Statistics for Business Decisions: Analysis and Solutions

This assignment solution provides a comprehensive analysis of statistical concepts relevant to business decision-making. It covers various topics, including descriptive statistics (mean, median, mode, quartiles, percentiles), inferential statistics (hypothesis testing, confidence intervals), probability, and regression analysis. The solution includes detailed calculations, interpretations, and conclusions for each question, demonstrating the application of statistical methods to real-world business scenarios. The assignment addresses topics such as joint probability, expected value, variance, hypothesis testing, and regression analysis. The solution incorporates statistical software output and detailed explanations, providing a valuable resource for students studying statistics for business.

Statistics for business
Decision

Table of Contents
Question 1........................................................................................................................................3
a. i. Mean, median and mode......................................................................................................3
ii. First and Third Quartile..........................................................................................................3
iii. Interpret the 90th Percentile....................................................................................................3
b. Presenting inferential statistics with examples.......................................................................4
Question 2........................................................................................................................................5
a. i. Preparing Joint probability rate............................................................................................5
ii. Probability that a student is 24-26-year-old............................................................................5
iii. Number of University applied to independent of student age...............................................5
b. Computing expected value and variance.................................................................................6
Question 3........................................................................................................................................6
i. Presenting hypothesis...............................................................................................................6
ii. Deciding the suitable test statistic...........................................................................................6
iii. Presenting the values.............................................................................................................6
iv. Describing the decision at 99% confidence level..................................................................7
v. Conclusion..............................................................................................................................7
Question 4........................................................................................................................................7
a. Hypothesis...............................................................................................................................7
b. Decision rule at 5% significance level....................................................................................7
c. Calculate the test statistic........................................................................................................7
d. Result......................................................................................................................................8
Question 5........................................................................................................................................8
a. Missing values.........................................................................................................................8
b. Annual credit card charges......................................................................................................9
c. Regression equation is fit........................................................................................................9
Question 6........................................................................................................................................9
a. Linear trend equation forecast the sales of face mask for October 2020................................9
b. Calculate the forecast sales difference by using Weighted average method........................10
Appendix........................................................................................................................................11

Question 1
a. i. Mean, median and mode
Results
42
66
67
71
78
62
61
76
71
67
61
64
61
54
83
63
68
69
81
53
65.9 Mean
66.5 Median
61 Mode
ii. First and Third Quartile
In order to calculate the quartile, put a formula =QUARTILE(array, quartile). By applying
the same, for 1st quartile (sum of data, 1st quartile) and for 3rd (sum of data, 3rd quartile) the
outcome comes 61 and 71 respectively.
iii. Interpret the 90th Percentile
Results
42
53
54

61
61
61
62
63
64
66
67
67
68
69
71
71
76
78
81
83
18
90th Percentile =
0.90*20
79.5
Interpretation: From the above table, it is interpreted that 90th percentile of 20 student’s
score is falls under 18 which is a whole number. That is why, as per standard formula, both 18th
and 19th scores are taken and then average is used. Then the outcome (79.5) clearly indicate that
the student scores 79.5% as per the set percentile.
b. Presenting inferential statistics with examples
Inferential statistics allows to make predictions from the data. Also, it is used to compare the
difference between treatment groups. Moreover, there are two main areas i.e. Estimating
parameter and Hypothesis test. Hence, it can be said that in order to make judgement of
probability that an observed difference between groups is a dependable, inferential statistics is
used.
For example, a mall wants to make a survey and pick an employee, ask to collect sample of
100 who like shopping. Then during this time, there are two option available to an employee.
First is to make a bar chart which contain yes or no answers which is known as descriptive
statistic. Next answer is such that around 75-80% of the population like to do shopping which is
known as inferential statistics.

Another example is such that to determine the exam marks of all the student in UK. Thus,
it is not feasible to measure the marks of population within UK. That is why, a sample is chosen
which used to represent the larger population of UK students. Thus, it will help to determine the
trend of marks of student within Universities.
Question 2
a. i. Preparing Joint probability rate
Joint Probability (AUB)
A
Yes
(B)
23 and under 207
24 - 26 299
27 - 30 185
31 - 35 66
36 and over 51
ii. Probability that a student is 24-26-year-old
By using the formula of probability = number of favourable outcome / Total number of
outcome, the chances of applying to more than 1 University is identified.
Probability = 299/678
= 0.44
iii. Number of University applied to independent of student age

