HI6007 Statistics for Business Decisions Online Tutoring
Question 1: Post Graduate Units
A: Mean, Median and Mode, Quartile and Percentile
- 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
- First & Third Quartile
First Quartile = (n+1) x 1/4 = (20 + 1)/4 = 5.25th = (5th + 6th)/2 = (61 + 61)/2 = 61
Third Quartile = (n+1) x 3/4 = (20 + 1) x (3/4) = 15.75th value = (15th + 16th)/2 = (71+71)/2 = 71
- 9oth Percentile
Computing the position L = K/100 x N, whereas N is the total number of observations
L = (90/100) x 20 = 18th value
Hence, the 90th 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
- 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 | ||
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.
- P(24-26/ Yes) = P(24-26 Yes / P Yes = 299/808 or (0.1482/0.4004) = 0.3700495
- 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
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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
Z0.01 = 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:
- Hypothesis: H0: µ = 838 v/s H1: µ 838 (one tail test)
- We used Z test statistics, because the sample size is more than 30 & population SD is given
- Test Statistics = Z-Cal = -2.4012 and P-Value = 0.0082
- We reject H0 if p-value α so p-value = 0.0082 01 so we reject H0
- 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.
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