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Statistical Decisions

We have seen that confidence intervals can be used to make inferences about a population parameter since confidence intervals represent a range of plausible values for the parameter. In this section we will expand these ideas to construct methods for statistical decision-making based on information contained in a randomly selected sample from the population.

Consider a situation in which a manufacturer currently has a 10% defective rate for one of its products. Suppose that you are assigned the task of improving the production process to reduce this rate. After studying the problem your workgroup propose a solution that will require an initial outlay of $500,000 for new equipment and retraining of production staff. However, before agreeing to these changes, the COO asks you undertake a pilot production run to answer two questions: (1) Will the new process actually reduce the defective rate? (2) How long will it take to recover the initial cost of implementing the new process through reduced warranty claims on the product? Suppose that after the pilot program is performed you find that in a production run of 400 units, 20 fail to meet specification. We will treat this set of 400 units as a random sample from the population of all units produced by this process and the answers to the two questions will be based on the information contained in the sample.

The first question represents a decision problem in which you are asked to decide between two actions: implement the new process or don't implement it. If the pilot study shows strong evidence that the new process will reduce the defective rate, then it will be implemented. If, however, the evidence provided by the pilot study is weak or shows no change, then the new process will not be implemented. These actions correspond to two possible sets of values for the defective rate: $\pi \ge 0.10$ and $\pi < 0.10$. The second set has the burden of proof for our decision. That is, we will only believe it, and take the corresponding action to implement the new process, if the evidence provided by the pilot study indicates strongly that this is true. In the language of statistics, the hypothesis that has the burden of proof is referred to at the alternative hypothesis.

This implies that we initially assume that nothing has changed, that is, the defective rate under the new process is the same as before, $\pi = 0.10$. This hypothesis is referred to as the null hypothesis since it is presumed to be true by default. It is possible that a random sample of 400 units will contain fewer than 10% defectives when $\pi = 0.10$, but we would not expect this proportion to be very much less than 0.10.

Our random sample of 400 units contained 20 defectives, a sample proportion of 0.05. This is less than what we would expect to see if the defective rate is the same as before, and so leads to some doubt that our initial assumption is true. Since an outcome of the pilot study that gave fewer than 20 defectives would provide even greater doubt about the initial assumption, this doubt can be expressed as the likelihood of observing a sample proportion that is at least as far below $\pi = 0.10$ as we have in our sample. That is, we must evaluate,

\begin{displaymath}
P(\widehat{p} \le 0.05)
\end{displaymath}

under the assumption that $\pi = 0.10$. We know from our derivation of confidence intervals for a population proportion that under this assumption, $\widehat{p}$ has approximately a normal distribution with mean $\pi = 0.10$ and s.d.

\begin{displaymath}
\sqrt{\pi(1-\pi)/n} = \sqrt{(.1)(.9)/400} = 0.015.
\end{displaymath}

Therefore, the doubt concerning the assumption that $\pi = 0.10$ is given by

\begin{displaymath}
P(\widehat{p} \le 0.05) \approx P(Z \le (.05 - .10)/.015) = P(Z \le -3.33) = 0.0004.
\end{displaymath}

Therefore, if we assume the defective rate using the new process is unchanged, then we might see a sample proportion of 0.05 or lower in a random sample of size 400, but the chance of that happening is only 0.0004. It would be much more reasonable to reject that assumption and believe that the defective rate with the new process is lower than 0.10. If we take the action to implement the new process based on this data, the probability we have taken the wrong action is only 0.0004.

Now that we believe the new process will reduce the defection rate, we can answer the COO's second question by constructing a confidence interval for the proportion of defectives using the new process,

\begin{displaymath}
.05 \pm 1.96\sqrt{(.05)(.95)/400} \ \Leftrightarrow\ .05\pm .011 \ \Leftrightarrow\ [.039,.061].
\end{displaymath}

Example. Currently the market share of a company's brand for a particular product in the DFW metropolitan area is 25%. An advertising campaign based on social media placements is run for one month. A random sample of 750 potential purchasers of this product contained 225 who planned to buy this company's brand. Do you believe the market share of this company's brand has increased? Why or why not? Construct a 90% confidence interval for this company's market share after the advertising campaign.

Solution The sample proportion of potential purchasers who plan to buy this brand is 225/750 = 0.30 which is higher than the current market share. Under the assumption that this market share has not changed, the probability of seeing a sample proportion of 0.30 or higher in a random sample of n=750 is given by

\begin{eqnarray*}
P(\widehat{p} \ge 0.30) &=& P(Z \ge (0.30 - 0.25)/sqrt{(0.25)(0.75)/750}\\
&=& P(Z \ge 3.16)\\
&=& 1 - 0.9992\\
&=& 0.0008.
\end{eqnarray*}

It is highly unlikely that this would happen, so we should believe that the market share has increased. A 90% confidence interval for the market share after this advertising campaign is

\begin{displaymath}
0.30 \pm (1.645)\sqrt{(0.3)(0.7)/750} \ \Leftrightarrow\ .30\pm .028 \ \Leftrightarrow\ [.272,.328].
\end{displaymath}



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Next: Hypothesis tests for a Up: Class Notes Previous: Other Estimation Problems
Larry Ammann
2014-12-08