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.