Stuck in interpreting p-value

Question

Bob and Daniel are arguing over the average cost of a particular chocolate bar in their local neighborhood. Bob believes that the average cost of a chocolate bar is at least $\$1.50$ while Daniel believes the contrary. To settle their argument, they head out to local stores and record each price they observe for this particular chocolate bar.

After visiting $10$ stores, they find that the average cost is $\$1.35$ with a standard deviation of $\$0.05$. Conduct a hypothesis test to contradict or confirm Bob's argument. You may assume that the cost of the chocolate bars are normally distributed.

My working:

$H_0$: $\mu < 1.50$ and $H_a$: $\mu \geq 1.50$.

The test statistic is,

$T = \frac{1.35-1.50}{\frac{0.05}{\sqrt{10}}} = -9.48$ with df = $9$.

This is where I'm stuck, usually we calculate the p-value in the direction of the null hypothesis. That is, $Pr(T\geq-9.48)$, which is effectively $1$. What am I missing here? Surely I'm wrong.

Answer 1

You should be calculating the $P(-9.48≥T)$ (the opposite of what you have done) and then work from there. You know thus that the p-value is very close to $0$, meaning that $H_0$ should be rejected. However, this is dependent on the confidence level, although, it is safe to say with such a low p-value, $H_0$ should be rejected to a reasonable degree of certainty. If this feels counter-intuitive, bear in mind that the standard deviation of the data collected is very small.

Answer 2

The T-value shows how unlikely it is to get the sample mean you got (1.35) when the true mean is 1.5 for this 2-tailed distribution. You want to $p=P(T<-9.48)$, what is the probability that you got this sample mean by chance. That p-value is less than 0.00001. So the null (that the true mean is at least 1.5) should be rejected (at any reasonable significance).

Answer 3

This should be rather a comment, but too long for one.

The formulation is misleading; I don't like it at all.

"Conduct a hypothesis test to contradict or confirm Bob's argument" implicitly asks to take Bob's opinion as the null. But why Bob's? In the first paragraph, the roles of Bob and Daniel are equivalent! So with this "statistical test", Daniel is discriminated: if the observed average is $\$1.50$ or even slightly less than that, Bob's opinion wins.

Well, one can try to set the significance level at something like $0.5$ to "establish the balance", but this would be even more illegal and misleading. In statistical testing, the roles of null and its alternative are very different! So you can't apply these procedures in the case where there's symmetry. In those cases, you may try to formulate it as a statistical decision making problem, but, I repeat, never as a statistical testing problem.

So somehow the author teaches us to apply statistical tests where they are not applicable.

To make this clearer, here is a formulation suitable for statistical testing.

Bob believes that the average cost of a chocolate bar is at least $\$1.50$. Daniel wants to change Bob's mind. To this end, he invites Bob to visit local stores and record each price they observe for this particular chocolate bar.

After visiting $10$ stores, they find that the average cost is $\$1.35$ with a standard deviation of $0.05$.

Design a hypothesis test for Daniel to change Bob's opinion. (You may assume that the prices of the chocolate bars are normally distributed.) Will the observed data be enough for that?

Answer 4

Your choice of hypothesis is just wrong. Note that you must choose the alternative hypothesis so that it is in favor of the sample. Here you must choose $H_a\colon \mu<1.5$ since $\bar x=1.35$.