Question

Exercises 7.9 to 7.12 give a null hypothesis for a goodness-of-fit test and a frequency table from a sample. For each table, find: (a) The expected count for the category labeled B. (b) The contribution to the sum of the chi-square statistic for the category labeled \(\mathrm{B}\). (c) The degrees of freedom for the chi-square distribution for that table. $$ \begin{aligned} &H_{0}: p_{a}=0.1, p_{b}=0.35, p_{c}=0.2\\\ &p_{d}=0.05, p_{e}=0.1, p_{f}=0.2\\\ &H_{a}: \text { Some }\\\ &\text { is }\\\ &p_{i}\\\ &\text { wrong }\\\ &\begin{array}{lccccc} \hline \mathrm{A} & \mathrm{B} & \mathrm{C} & \mathrm{D} & \mathrm{E} & \mathrm{F} \\ 210 & 732 & 396 & 125 & 213 & 324 \\ \hline \end{array} \end{aligned} $$

Short answer

The expected count for category B is 700. The contribution to the Chi-square statistic for category B is 1.389. The degrees of freedom for the Chi-square distribution for the table is 5.

Step-by-step solution

Find the Expected Count for Category B

In a Chi-Square Goodness-of-Fit test, the expected count of an event is calculated by multiplying the total count by the expected probability of that event. The given null hypothesis \(H_{0}\) gives expected probabilities \(p\) for each category. The total count of requests is the sum total of all categories, which equals \(210+732+396+125+213+324=2000\). Thus, the expected count \(O_{B}\) for the category labeled B is calculated by multiplying the total count by the expected probability \(p_{B}\) for category B: \( O_{B} = (total count) * p_{B} = 2000 * 0.35 = 700\)

Calculate the Contribution to Chi-square Statistic for Category B

The contribution to the Chi-square statistic (\(X^{2}\)) for category B is calculated using the formula: \(X_{B}^{2} = (O_{B} - E_{B})^{2} / E_{B}\) where \(O_{B}\) is the observed count for category B (given as 732 in the table) and \(E_{B}\) is the expected count for category B (calculated in Step 1 as 700). Plugging these values into the equation gives: \( X_{B}^{2} = (732 - 700)^{2} / 700= 1.389\)

Find the Degrees of Freedom for the Chi-Square Distribution

The degrees of freedom for a Chi-square distribution are determined by the number of categories minus 1. In this case, the table has six categories (A, B, C, D, E, and F), so the degrees of freedom (\(df\)) are : \(df = num. categories - 1 = 6 - 1 = 5\)

Key concepts

Expected Count

The expected count in a Chi-Square Goodness-of-Fit test is critical for comparing observed data with what we would theoretically anticipate under the null hypothesis. It's calculated by taking the total number of observations and multiplying it by what we'd expect the proportion of that category to be if the null hypothesis were true.

For instance, if a die is fair, we'd expect each number to come up about one-sixth of the time in a large number of rolls. If we rolled the die 600 times, the expected count for each number would be 100 rolls. In a statistical test, we apply the same principle to see if our observed data significantly deviates from these expected counts, which might suggest that the die is not fair after all.

In the provided exercise, the total number of observations was 2000, and the expected proportion of category B, according to the null hypothesis, was 0.35. Thus, the expected count for category B was calculated to be 700, highlighting how frequent we would expect category B to occur purely by chance if the null hypothesis held true.

Chi-Square Statistic

The chi-square statistic is a measure used to evaluate how well observed outcomes fit expected outcomes under the null hypothesis. It's calculated by summing the squares of the differences between the observed (\(O\)) and expected (\(E\)) counts, divided by the expected counts for all categories.

Mathematically, we express this as \( \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \), where \(O_i\) and \(E_i\) are the observed and expected counts for category \(i\), respectively. A higher chi-square value indicates a greater discrepancy between the observed and expected data, which can lead us to question the validity of the null hypothesis.

In our exercise, we saw a contribution to the chi-square statistic for category B calculated to be 1.389. This value is just one part of the overall chi-square statistic, which would be completed by calculating similar contributions for each category and summing them up.

Degrees of Freedom

Degrees of freedom are an essential concept in statistics, representing the number of values in a calculation that are free to vary. When conducting a Chi-Square Goodness-of-Fit test, the degrees of freedom are determined by the number of categories we are examining minus one.This subtraction accommodates the constraint that the total observed frequencies must equal the total expected frequencies. The degrees of freedom affect the shape of the chi-square distribution, which is used to determine the p-value of the test. In general, the more degrees of freedom, the closer the distribution will look to a normal distribution.

In the context of the exercise, the chi-square distribution is based on six categories, meaning the degrees of freedom we calculated were 5 (\(6 - 1 = 5\)). This figure is then used to reference the chi-square distribution table or a computational tool to assess the likelihood of observing a chi-square statistic as extreme as our calculation, assuming the null hypothesis is true.

Null Hypothesis

The null hypothesis (\(H_0\)) in a statistical test is a statement of no effect or no difference. It's the default assumption that there is no relationship between two measured phenomena. In a Chi-Square Goodness-of-Fit test, we use the null hypothesis to specify the expected probabilities of the different categories if the hypothesis is true.

To test the null hypothesis, we compare the observed data with what we would expect under this assumption. If the observed data are significantly different from the expected data, we may have enough evidence to reject the null hypothesis in favor of the alternative hypothesis (\(H_a\)), which suggests that at least one of the expected probabilities does not hold true.

In our exercise, the table provided the expected probabilities for each category under the null hypothesis. A deviation from these probabilities in our observed data might indicate that the assumption of the null hypothesis - that the data fits these expected probabilities - could be incorrect.