HI6007 Statistics for Business Decisions Assignment Help

Question 1: Post Graduate Units

A: Mean, Median and Mode, Quartile and Percentile

1. Mean, Median & Mode

Mean = E(x) = ∑X/n = 1284/20 = 65.9 students’ results

Median = arranging the data in ascending order as follows;

“42

53

54

61

61

61

62

63

64

66

67

67

68

69

71

71

76

78

81

83”

In case of the even data, the middle two values are added and divided by two to get the middle value.

{(n+1)/2}^th value = 66 + 67 / 2 = 66.5

Mode = the most recurring number (high frequency) is the mode = 61

2. First & Third Quartile

First Quartile = (n+1) x 1/4 = (20 + 1)/4 = 5.25^th = (5^th + 6^th)/2 = (61 + 61)/2 = 61

Third Quartile = (n+1) x 3/4 = (20 + 1) x (3/4) = 15.75^th value = (15^th + 16^th)/2 = (71+71)/2 = 71

3. 9oth Percentile

Computing the position L = K/100 x N, whereas N is the total number of observations

L = (90/100) x 20 = 18^th value

Hence, the 90^th percentile is 78

B: Inferential Statistics

Inferential statistics take the data from a sample and then makes the inference about the larger population from which sample is taken. The most common inferential statistics include confidence intervals, regression analysis and hypothesis testing. For example, one might stand in a mall and ask questions from a sample of 100 people about whether they like to shop at Sephora or not.

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Question 2: Holmes Institute

A: Joint Probability

1. Table

	Yes	No	Total
≤ 23	207	201	408
24-26	299	379	678
27-30	185	268	453
31-35	66	193	259
≥ 36	51	169	220
Total	808	1210	2018

 

	Y
	Yes	No	Total
≤ 23	207/2018 = 0.1026	201/2018 = 0.996	408/2018 = 0.2022
24-26	299/2018 = 0.1482	379/2018 = 0.1878	678/2018 = 0.3660
X   27-30	185/2018 = 0.0917	268/2018 = 0.1328	453/2018 = 0.2245
31-35	66/2018 = 0.0327	193/2018 = 0.0956	259/2018 = 0.1283
≥ 36	51/2018 = 0.0253	169/2018 = 0.0837	220/2018 = 0.1090
Total	808/2018 = 0.4004	1210/2018 = 0.5996	2018/2018 = 1

Every column represents the joint probabilities.

2. P(24-26/ Yes) = P(24-26 Yes / P Yes = 299/808 or (0.1482/0.4004) = 0.3700495
3. P(≤ 23/ Yes) = 207/808 = 0.2562

Since, P (≤23/Yes) ≠ P (X≤23). So, events are not independent.  

B: Expected Value and Variance

X	P(X)	X x P(X)	X2 x P(X)
10	0.05	0.5	5
20	0.1	2	40
30	0.1	3	90
40	0.2	8	320
50	0.35	17.5	875
60	0.2	12	720

 

	P(X)	∑X x P(x)	∑X2 x P(x)
Total Sum	1	43	2050

 

Mean = ∑X x P(X) = 43

Variance          = E(X²) – [E(X)]²

                        = ∑x² P (x) – [∑x P (x)]²

                        = 2050 – (43)²

                        = 2050 – 1849

                        = 201

 

 

Question 3: Drugs in Country X

Aim: Carrying out the hypothesis testing

Given: Sample Size = n = 60; Sample Mean =  = 745

Parameter: Let µ be the population mean, so µ is the annual average expenditure per person in the North East Region so  = 838

Let µ be the annual average expenditure from Midwest

Hypothesis:

H₀ = the annual average expenditure per person from Midwest doesn’t differ significantly from the expenditure in the Northeast Region

H₁ = the annual average expenditure per person from Midwest is lower than in the Northeast region

Symbolically,

H₀ = µ =      v/s       H₁ = µ  

H₀ = µ = 838   v/s       H₁ = µ  838

Here as population SD is given, so Z-test will be used for testing the hypothesis

Test Statistic:

Z calculated =

So under H₀, µ = 838

Z calculated =  = -2.4012

Decision: Reject H₀ if Z-cal  Zα

Here 1 – α = 99% = 0.99 = α = 0.01

Z_0.01 = 2.32

So here Z-Cal = -2.4012  -2.32

So we reject H₀

Conclusion:

The population annual average expenditure per person in Midwest may be lower than in the Northeast region

Answers:

1. Hypothesis: H₀: µ = 838 v/s        H₁: µ  838 (one tail test)
2. We used Z test statistics, because the sample size is more than 30 & population SD is given
3. Test Statistics = Z-Cal = -2.4012 and P-Value = 0.0082
4. We reject H_0­ if p-value α so p-value = 0.0082  01 so we reject H₀
5. There is a sufficient evidence to conclude that the population annual average expenditure per person in Midwest may be lower than in the Northeast region.

Question 4: Gasoline

A: Hypothesis Testing

To test:

H₀ = µ1 = µ2 = µ3 … µn = 0 i.e. there is no difference in treatment means

H₀ = µ1 = µ2 = µ3 … µn ≠ 0 i.e. some means are different

B: Decision Rule

Test criteria:

Reject Null Hypothesis if f > f (a, df, N-a)

Critical values f (0.05, 2, 27) = 3.3541 (from the F-table)

Reject H_0, if f > 3.3541

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