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Managerial Statistics
WEB CASE
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1). Data Collection procedures followed CCACC toform its
conclusions are flawed, because theyvefollowed Convenience
Sampling
They should have followed Simple Random Samplingto make their
conclusions more accurate. Alsosample size used should be more than
5 units.
2) (a) Oxford Os mean weight:
(360.4 + 361.8 + 362.3 + 364.2 +371.4 )/5
=(1820.1)/5
= 364.02
Oxford Alpine frosted flakes mean weight:
Chapter 7
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(b)
Parameters Oxford Os Alpine
368 368
364.02 368.04
15 15
Z -0.59 0.006
Z(Critical) -1.64 -
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3). It will be incorrect to conclude about the filling process
of the
two cereals, as the conclusion is drawn using biased samples of
5units each.
4). Yes, the request is reasonable because allegations is
not
backed with ample evidence.
5). Yes, such techniques can be used if we take an unbiased
sample by using simple random sampling. In such scenario, if
there is any cheating case it can be proved by CCACC.
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Oxford Cereal chapter 9
The Null &Alternate hypothesisare as follows
H0 : < 368
H1 : > 368
As depicted the p value0.07
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Oxford Cereal chapter 9
The confidence level for themean weight of the box contents
is
367.16 < < 373.71
This implies that the contents ofthe box could weigh anything
upto 373.71 grams,hence Oxfordcereals claim that many boxes
contain more than 368 grams isnot surprising and is true.
Possibility that the publicexperiment and the CCACCclaim can
both be right exists
because the sample size taken inboth cases could have been
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Q. On a recent purchase of a new sample of cereal boxes--ten
produced atPlant #1 and ten produced at Plant #2. It was discovered
that the average
weight of Plant #1 boxes was 372.44 grams, but that the average
weight ofPlant #2 was 365.64 grams, less than the 368 grams
claimed.
Even after the recent public experiment about cereal box
weights, theCCACC remains convinced that Oxford Cereals has mislead
public.
1. Do the CCACCs results prove that there is a statistical
difference in themean weights of cereal boxes produced at Plant# 1
and 2?
2. Perform the appropriate analysis to test the CCACCs
hypothesis. Whatconclusions can you reach based on the data?
Chapter 10- OXFORDCEREALS
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Samples are randomly and independently drawn.
Populations are normally distributed
Population variances are unknown
If population variances are equal, the pooled-variance t test is
appropriate
Here, use a separate-variance t test, which includes the two
separate sample variancesin the computation of the test
statistic
We have 2 sets of data where the variances are unknown, so we
have 2 options here.We can either apply POOLED-VARIANCE t-test or
SEPARATE VARIANCE t-test.
So we will apply F-test here to decide for the same.
Solution
2
2
2
1
S
SF =
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DATA
Plant 1 Plant 2
375.28 364.51
368.19 367.95356.98 368.62
366.1 370.25
379.84 345.45
365.51 359.36376.39 379.64
398.66 378.71
353.09 358.14
384.37 363.74
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From the F-test we can see that variances are different,
so we will apply separate variance t-test.
F-testF-Test Two-Sample for Variances
Variable 1 Variable 2
Mean 372.441 365.637
Variance 180.8843 101.1672
Observations 10 10
df 9 9
F 1.787974
P(F
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t-Test: Two-Sample Assuming Unequal Variances
Variable 1 Variable 2
Mean 372.441 365.637
Variance 180.884277 101.167201
Observations 10 10
Hypothesized Mean Difference 0
df 17t Stat 1.2811504
P(T
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Hypothesis:
From the results, we can see that the t-value is less than
the t-critical, so we can accept our hypothesis.
Interpretations
Two-tail test:
H0: 1 = 2H1: 1 2
i.e.,
H0: 1 2 = 0H1: 1 2 0
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THANK YOU.
