An Assignment
On
Applied Statistics
Prepared for
Gazi Mohammad Hasan Jamil
Associate Professor
Department of Finance
Faculty of Business Studies
University of Dhaka
Prepared by
Group No: 07
BBA 24th Batch
Section: C
Department of Finance
Faculty of Business Studies
University of Dhaka
Date of Submission: October 17, 2019
Binomial Probability Distribution
It is reported that 20% of students are exclusively using smart phones for playing online games.
In a sample of 12 students,
a) Find the probability that no one uses a smart phone exclusively for playing online games.
b) Find the probability that exactly 2 students use a smart phone exclusively for playing online
games.
c) Find the probability that exactly 5 students use a smart phone exclusively for playing online
games.
We will solve this problem by using excel software. Here are the procedures for finding the
solution:
a)when no one uses, success = 0, probability = .20
The result is 0.069 which perfectly matches with the result of Table B.1 .
b) The probability that exactly 2 students use a smart phone exclusively for playing online games
is :
The result is 0.283 .
c) The probability that exactly 5 students use a smart phone exclusively for playing online games:
The result is 0.05 .
Poisson Probability Distribution
It is estimated that 1.5% of the students using smart phones for their study purpose only. What is
the probability that out of 1200 students, at least 25 students are using their smart phones for
study purpose only?
Solution:
The result indicates that 2.4% of 25 students are using smart phones for their study purpose only.
Normal probability distribution
Suppose customers of retail shop Agora have a mean expenditure of 500, standard deviation of
expenditure is 100 and this follows normal probability distribution .We can get answer of
following questions from this information:
a) What is the probability of selecting a customer whose expenditure is less than 600?
b) What is the probability of selecting a customer whose expenditure is more than 600?
c) What is the probability of selecting a customer whose expenditure is between 400 and 600?
Uniform probability distribution
Retail shop Agora offer their own membership card to their customers. A customer with a
membership card gets a 10% discount on purchase .Everyday they serve customer with
membership card which ranges from 12 to 20 .This process follows a uniform distribution.
Suppose we want to ask following questions:
a)What is the mean number of customer?
b) What is the standard deviation of this process?
c) What is the likelihood that number of customers will be less than 15?
d) What is the likelihood that number of customers will be more than 18?
Chapter 11 & Chapter 12
In a business/ school, there are 4 departments namely Finance, Accounting, Marketing & Management. A
sample of the CGPA of the 1st year students is given below.
Finance Accounting Marketing Management
2.49 2.58 3.59 3.11
2.25 3.43 3.06 3.33
3.72 3.80 2.79 2.76
3.79 3.69 2.85 2.59
3.62 3.81 3.24 3.40
2.20 3.40 2.03 3.51
3.14 3.80 3.90 2.78
2.02 3.45 3.34 3.88
3.29 3.22 2.25 3.19
3.13 2.98 3.91 3.75
2.19 2.61 2.98 2.76
2.15 3.91 2.92 2.50
2.29 2.68 3.99 3.80
3.56 3.67 2.79 2.30
2.95 3.14 3.88 3.78
3.21 3.12 3.83 2.02
2.12 3.88 2.75 2.62
3.67 3.42 3.44 3.52
3.99 3.18 2.89 3.05
2.38 2.69 2.42 3.40
3.47 2.27 2.67 2.59
2.09 3.85 2.58 3.17
3.43 2.38 2.01 2.82
2.87 3.95 3.94 2.93
3.62 2.44 2.17 3.57
3.52 3.83 3.77 2.37
3.78 2.54 3.79 3.68
2.53 3.73 2.91 2.78
3.53 2.34 2.49
3.62 3.93 3.94
3.92 2.96
3.75 3.80
2.17
3.75
3.90
1. At the 0.05 significance level, can we conclude that the mean CGPA of Finance department is equal
to marketing department?
2. At the 0.05 significance level, can we conclude that the mean CGPA of all the departments are
equal?
