Section 11.4 Choosing the Sample Size
One of the more important contributions of Statistics to research is providing guidelines for the design of experiments and surveys. A well planed experiment may produce accurate enough answers to the research questions while optimizing the use of resources. On the other hand, poorly planed trials may fail to produce such answers or may waste valuable resources.
Unfortunately, in this book we do not cover the subject of experiment design. Still, we would like to give a brief discussion of a narrow aspect in design: The selection of the sample size.
An important consideration at the stage of the planning of an experiment or a survey is the number of observations that should be collected. Indeed, having a larger sample size is usually preferable from the statistical point of view. However, an increase in the sample size typically involves an increase in expenses. Thereby, one would prefer to collect the minimal number of observations that is still sufficient in order to reach a valid conclusion.
As an example, consider an opinion poll aimed at the estimation of the proportion in the population of those that support a specific candidate that considers running for an office. How large the sample must be in order to assure, with high probability, that the percentage in the sample of supporters is within 0.5% of the percentage in the population? Within 0.25%?
A natural way to address this problem is via a confidence interval for the proportion. If the range of the confidence interval is no more than 0.05 (or 0.025 in the other case) then with a probability equal to the confidence level it is assured that the population relative frequency is within the given distance from the sample proportion.
Consider a confidence level of 95%. Recall that the structure of the confidence interval for the proportion is \(\hat P \pm 1.96 \cdot \{\hat P (1-\hat P)/n\}^{1/2}\text{.}\) The range of the confidence interval is \(1.96 \cdot \{\hat P (1-\hat P)/n\}^{1/2}\text{.}\) How large should \(n\) be in order to guarantee that the range is no more than 0.05?
The answer to this question depends on the magnitude of \(\hat P (1-\hat P)\text{,}\) which is a random quantity. Luckily, one may observe that the maximal value of the quadratic function \(f(p) = p (1-p)\) is 1/4. It follows that
1
The derivative is \(f'(p) = 1-2p\text{.}\) Solving \(f'(p)=0\) produces \(p=1/2\) as the maximizer. Plugging this value in the function gives \(1/4\) as the maximal value of the function.
\begin{equation*}
1.96 \cdot \{\hat P (1-\hat P)/n\}^{1/2} \leq 1.96 \cdot \{0.25/n\}^{1/2} = 0.98/\sqrt{n}\;.
\end{equation*}
Finally,
\begin{equation*}
0.98/\sqrt{n} \leq 0.05 \quad \Longrightarrow \quad \sqrt{n} \geq 0.98/0.05 = 19.6
\quad \Longrightarrow \quad n \geq (19.6)^2 = 384.16\;.
\end{equation*}
The conclusion is that \(n\) should be larger than 384 in order to assure the given range. For example, \(n=385\) should be sufficient.
If the request is for an interval of range 0.025 then the last line of reasoning should be modified accordingly:
\begin{equation*}
0.98/\sqrt{n} \leq 0.025 \quad \Longrightarrow \quad \sqrt{n} \geq \frac{0.98}{0.025} = 39.2
\quad \Longrightarrow \quad n \geq (39.2)^2 = 1536.64\;.
\end{equation*}
Consequently, \(n=1537\) will do. Increasing the accuracy by 50% requires a sample size that is 4 times larger.
More examples that involve selection of the sample size will be considered as part of the homework.
