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

a) Since the test statistic value of 19.0 is greater than the critical value of 16.27, reject the null hypothesis. There is evidence at a 0.001 level of significance to suggest that the nut mixture differs from the advertised proportions. \nb) Yes, the chi-square test can be used for a sample size of 40, as all the expected counts are above 5.

Step-by-step solution

Check Critical Chi-Square Value for Significance Level

Use a chi-square distribution table to find the critical chi-square value. For a significance level of 0.001 and 3 degrees of freedom (4 - 1, because there are 4 types of nuts), the critical chi-square value is 16.27.

Compare Test Statistic to Critical Value

Given that the test statistic value (\(X^{2}\)) is 19.0, this is greater than the critical value. So the conclusion is to reject the null hypothesis. This means that there is evidence at a 0.001 level of significance, suggesting that the nut mixture differs from the advertised proportions.

Apply Chi-Square Test to Different Sample Size

For a sample size of 40, check each expected count. The expected count for each nut type is calculated by multiplying the sample size by the respective proportion. This gives: \nType 1 -> 40 * 0.4 = 16, \nType 2 -> 40 * 0.3 = 12, \nType 3 -> 40 * 0.2 = 8, \nType 4 -> 40 * 0.1 = 4 \nAll of these values are above 5, thus meeting the requirement for a Chi-square test (expected count for each category should be 5 or more). Consequently, a chi-square goodness-of-fit test can be used.

Key concepts

Critical Chi-Square Value

When conducting a chi-square goodness-of-fit test, the critical chi-square value is essential in making a statistical decision. It's the benchmark against which the test statistic is compared. This value is determined based on the desired level of significance, which reflects the researcher's tolerance for error. For instance, a significance level of 0.001 means we allow just a 0.1% chance of wrongly rejecting the null hypothesis.

Using a chi-square distribution table or software, we can find the critical value that corresponds to this significance level and the degrees of freedom for our test. If our calculated test statistic exceeds this critical boundary, we move to reject the null hypothesis, implying that the observed data does not fit the expected distribution well. In the case of our exercise, the critical value is 16.27 for 3 degrees of freedom, and since our test statistic of 19.0 is greater, we reject the null.

Null Hypothesis

The null hypothesis in a chi-square goodness-of-fit test expresses the assumption that there's no significant difference between the observed frequencies and the expected frequencies. It's the default position that holds the observed outcomes are likely the result of random chance.

In the context of the exercise, the null hypothesis claims that the actual distribution of nuts in the packages aligns with the advertised percentages. By calculating the chi-square test statistic and comparing it to the critical chi-square value, we can make an informed decision on whether to maintain this hypothesis or to reject it in favor of an alternative hypothesis that suggests a significant discrepancy.

Expected Count

Expected count is a cornerstone concept in the chi-square goodness-of-fit test, representing the frequency we would expect in each category if the null hypothesis were true. To calculate the expected counts, we multiply the total sample size by the expected proportion for each category.

In our nuts mix exercise, if we have a sample of 200 nuts, the expected count for Type 1 nuts is 200 multiplied by 40%, giving us 80 nuts. It's important to note that the reliability of the chi-square test results partly hinges on these expected counts being sufficiently large. Typically, each expected count should be 5 or greater to ensure that the chi-square distribution is an appropriate model for the sampling distribution.

Degrees of Freedom

Degrees of freedom in a chi-square test refer to the number of categories reduced by the number of parameters estimated from the data. Generally, it's the count of values that are free to vary as we estimate the characteristics of a population from a sample.

In chi-square tests, degrees of freedom are calculated as the number of categories minus 1 (df = k - 1). This subtraction is due to the constraint that the expected frequencies must sum up to the sample size. For the nut package example, with four types of nuts, our degrees of freedom would be 3 (4-1). This information is crucial when using statistical tables or software to determine the critical chi-square value for making decisions based on our test statistic.