Chapter 15, Problem 15.32E

(a)

To determine

To find: the probability that I specific test shows a difference that is statistically significant at the 5% level.

Answer to Problem 15.32E

The probability that exactly 1 out of 77 cases yields a positive result is $0.078$ .

Explanation of Solution

Given:

The poll stayed open for several weeks in October 2012. Of the 1290 vote cast, 864 were “yes” response.

Calculation:

To solve for the probability that exactly 1 out of 77 tests yield a statistically significant result, you will need to reach back to chapter 12 and model this as a binomial probability. Use the binomial probability equation $P\left(X=k\right)=\left(\begin{array}{c}n\\ k\end{array}\right)p^k{\left(1-p\right)}^{n-k}$ where you are solving for 1 out of the 77 cases which means $\left(P\left(X=1\right)\right)$ and $\left(n=77\right)$ and the probability is $\left(p=0.05\right)$ to represent a 5% significance level.

Here is the initial setup. Remember that the value for $\left(k=X\right)$ so $\left(k=1\right)$ since $\left(X=1\right)$ . For simplicity, the binomial coefficients $\left(\begin{array}{c}n\\ k\end{array}\right)$ will be left out for now.

  $\begin{array}{l}P\left(X=5\right)=\left(\begin{array}{c}n\\ k\end{array}\right)p^k{\left(1-p\right)}^{n-k}\hfill \\ =\left(\begin{array}{c}n\\ k\end{array}\right){\left(0.05\right)}^1{\left(1-0.05\right)}^{77-1}\hfill \\ \begin{array}{l}=\left(\begin{array}{c}n\\ k\end{array}\right)\left(0.05\right){\left(0.95\right)}^{76}\\ =\left(\begin{array}{c}n\\ k\end{array}\right)\left(0.05\right)\left(0.020277\right)\\ =\left(\begin{array}{c}n\\ k\end{array}\right)\left(0.0010139\right)\end{array}\hfill \end{array}$

Now add the information for the binomial coefficients $\left(\begin{array}{c}n\\ k\end{array}\right)$ . Recall that the binomial coefficient calculated with the following equation

  $\begin{array}{l}\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{n!}{k!\left(n-k\right)!}\\ =\left(\begin{array}{c}n\\ k\end{array}\right)\left(0.0010139\right)\\ =\left(\frac{n!}{k!\left(n-k\right)!}\right)\left(0.0010139\right)\\ =\left(\frac{77!}{1!\left(77-1\right)!}\right)\left(0.0010139\right)\\ =\left(\frac{77!}{1!\times 76!}\right)\left(0.0010139\right)\\ =77\times \left(0.0010139\right)\\ =0.078\end{array}$

The probability that exactly 1 out of 77 cases yields a positive result is $0.078$ .

Conclusion:

Therefore, the probability that exactly 1 out of 77 cases yields a positive result is $0.078$ .

(b)

To determine

To prove: it is not surprising that 2 of the 77 tests were statistically significant at the 5% level.

Answer to Problem 15.32E

It is not surprising that 2 of the 77 tests yielded significant results.

Explanation of Solution

Calculation:

Because they ran 77 tests at a 5% significance level, then you would expect to see about 4 of these 77 tests have a significant result based on the following logic.

  $\begin{array}{l}0.05=\frac{x}{77}\hfill \\ 77\times 0.05=x\hfill \\ 3.85=x\hfill \end{array}$

It is similar to rolling double 6's in craps. You expect this to occur 1 out of 36 times which means

It has a significance level of $\left(\frac{1}{36}=0.0278\right)$ . Let's say someone plays craps 144 times, you expect to see this unlikely event to occur approximately four times since...

  $144\times 0.0278=4$

Thus it is not surprising that 2 of the 77 tests yielded significant results.

Conclusion:

Therefore, it is not surprising that 2 of the 77 tests yielded significant results.