.
Interpretation: From the above trend line, it is interpreted that as the age increases, the number
of people did not apply to more than 1 University. Therefore, it is a reflected that Universities are
dependent to student age group because if people become older, they did not enrol within any
Universities.
b. Computing expected value and variance
x f(x)
10 0.05
20 0.10
30 0.10
40 0.20
50 0.35
60 0.20
Expected
value 43
Variance 350
Question 3
Sample mean = $745
Sample size (n) = 60
Standard deviation (ϭ) = $300

i. Presenting hypothesis
Null Hypothesis (H0): The population annual expenditure for prescription drugs per person is not
lower in the Midwest than in the Northeast
Alternative Hypothesis (H1): The population annual expenditure for prescription drugs per
person is lower in the Midwest than in the Northeast
ii. Deciding the suitable test statistic
Z = 745 – 838 / 300/ udrt 60
Z= -2.40
Hence, test statistic is -2.40
iii. Presenting the values
P value = P (Z < 2.40)
= 0.0082
Thus the p value is 0.0082
iv. Describing the decision at 99% confidence level
The confidence level is 0.01 and p value is 0.0082. So, as per the standard criteria it is
analysed that p < 0.01. that is why, null hypothesis is rejected over others.
v. Conclusion
By applying the relevant formula, it has been concluded that the population annual
expenditure for prescription drugs per person is lower in the Midwest than in the Northeast.
Question 4
a. Hypothesis
Null Hypothesis (H0): There is no significant difference between the mean value of price of
gasoline for three bands.
Alternative Hypothesis (H1): There is a significant difference between the mean value of price of
gasoline for three bands.

b. Decision rule at 5% significance level
As per the standard criteria, if the p value of the data is higher than 0.05 then null
hypothesis is accepted which is clearly reflected that there is no difference between the mean
values.
c. Calculate the test statistic
In order to calculate the values, different steps are followed which are as mention below:
Step 1:
In excel, click on data, then data analysis table.
Step 2:
In data analysis table, click on anova single factor table which helps to identify the p
value.
Therefore, by applying the same, it is analysed that p value of the given data is 0.36.
d. Result
By applying relevant test, it has been identified that p value (0.36) is greater than 0.05 (as
output attached in appendix 1), that is why, alternative hypothesis is rejected. Therefore, there is
no relationship between the prices of any of the three gasoline bands.
Question 5
a. Missing values
B = sample size = 50
K = 3
C= N-K
= 50- 3
C = 47
SSE / SSB (D)= ?
SST= 42699148.82
SSW = 7448393.148
SST = SSB + SSW
42699148.82 = D + 7448393.148
35250755.67 = D

E= SSB /(k-1)
E = 35250755.67 / (3-1)
E = 35250755.67 / 2
E = 17625377.84
F = SSW / (N-K)
F = 7448393.148 / (50 - 3)
F = 7448393.148 / 47
F = 2482797.71
G= E / F
G = 17625377.84 / 2482797.71
G = 7.098
T stat (H) = Coefficient / Standard error
= 33.1330 / 3.9679
H = 0.083
A= 1- SSE /SST
A = 1 - 35250755.67 / 42699148.82
A = 0.82
b. Annual credit card charges
The regression equation is Y = a + bx
So, a = 1304.9048
B = 3.6979
Then the credit charges = 1304.9048 + 3.6979 * 40000
= 1304.9048 + 147916
= 149220.90

c. Regression equation is fit
Yes, there is a direct relationship between the significant factor and as the p value is lower
than 0.05 that is why, the regression equation is clearly fit to data. Therefore, p<0.05 then
alternative hypothesis is accepted and that is why, there is a significant difference between the
consumer characteristics that can be used to predict the amount charged by credit card users.
Question 6
a. Linear trend equation forecast the sales of face mask for October 2020
b. Calculate the forecast sales difference by using Weighted average method
sales (in $)
April 17000
May 18000
June 19500
July 22000
August 21000
September 23000 22200
By applying the method of Weighted average method = 0.2* sales of July (22000) + 0.3* sale of
August (21000) + 0.5* sale of September (23000) = 22200

Appendix
Anova: Single
Factor
SUMMARY
Groups Count Sum Average Variance
A 10 38.11 3.811 0.003721
B 10 38.46 3.846 0.005382
C 10 38.5 3.85 0.004244
ANOVA
Source of
Variation SS df MS F P-value F crit
Between Groups 0.009207 2 0.004603 1.034629 0.369022 3.354131
Within Groups 0.12013 27 0.004449
Total 0.129337 29