Question

Packages of mixed nuts made by a certain company contain four types of nuts. The percentages of nuts of Types \(1,2,3,\) and 4 are advertised to be \(40 \%, 30 \%, 20 \%,\) and \(10 \%,\) respectively. A random sample of nuts is selected, and each one is categorized by type. a. If the sample size is 200 and the resulting test statistic value is \(X^{2}=19.0,\) what conclusion would be appropriate for a significance level of \(0.001 ?\) b. If the random sample had consisted of only 40 nuts, would you use the chi- square goodness-of-fit test? Explain your reasoning.

Short answer

For a sample of 200 nuts, 1,000 significance level, and a test statistic value of 19.0, the conclusion is that the proportions of nuts are not as advertised. Chi-square goodness-of-fit test is not appropriate for a sample size of 40 because the expected frequency of at least one type of nut is less than 5.

Step-by-step solution

Calculate expected frequencies

The expected frequency for each type of nut can be calculated by multiplying the sample size with the percentage of each type of nut. Therefore for Type 1 nuts, the expected frequency is \(40\% * 200 = 80\). Following the same calculation for Type 2, Type 3 and Type 4 gives 60, 40 and 20 respectively.

Conduct Chi-square goodness-of-fit test

Now, the computed value of the test statistic, \(X^{2}=19.0,\) compares the observed frequencies in the sample with the expected frequencies mentioned above. The appropriate conclusion is found by comparing the calculated test statistic with the critical value from the chi-square distribution with degrees of freedom \(df = n - 1\), where \(n\) is the number of categories (in this case, the types of nuts). Since we have 4 types of nuts, \(df = 4 - 1 = 3\). If the calculated test statistic is greater than the critical value, we reject the null hypothesis, indicating that the proportions of nuts are not as advertised. Using a standard chi square table, a \(X^2\) value of 19 exceeds the critical value in the \(X^2\) distribution for \(df = 3\) at the .001 significance level, indicating that the null hypothesis should be rejected.

Assess Appropriateness of the Test for Smaller Sample size

For a sample size of 40, using the chi-square goodness-of-fit test may not be appropriate since the test requires that expected frequencies should be at least 5 for all categories. In this case, if we calculate the expected frequencies for a sample size of 40, the expected frequency for the 4th type of nut would be \(10\% * 40 = 4\), which is less than 5. So, we cannot use the chi-square goodness-of-fit in this case.