Chapter 23, Problem 33E

(a)

To determine

To find out what kind of error did Hopps make.

Answer to Problem 33E

Hopps make Type I error.

Explanation of Solution

In the question, it is given that YvonHopps ran an experiment to test optimum power and time settings for microwave popcorn. His goal was to find a combination of power and time that would deliver high quality popcorn with less than $10\%$ of the kernels left unpopped, on average. He concluded that this popping method achieved the $10\%$ goal. Thus, we have the hypotheses that:

  $\begin{array}{l}{H}_0:\mu =10\%\\ H_1:\mu <10\%\end{array}$

The Type I error is when we reject the null hypothesis, when it is true. And Type II error is when we fail to reject the null hypothesis when it is false.

Therefore, if he concludes that the $10\%$ goal is achieved while the goal was not achieved then we rejected the null hypothesis while it was true and thus we made a Type I error.

(b)

To determine

To explain does this provide evidence that he met his goal of an average of no more than $10\%$ uncooked kernels.

Answer to Problem 33E

Yes, this provide evidence that he met his goal of an average of no more than $10\%$ uncooked kernels.

Explanation of Solution

In the question, it is given that YvonHopps ran an experiment to test optimum power and time settings for microwave popcorn. His goal was to find a combination of power and time that would deliver high quality popcorn with less than $10\%$ of the kernels left unpopped, on average. He concluded that this popping method achieved the $10\%$ goal. Thus, we have the hypotheses that:

  $\begin{array}{l}{H}_0:\mu =10\%\\ H_1:\mu <10\%\end{array}$

  $n=8$

The mean is:

  $\begin{array}{l}\overline{x}=\frac{7+13.2+10+6+7.8+2.8+2.2+5.2}{8}\\ =6.775\end{array}$

The standard deviation is:

  $\begin{array}{l}s=\sqrt{\frac{{(6-6.775)}^2+\mathrm{....}+{(5.2-6.775)}^2}{8-1}}\\ =3.637\end{array}$

The value of test statistic is:

  $\begin{array}{l}t=\frac{\overline{x}-{\mu }_0}{\frac{s}{\sqrt{n}}}\\ =\frac{6.775-10}{\frac{3.637}{\sqrt{8}}}\\ =-2.51\end{array}$

The degrees of freedom is:

  $\begin{array}{l}df=n-1\\ =8-1\\ =7\end{array}$

The P-value is then:

  $0.01<P<0.025$

If the P-value is smaller than the significance level, then the null hypothesis is rejected, then

  $P<0.05=5\%\Rightarrow \text{Reject }{H}_0$

Thus, there is sufficient evidence to support the claim that his goal of an average of no more than $10\%$ uncooked kernels.