# Hypothesis testing

A bottler of iced tea wishes to ensure that an average of 16 ounces of tea is in each bottle. In order to analyze the accuracy of the bottling process, a random sample of 150 bottles is taken. Using a t-distributed test statistic of -1.09 and a 5% level of significance, the bottler should: A) reject the null hypothesis and conclude that bottles contain an average 16 ounces of tea. B) reject the null hypothesis and conclude that bottles do not contain an average of 16 ounces of tea. C) not reject the null hypothesis and conclude that bottles contain an average 16 ounces of tea. D) not reject the null hypothesis and conclude that bottles do not contain an average of 16 ounces of tea. Your answer: C was correct! Ho:µ = 16; Ha: µ ≠ 16. Do not reject the null since |t|=1.09 < 1.96 (critical value). -------------------------------------------------------------------------------------------------------------- Although I got the question correct, but I don’t really agree with their interpretation, I thought when you fail to reject the null, the results are NOT statistically significant, so how can you conclude that the bottles do contain an avg of 16 ounces of tea…but on the other hand, when you try to test such a scenario, where your “hope for” condition is miu= some value, how do you do that to achieve statistical significance? any insights?

Yeah, the right answer is E) not reject the null hypothesis because there is not enough evidence that the mean is different from 16 ounces at the 5% level of significance.

Joey, so how do you test for this, when you try to test a scenario where your “hope for” condition is miu= some value, how do you do that to achieve statistical significance? I.e. is it possible to have the alternative hypothesis be miu= some value rather than miu not equal to some value?

It’s a philosophical issue. What do you mean by 16 oz? Is 16.00001 = 16 oz? You can do a C.I. and conclude with whatever prob you want that the mean is arbitrarily close to 16.