
Underfilled Soft Drink Bottles?
A distributor of soft drinks believes that the machine
dispensing soda into bottles is, on average, underfilling
the bottles. To test this claim some bottles will be
randomly sampled and the contents measured. Because of
process variation, each bottle has a slightly different
amount of soda in it. The target is a mean fill of 300
ml; the distributor suspects the mean fill is below 300
ml. Is the convincing evidence for an average underfill
(if so, the machine must be shut down and recalibrated --
at considerable expense).
The distributor wishes to test the following
hypotheses:
H0: m
=
300 Ha:
m < 300
where m is
thepopulation mean fill of all bottles.
A random sample of 6 bottles is obtained;
here are the amounts of soda (ml) in each bottle: 398.4,
296.7, 300.0, 297.9, 299.2, 296.0.
Before proceeding with any statistical
inference, it's best to take a look at the data --
graphically. Here's a dotplot (which is sufficient given
such a small sample size).

Nothing irregular here. However, since
the sample is quite small, we need some assurance that
the sampled population has a normal distribution. A
histogram is useless in assessing such a claim -- there's
too little data for a histogram to indicate much of
anything. So, we obtain a normal probability plot.

This plot indicates by its linearity that
an assumption of normally distributed data is a
reasonable one.
Software is used to compute the
"test statistic" t = (x-bar - m) / [s / sqrt(n) ]. Here's what I
obtained.
T-Test of the Mean
|
| Test of m
= 300.000 vs m <
300.000 |
Variable
|
N
|
Mean
|
StDev
|
SE Mean
|
T
|
P
|
Fill
|
6
|
298.033
|
1.503
|
0.614
|
-3.21
|
0.012
|
The sample mean is 298.033; the sample
standard deviation is 1.503. The standard error of the
sample mean (SE Mean) is the denominator of the
t-statistic given by s / sqrt(n) -- here it's 1.503 /
sqrt(6) = 0.614. Then
t = (298.033 - 300)/0.614 = -3.21.
If the significance level is 0.01 then H0
is rejected when t < -t.01 = -3.365.
Consequently H0 is not rejected at
level a = 0.01.
Note that at significance level 0.05 then
H0 is rejected (because -t.05
= -2.015).
The P-value is 0.012, indicating that for
all a > 0.012 the decision
is reject H0
and for all a < 0.012 the
decision is do not reject H0.
|