Polluted well water: binomial distribution problem.

Question

My probability textbook presents the following example problem on binomial distributions:

It is conjectured that an impurity exists in $30$% of all drinking wells in a certain rural community. In order to gain some insight into the true extent of the problem, it is determined that some testing is necessary. It is too expensive to test all of the wells in the area, so 10 are randomly selected for testing.

And one of the questions asks:

The notion that 30% of the wells are impure is merely a conjecture put forth by the area water board. Suppose 10 wells are randomly selected and 6 are found to contain the impurity. What does this imply about the conjecture? Use a probability statement.

I solved this problem on my own by calculating the probability that $6$ wells are found to contain the impurity. Let $X$ denote the number of impure wells found. Then:

$P(X=6)=b(6; 10, 0.3) =$ $10 \choose 6$ $(0.3)^6(0.7)^4=0.0368 = 3.68\%$

So this is how I interpret this result:

If we assume the conjecture is in fact correct, then it would imply that the probability of finding $6$ wells that are impure out of a random sample of $10$ is only $3.68\%$, which is quite low. This suggests the conjecture is inaccurate.

Here is how they approached the problem:

Here's what I'm wondering: why did they choose to consider the probability that $X \geq 6$?

Answer 1

We are trying to assess whether or not having $6$ impure wells is believable, under the assumption that the actually proportion is $30$%. If having $6$ impure wells were hard to believe, then having more than $6$ would be even harder. So one should count the probability of $6$ or more, not just $6$.

Perhaps it will help if you consider another example which illustrates the issue with only calculating the probability of one specific value: I want to know whether I should believe that a coin is fair. So I assume it is fair, toss it $1000$ times, and observe that it comes up heads $500$ times. Under the assumption of fairness, I calculate that the probability of this outcome is $0.025$. This is very small, so I do not believe the coin is fair.

The conclusion is clearly wrong! I hope this illustrates the problems that may arise with calculating probabilities for single outcomes only.

Answer 2

This is the well-known procedure for a hypothesis test. You are trying to determine if the value $X=6$ lies in the critical region (usually defined as the upper or lower $5$% of the distribution). You can determine this to be the case in this example if $p(X\geq6)<0.05$ which is true.

We conclude therefore that there is evidence that the percentage of polluted wells is greater than $30$% at the "$5$% level" of significance.

The individual probability $p(X=6)$ is of no importance. However the occurrence of 6 polluted wells in the sample can be regarded as very unlikely if the proportion is only $30$%.