Chocolate Packet Weight Analysis

This assignment involves a comprehensive analysis of chocolate packet weights for different cocoa percentages (70% and 85%). Students perform hypothesis tests to determine if there's a significant difference in average weight between the types, construct 99% confidence intervals, and provide recommendations for process target adjustments based on the findings.

STATISTICAL ANALYSIS TO SUPPORT DECISION MAKING
STAT1060 – ASSESSMENT 2
Student id
[Pick the date]

1. The total minutes with respect to each of cause is shown below:
Cocoa feeder
issue Excess of sugar
coating Mixer problem Operator
missing Packaging issue Sugar feeder
issue
0
20
40
60
80
100
120
140
120
15
40
20 25
95
Column Chart : Cause vs Time
Cause
Time (min)
Based on the above graph, it is apparent that the two longest delays are on account of feeder
issues related to cocoa and sugar and hence these need to be the priority areas as par as the
management is concerned and ensuring that these two issues are minimal would enhance the
overall productivity and minimize wastage of limited resources.
2. Process target mean = 156 g
Standard deviation = 2 g
(a) Packet weights are normally distributed and thus, the percentage of packets which that
are underweight would be calculated with the help of empirical rule.
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Empirical Rule
Nearly 99.7% of data would fall between 3 standard deviation above and below of the mean.
¿ mean ±3(Standard deviation)
¿ 156 ± 3 ( 2 )
¿ 156 ± 6
¿ 150 162
It can be seen that all the values fall within this range and hence, (100-99.7) = 0.3% of the
packets would be underweight.
(b) The output from statstar.io in order to find the process target is shown below:
Relevant graph
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Therefore, the process target is computed as 153.29 g determined through the use of trial and
error method.
3. The contingency table is highlighted below:
(a) This relationship would be important to know since if gender preferences do tend to exist,
then the promotional campaign would be designed in such a matter that the gender which
tends to have a liking or affinity for the given type of chocolate would be targeted and the
campaign would be designed keeping in mind the target audience. Further, alternative
products which are more appealing to the gender which does not have a liking towards dark
chocolate may be looked at.
(b) Considering that sample has been chosen from the employees, it might be the case that the
sample is not random and thus the ones who might be willing to participate may have a liking
towards dark chocolate. Also, it might be possible that the responses of certain employees
may not be correct as the employer is conducting the survey. Hence, it is highlight likely that
the percentage of people in general who like dark chocolate may be overestimated.
(c) Level of significance =5%
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The hypotheses are presented below:
Null hypothesis H0 :Liking for the dark chocolate is independent of the gender of the employee.
Alternative hypothesis H1 :Liking for the dark chocolate is dependent on gender of the employee.
Chi square test would be used to test the hypothesis.
Value of chi square X2 =6.437
Degree of freedom ¿ ( r −1 ) ( c −1 )= ( 2−1 ) ( 2−1 )=1
Probability density function
X =X2 (1)
P ( X ≥6.437 )
The p value corresponding to chi square statistics and degree of freedom comes out to be
0.011176.
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It can be said that p value is lower than level of significance and hence, sufficient evidence is
present to reject the null hypothesis. Therefore, alternative hypothesis would be accepted. Thus,
it can be concluded that liking for the dark chocolate is dependent on gender of the employee.
Relevant graph
4. (a) Graphical display for the 100 packets for 70% cocoa is highlighted below:
152 - 153.19
153.2 - 154.39
154.4 - 155.59
155.6 - 156.79
156.8 - 157.99
158 - 159.19
159.2 - 160.39
160.4 - 161.59
0 5 10 15 20 25 30
Bar Chart - 70% Cocoa
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It is apparent from the above that the distribution seems to be skewed towards the right and
hence the underlying distribution would not be normal. The prevalence of the skew on the right
is visible from the right tail and existence of very high values i.e. exceeding 161g. It is apparent
