An aspirin manufacturer ﬁlls bottles by weight rather than by count. Since each bottle must contain 100 tablets the average weight per tablet must be 5gr. One tablet from each bottle from a very large sample of bottles is weighed. The average weight of the sample was found to be 4.87gr with a (sample) standard deviation of 0.35gr. Is the information provided on a bottle’s label indicating that a bottle contains 100 tablets misleading?

Question

anonymous · Answer

So I tries solving for p-value using a two-tailed z test following an example from my stats book. I used the equation:

$$z = \frac{ x - \mu }{ s/\sqrt{n} } = \frac{ 4.87-5 }{ 0.35/\sqrt{100} } = -3.71$$ The p-value for a two-tailed z test is supposed to be:

$$2[1-\phi(\left| z \right|)]$$ But when I use the standard normal curve area chart to find 3.7 it doesn't even give me a value. I'm wondering if I used the wrong test or I'm just solving this the wrong way.

mww · Answer

One thing I must ask. Why are you using a z-table when you don't have the population statistic? You need to use Student's t tables. so really it's not Z but t

mww · Answer

in any case yes you needed to use a one sample t-test.
What you get is often not in the tables because it's such an extreme value so the p value is very small, smaller than 0.001. Since the t value is negative, your aspirin bottle is very likely to be smaller than the the theoretical value, so fewer than 100 tablets is not surprising. The alternative hypothesis is supported.

mww · Answer

If you use this table (99df for n = 100) you see this is not even on the table. No biggie, to increase to 3.71 you must also reduce the p value so it becomes even more significant!

mww · Answer

^ https://www.easycalculation.com/statistics/t-distribution-critical-value-table.php