Descriptive Statistics:
Finance Accounting Marketing Management
Mean 3.020407 Mean 3.29186 Mean 3.12167 Mean 3.070021
Standard 0.118989 Standard 0.099953 Standard 0.111023 Standard 0.096844
Error Error Error Error
Median 3.177232 Median 3.428843 Median 2.97284 Median 3.079952
Mode 3.62294 Mode #N/A Mode #N/A Mode 2.764886
Standard 0.651728 Standard 0.591328 Standard 0.628044 Standard 0.512451
Deviation Deviation Deviation Deviation
Sample 0.424749 Sample 0.349669 Sample 0.394439 Sample 0.262606
Variance Variance Variance Variance
Solution to 01
The Excel commands for the two-sample t-test
T-Test: Two-Sample Assuming Equal Variances
Select the Data tab on the top menu. Then, on the
Far right, select Data Analysis. Select t-Test: Two
Sample Assuming Equal Variances, and then click
OK.
In the dialog box, indicate that the range of Variable
1 is from B2 to B32 and Variable 2 from D2 to
D34, the Hypothesized Mean Difference is 0, click
Labels, Alpha is 0.05, and the Output Range is
J3. Click OK.
Excel Output
t-Test: Two-Sample Assuming Equal Variances
Finance Marketing
Mean 3.020407 3.12167
Variance 0.424749 0.394439
Observations 30 32
Pooled Variance 0.409089
Hypothesized Mean Difference 0
df 60
t Stat -0.62299
P(T<=t) one-tail 0.267826
t Critical one-tail 1.670649
P(T<=t) two-tail 0.535652
t Critical two-tail 2.000298
Test process:
Step 1: State the null hypothesis and the alternate hypothesis.
H0: µF= µM
H1: µF≠ µM
Step 2: Select the level of significance.
We selected the .05 significance level.
Step 3: Determine the test statistic.
t test
Step 4: Formulate a decision rule.
Reject H0 if t< -2.000298 or t> 2.000298
Step 5: Make the decision regarding H0
Fail to reject H0
Step 6: interpret the result.
The mean CGPA of Finance department is equal to marketing department.
t-Test: Two-Sample Assuming Unequal Variances
Select the Data tab on the top menu. Then, on the
far right, select Data Analysis. Select t-Test: Two
Sample Assuming Equal Variances, and then click
OK.
In the dialog box, indicate that the range of Variable
1 is from B2 to B32 and Variable 2 from D2 to
D34, the Hypothesized Mean Difference is 0, click
Labels, Alpha is 0.05, and the Output Range is
J3. Click OK.
Excel Output
t-Test: Two-Sample Assuming Unequal Variances
Finance Marketing
Mean 3.020407 3.12167
Variance 0.424749 0.394439
Observations 30 32
Hypothesized Mean Difference 0
df 59
t Stat -0.62223
P(T<=t) one-tail 0.268093
t Critical one-tail 1.671093
P(T<=t) two-tail 0.536185
t Critical two-tail 2.000995
Test process:
Step 1: State the null hypothesis and the alternate hypothesis.
H0: µF= µM
H1: µF≠ µM
Step 2: Select the level of significance.
We selected the .05 significance level.
Step 3: Determine the test statistic.
t test
Step 4: Formulate a decision rule.
Reject H0 if t< -2.000995or t> 2.000995
Step 5: Make the decision regarding H0
Fail to reject H0
Step 6: interpret the result.
The mean CGPA of Finance department is equal to marketing department.
Solution to 02
The Excel commands for the one-way ANOVA
Select the Data tab on the top menu. Then, on
the far right, select Data Analysis. Select ANOVA:
Single Factor, then click OK.
In the subsequent dialog box, make the input
Range B2: E37, click on Grouped by Columns,
Click on Labels in first row, the Alpha text box
is .05, and finally select Output Range as I3 and
Click OK.
Excel output
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 1.37874 3 0.45958 1.277655 0.285167 2.679535
Within Groups 43.52441 121 0.359706
Total 44.90315 124
Test process
Step 1: State the null hypothesis and the alternate hypothesis.
H0: µF= µa = µMK= µMg
H1: Not all treatment means are the same.
Step 2: Select the level of significance.
We selected the .05 significance level.
Step 3: Determine the test statistic.
F test
Step 4: Formulate a decision rule.
Reject H0 if F>2.679535
Step 5: Make the decision regarding H0
Fail to reject H0
Step 6: Interpret the result.
The mean CGPA of all the departments are equal