that the minimum level is about 4g lower than the process target of 156g. However, the higher
level is more than 5.5 g higher than the process target of 156 g.
(b) Let x=150 , μ=156 ,σ =2
z= ( x −μ
σ )
z= (150−156
2 )
z=−3
Probability would be determined with the help of NORMSDIST ().
P( Z ←3)=NORMSDIST ( −3 ) =0.00135
Considering the miniscule probability of the packet being underweight, it is apparent that the
process target should be fixed at a lower value and the current level of 156g seems excessively
high.
(c) Level of significance = 1%
The hypotheses are presented below:
Null hypothesis H0 :μ=156
Alternative hypothesis H1 :μ ≠ 156
It is apparent that sample size is higher than 30 but still z statistic would not be used as the
standard deviation of the population is assumed to be unknown. Hence, the appropriate
alternative would be T statistic.
Sample mean=155.948
Standard deviation=1.888
Computed T statistic ¿
( 155.948−156
1.888
√ 100 )=−0.278
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The p value corresponding to t score above with 99 degrees of freedom is 0.7815.
It can be seen that p value is higher than level of significance and therefore, fails to reject the null
hypothesis. Hence, the process target of 156g was achieved during the weekly production.
(d) 99% confidence interval
It can be interpreted with a 99% confidence or 0.99 probability that the average weight of the
70% cocoa 150 g packets would lie between 155.45 g and 156.44 g.
e) Part c and Part d are related since the null hypothesis in part c would have been rejected only
if the sample mean would not have lied in the 99% confidence interval that has been determined
in part d. However, since the sample mean did lie in the 99% confidence interval, hence null
hypothesis was not rejected.
5. (a) Graphical representation to compare 70% and 85% cocoa chocolate packet’s weight
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149..01 -
151 151.01 - 153 153.01 - 155 155.01 - 157 157.01 - 159 159.01 - 161 161.01 - 163
0
5
10
15
20
25
30
35
40
45
50 Bar Chart 70% cocoa
85% cocoa
Weight (g)
Frequency
Both the distributions seem to be skewed on the right side and hence are not normal. Further for
85% cocoa 150g packet, it seems that being underweight is a likely possibility while for the 70%
cocoa 150g packet, the likely possibility is being overweight. The mean weight of the 150g
packet of 85% cocoa is clearly lesser than the corresponding mean of 75% cocoa.
(b) Level of significance = 1%
The hypotheses are presented below:
Null hypothesis H0 :μ1−μ2=0
Alternative hypothesis H1 :μ1−μ2 ≠ 0
The two tailed t test with equal variance is taken into account to test the validity of the claim.
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It can be seen from the above output that p value for two tailed comes out to be zero. Therefore,
it can be said that p value is lower than level of significance and hence, sufficient evidence
present to reject null hypothesis and to accept alternative hypothesis. Thus, it would be fair to
conclude that statistically significant difference exists in average mean weight between 85%
cocoa and 70% cocoa 150 g packets.
(c)Hypothesis testing needs to be performed using the confidence interval approach.
The hypotheses are presented below:
Null hypothesis H0 :μ1−μ2=0
Alternative hypothesis H1 :μ1−μ2 ≠ 0
The computation of the 99% confidence for the difference in means is as indicated below.
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Hence, from the above, it is apparent that the 99% confidence interval for the mean difference
amounts to (2.63g, 4.04g). Clearly, zero which is the hypothesized value for the given hypothesis
is not included in the 99% confidence interval and hence the null hypothesis would be rejected.
Therefore, it would be fair to conclude that the difference between the means of the 70% cocoa
and 85% cocoa is not equal to zero.
6) Based on the above analysis, it is apparent that the management would aim at fixing the
process target more rationally so that the overall production cost would fall. This is because, the
current process target especially for 70% cocoa seems to be quite high with sample starting from
more than 152 g while the actual quantity printed on the packet is only 150g. Hence, by reducing
the process control target, this surplus amount in each packet can be curtailed and hence the cost
could be lowered.
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