Strayer University
Topic: Bottle Company Case Study
MAT 300 – Statistics
Professor Jonathan Nukpezah
Question #1: This question calculates the mean, median and Standard Deviation for the bottling case study:
Mean
Mean =14.87
Note: (Calculation of Mean=14.5+14.6+14.7+14.8+14.9+15.3+14.9+15.5+14.8+15.2+15+15.1+15+14.4+15.8+14+16+16.1+15.8+14.5+14.1+14.2+14+14.9+14.7+14.5+14.6+14.8+14.8+14.6) / 30)
Median =14,14,14.1,14.2,14.4,14.5,14.5,14.5,14.6,14.6,14.6,14.7,14.7,14.8,14.8,14.8,14.8,14.9,14.9,14.9,15,15,15.1,15.2,15.3,15.5,15.8,15.8,16,16.1
So Mean = 14.87
Standard Deviation
SD=0.550
Note: (Calculation for SD= Since mean = 14.87
So now Variance = 0.302
Standard deviation =
So by putting values we get Standard Deviation = 0.550)
Question #2: The calculation for the can be done in the following manner (Prof. McGahagan, 2012). The standard error is 0.001, the margin of error is 5% as the confidence interval is set to 95%
So
Confidence interval = 14.87±0.001
So the upper limit is 14.871
Lower limit is = 14.869
Question # 3: For hypothesis testing
H0 = null hypothesis
H1= alternate hypothesis
For the case study, let’s consider
H0 = bottles contain less than 16 ounces
So, H0= µ<16
H1= bottles contain 16 ounces=µ= 16
Applying the test, putting mean value calculated above and other values
We get
Z= 14.87-60/0.550/√30
= -451 approx
So -451< -14.689
The null hypothesis rejected in the above calculations. The quantity in the bottle is the same is it was advertised. (Martin 2006)
Question #4: As a manager I suggest that the:
Bottles do not contain less than 16 ounces. The use of standardized plants is the assured that the bottles are filled as per the requirements and the chances of low level of liquid is less in this case. This might be able to happen because of the bubbles originate from the carbon dioxide.
The suggested method to this issue is