Business Statistics Module 6 Assignment: Data Analysis and Statistics

This document provides a complete solution to a Business Statistics module 6 assignment. The assignment covers several key statistical concepts, including the influence of package types on Chi-square statistics, the comparison of p-values and Cramer's V for standardization, and the calculation of average Chi-square values. The solution then delves into probability calculations, such as determining the probability of p-values being below a certain threshold, and analyzing the association between package type and location based on p-values. Further, the assignment explores the application of binomial and normal models to analyze damaged machines, including hypothesis testing and the appropriateness of each model. Finally, the document addresses data analysis for contact lenses and stock movement, determining the suitability of Chi-square tests in various scenarios and providing recommendations for data collection and analysis. The solution includes references to relevant statistical resources.

Business Statistics Module six Assignment
Student’s Name
Institution Affiliation

Question One
a. Influence of a large number of package type on the value of Chi-square
statistics
According to Stine& Foster (2014), large sample increases the Chi- Square statistics. Therefore,
a large number of package types will result in a large value of Chi- square statistics.
b.
When the independency of null hypothesis holds, chi-squared statistics are not directly compared
between two tables of different dimensions, as they are comparable only with contingency table,
for example, 2x2 contingency table, due to fact that dissimilar dimensional tables produce
different results. This rule is applied to avoid a rejection of the null hypothesis when it’s
statistically significant.
c. Explaining whether Cramer’s V be used instead of p-values to standardize the
results.
For standardization of the results, Cramer’s V will not be employed instead of p-value as it’s
only indicate the effect size of the Chi-square statistics as opposed to p-value which shows
whether Chi-square statistics is statistically significant or not.
Advantages of p-values over Cramer’s’ V
P-values help in decision making about the null hypothesis. The rejection or the adoption of
null hypothesis depends on the p-value which p-values p-value which is subjected to a specified
degree of significance say 5%. When the computed p-value is below this level of
significance null hypothesis is reject and above this level null hypothesis is adopted(Ruppert,
2014). The Cramer’s V on the other hand will be used to determine the effect size of the

statistics, which helps in determination of the accuracy of test statistics. The Cramer V
below 0.20 indicates small effect, above 0.20 shows medium effect and, 0.70 and above shows
large effect (Stine& Foster (2014).
Disadvantages of p-values
When the p-values is higher than the significance level, the results of the Chi-square
statistics is not efficient and the Cramer’s V will show small effects which indicate
decreases importance of Chi-square.
d. Average value of 69 chi-squared statistics, given that of the 650 products, 69 comes
in 5 types of packaging.
The average will be given by
Average= Number of products
Number of types of packaging= 69
5 =13.80
Therefore, the average value of 69 Chi-Square Statistics is 13.80
Question Two
a.
The probability of a single p-value been above 0.05 is 0.95, therefore the probability for the 50 p-
values been less than 0.05 will be given
prob( p<0.05)=1−0.95n , wheren is he number of products
1−0.9550=0.9231
The number of p-values which are less than 0.05 will be given by
np=50∗0.9231=46.15≈ 46 pvalues

b.
This means there should be 1 to 50 p-values which are less than 0.01
The probability of a single p-value been above 0.01 is 0.99, suggesting that the probability for the 50
p-values been less than 0.01 will be computed as.
Prob ¿
c.
If the smallest value of p-value is less than 0.01, the package type and location will be
associated as 0.01 is less than 0.05 significant level
Question Three
Data: Damaged Machines
a.
Data supply sufficient evidence not to accept the null hypothesis. Below is proof of this
argument.
Binomial model
The p-value in the binomial model indicates probability.
Information provided:
Number of successful trials is exactly 5 dents
Total Number of independent trials is 60 washers
The probability of success in each trial is 2 %( 0.02)

Using the above information the probability of obtaining 5 dents can be determined using
Microsoft Excel Binomial formula as follows.
=BINOM.DIST (5, 60, 0.02, FALSE) = 0.005753
The binomial probability is 0.005753, which is less than 0.05, thus the null hypothesis is
rejected. This clearly shows that the data supply sufficient evidence to reject the null hypothesis.
b. Necessary assumptions for using the binomial model
Is that the assessment should yield independent results with the constant percentage that
maintain damaged machines to a certain level for the determination of defects.
c. Normal model
Here the probability( p) of finding a successful dent will be the subject of the test.
The hypothesis to be tested using z-statistics
H0 : p ≤ 0.02
H1 : p>0.02
First, Computation of sample probability ( ^p)
This will be given by,
^p= Number of successiful trials
Total Number of indendepent trials
¿ 5
60 = 1
12 =0.083
Second, computation of z-score
z= ^p− p
√ pq
n

¿ 0.083−0.02
√ 0.02 ( 1−0.02 )
60
¿ 0.063
0.018 =3.49
Thus the z-score is 3.49
The probability of z-score from z-table
P ( z ≥ 3.49 )=1−P ( z <3.49 )
¿ 1−0.99976=0.00024
Therefore, the computed p is 0.00024, which is less than 0.05, thus the null hypothesis be
rejected. This implies that p is greater than 0.02
d. The test procedure that should be used.
The binomial test procedure is more appropriate compared to normal test since the probability of
successful trials (0.02) is very small. Also, the success/ failure condition is not met. Success
condition, n∗p=60∗0.02=1.2 failure condition, n∗q=60∗0.98=58.8,
Question Four
Data: New Contact Lens
a.
It should be used for either left or right in every patient. The choice of the eye should be done
randomly to avoid biasness.
b.

The data obtained for the comfort level with new materials lens should be analyzed and the
results should be compared with the analysis results of the comfort level of the old material lens.
Question Five
Data: Stock Movement
For comparison purpose in the Chi-squared test, two categorical variables have to be
independent. In the stock market case, stocks up and downs, and days are dependent on each
other, thus the data provided is not appropriate for employing Chi-square test of independence.

References
Stine, R., & Foster, D. (2014). Statistics for Business: Decision Making and. Addison-Wesley
SOFTWARE-JMP.
Ruppert, D. (2014). Statistics and finance: an introduction. Springer.