 61% of all women aged 25 to 54 are married and are participating in the labour force. What is the probability that a randomly selected woman aged 25 to 54:

is married or is participating in the labour force
is married or is participating in the labour force but not both
is neither married nor participating in the labour force?

Given P(A) = .40, P(B) = . 25, P(C) = .35, P(B | A) = . .25 and P(A | C) = .80, solve the following.

P(A ÇB) = ___
P(A ÇC) = ___
If A and B are independent, P(A ÇB) = ___
If A and C are independent, P(A ÇC) = ___

Use the values in the contingency table to solve the equations given.

	Variable 1
Variable 2	E	F
A	85	75
B	40	55
C	40	85
D	95	25

P(F) = ___
P(B ÈF) = ___
P(D ÇF) = ___
P(B | F) = ___
P(A ÈB) = ___
P(B ÇC) = ___
P(F | B) = ___
P(A | B) = ___
P(B) = ___
Based on your answers to these calculations, parts a, d, g and i, are variables 1 and 2 independent? Why or why not?

Answer:

Sample size = 100000/200 = 500

k = (total sample)/(size of random sample) = 3500/175 = 20

sampling process will start between 1^st and k^th employee (including

Sorted List =[146, 204, 280, 298, 320, 356, 445, 450, 470, 786, 800, 820, 849, 918, 957, 964]

20^th Percentile(index) = 0.20 * 16 = 3.2(index). Therefore 20^th Percentile = 280.

60^th Percentile (index) = 0.60 * 16 = 9.6(index). Therefore 60^th Percentile = 786.

Q1 is the median (the middle) of the lower half of the data, and Q3 is the median (the middle) of the upper half of the data

Quartile Q1: 309

Quartile Q2: 460

Quartile Q3: 834.5

IQR = 834.5 – 309 = 525.5

Sample variance (s2)= (xi-x)2(n-1)

x_i= a number in data set

x = mean (average)

n = total number of samples in dataset

In the given example,

n = 8

x = (4+3+0+5+2+9+4+5)/8 = 4

Sample variance (s2) = (xi-x)2(n-1) = (0+1+16+1+4+25+0+1)/7 = 48/7 = 6.8571

Sample standard deviation (S_x) = (xi-x)2(n-1) = 6.8571 = 2.6186

Using Emperical Rule,

For 68% of data fall:

Time Range 1=419 -27=392 Minutes.

Time Range 2 = 419+27 = 446 Minutes

For 95% of data fall:

Time Range 1=419 –(2*27)=365 Minutes.

Time Range 2 = 419+(2*27) = 473 Minutes

For 99.7% of data fall:

Time Range 1=419 –(3*27)=338 Minutes.

Time Range 2 = 419+(3*27) =500 Minutes

n(A) = 42+54+21 = 117

P(A) = n(A)/n = 117/300 = 0.39

n(Z) = 21+39+12 = 72

P(Z) = 72/300 = 0.24

P(A ∩ X) = 42/300 = 0.14

P(B ∩ Z) = 39/300 = 0.13

P(A ∪ C) = (n(A)+n(C))/n = (117+66)/300 = 0.61

P(A ∩ B) = n(A ∩ B)/n = 0/300 = 0

	D	E
A	0.32	0.16
B	0.20	0.12
C	0.16	0.04

P(A) = 0.32+0.16 = 0.48

P(E) = 0.16+0.12+0.04 = 0.32

P(A ∩ E) = 0.16

P(A ? E) = P(A ∩ E)/P(E) = 0.16/0.32 = 0.5

Let A = the event that a woman randomly selected participate in labor force, and B = the event that a woman randomly selected is married, and ~X = complement of X.

P(A or B) = P(A) + P(B) - P(A and B) = .75 + .78 - .61 = .92

P[(A and ~B) or (B and ~A)] = P(A and ~B) + P(B and ~A) = P(A) - P(A and B) + P(B) - P(B and A) = .75 - .61 + .78 - .61 = .14 + .17 = .31
P(~A and ~B) = P(~(A or B)) = 1 - P(A or B) = 1 - .92 = .08

P(B | A) = P(A ∩ B)/P(A)

P(A ∩ B) = P(A)*P(B | A) = (0.40)*(0.25) = 0.1

P(A | C) = P(A ∩ C)/P(C)

P(A ∩ C) = P(C)*P(A | C) = (0.35)*(0.80) = 0.28

If A and B are independent, P(A ∩ B) = 0

d)If A and C are independent, P(A ∩C) = 0

Ans:

	Variable 1
Variable 2	E	F
A	85	75
B	40	55
C	40	85
D	95	25

P(F) = 240/500 = 0.48
P(B ∪F) = 280/500 = 0.56
P(D ∩F) = 25/500 = 0.05
P(B | F) = P(B ∩F)/P(F) = (55/500)/(240/500) = 55/240 = 0.23
P(A ∪B) = P(A)+P(B) = 255/500 = 0.51
P(B ∩C) = n(B ∩C)/n = 0
P(F | B) = P(B ∩F)/P(B) = 55/95 = 0.58
P(A | B) = P(A ∩B)/P(B) = 0
P(B) = 95/500 = 0.19
Based on your answers to these calculations, parts a, d, g and i, are variables 1 and 2 independent? Why or why not?I guess yes!