Question

If one can containing 500 nuts is selected at random from each of three different distributors of mixed nuts and there are, respectively, 345, 313, and 359 peanuts in each of the cans, can we conclude at the 0.01 level of significance that the mixed nuts of the three distributors contain equal proportions of peanuts?

Answer 1

Hmm, I think this involves carrying out the test for difference in proportions three times, but I could be wrong. Just what my intuition tells me.

For each sample, you have estimated population proportions
\[\hat{p}_1=\frac{345}{500}=0.69\\
\hat{p}_2=\frac{313}{500}=0.626\\
\hat{p}_3=\frac{359}{500}=0.718\]
The three tests have null/alternative hypotheses as follows:
\[\begin{array}{c|c|c}
\text{Test}&\text{Null}&\text{Alternative}\\
\hline
1&\hat{p}_1-\hat{p}_2=0&\hat{p}_1-\hat{p}_2\not=0\\
\hline
2&\hat{p}_1-\hat{p}_3=0&\hat{p}_1-\hat{p}_3\not=0\\
\hline
3&\hat{p}_2-\hat{p}_3=0&\hat{p}_2-\hat{p}_3\not=0\\
\end{array}\]
The test statistic is
\[Z=\frac{\hat{p}_A-\hat{p}_B}{\hat{P}(1-\hat{P})\left(\dfrac{1}{n_A}+\dfrac{1}{n_B}\right)}\]
where \(\hat{P}\) represents the total proportion of peanuts across both samples, i.e.
\[\hat{P}=\frac{\#_A+\#_B}{n_A+n_B}\]
The critical value for a test at 0.01 significance is about 2.575, which means if the magnitude of the test statistic is greater than 2.575, you would reject the null hypothesis.
\[\text{If }|Z|<2.575,\text{ you accept the null hypothesis.}\\
\text{Otherwise, if }|Z|>2.575,\text{ you reject it.}\]

Answer 2

\(A\) and \(B\) are used to denote any two different samples. \(n_A\) and \(n_B\) are the same for this problem, since all cans contain a total of 500 nuts.