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# BOTTLING COMPANY CASE STUDY (Math Problem Sample)

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Perform calculations. source..

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BOTTLING COMPANY CASE STUDY
Student’s Name:
Professor’s Name:
Course title:
Date:
Question 1: Mean, median, and standard deviation for ounces in the bottles
Solution:
The sampled values are given as:
14.23, 14.32, 14.98, 15.00, 15.11, 15.21, 15.42, 15.47, 15.65, 15.74, 15.77, 15.80, 15.82, 15.87, 15.98, 16.00, 16.02, 16.05, 16.21, 16.21, 16.23, 16.25, 16.31, 16.32, 16.34, 16.46, 16.47, 16.51, 16.91, 16.96
* Mean
Sample Mean=1n i=1i=nxi
=130i=1i=30xi
=130{14.23+14.32+…+16.91+16.96}
=130×475.62
=15.854
* Median
Since size of the sample = 30, an even number, then;
The median will be given by average of the two meddle observations, that is, [n2]th and [n2+1]th
Therefore, median = [n2]th + [n2+1]th 2
[n2]th=302
15thobservation =15.98
And
[n2+1]th=16thobservation=16.00
Median=15.98+16.002
=15.99
* Standard deviation
standard deviation s= i=1i=30(xi-x)2n-1
But x̅=15.854 and n=30
s= (14.23-15.854)2 +…+(16.96-15.854)230-1
s=(-1.624)2+…+(1.106)2 29
s=12.68629
s=0.6614
Question 2: A 95% Confidence Interval for the ounces in the bottles
Solution:
Confidence interval will be given by the summation of mean and margin error
confidence Interval=X±Zsn whereby Zsn is the margin error,
But X̅=15.854, s=0.6614 and n=30
At 95% confidence interval Z=1.96
=15.854±1.960.661430
=15.854±0.237
=(15.617, 16.091)
Therefore, we are 95% confident that population mean falls between 15.617 and 16.091
Question 3: Hypothesis test
Solution:
Set null hypothesis (H0) and alternative hypothesis (H1)
Note: The null hypothesis is usually a negative statement to the claim. In this case, the null hypothesis should suggest that bottles of the product do not contain less than 16 ounces. On the other hand, the alternative hypothesis is the claim we wish to test, that is, is mean volume less than 16?
H0:u0=16 ounces
H1:u1<16 ounces
Taking 95% as the acceptance level, then significance level would be 5% (0.05)
For one tailed test (lower-tail in this case), δ=0.052=0.025
Using normal distribution table to identify rejection region, we find Zδ=Z0.025=1.645
Calculating the test statistic;
Zc=x-u0s/√n
x=15.854, u0=16, s=0.6614 and n=30
Zc=15.854-160.6614/√30
=-0.1460.12075
=-1.2091
Comparing tabulated and calculated values
Zδ=-1.645<ZC=-1.2092
28289261155710390525113447
Rejection
66675041275region
2000250110490
11239501600202000250160020010096492667001114425-19051114425-1905005143508572500428625150495428625112395409575150495
390525-635 -1.645 Zc= -1.2091 0 00 -1.645 Zc= -1.2091 0
4743456273808553451351280
Acceptance region
From the sketch, the calculated value Zc falls within the accep...

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