Goodness of Fit Test: Videos & Practice Problems

A goodness of fit test is a statistical method used to determine if the observed frequencies in a dataset align with the expected frequencies based on a specific distribution. This test is particularly useful when assessing whether a die is fair, for example, by comparing the actual results of rolling a die multiple times against the theoretical expectation of a uniform distribution.

In a typical scenario, you might roll a six-sided die 60 times and record the observed frequencies of each outcome (1 through 6). The null hypothesis (H₀) posits that the observed frequencies match the expected frequencies, which, under the assumption of a fair die, would be 10 for each outcome (since 60 rolls divided by 6 outcomes equals 10). The alternative hypothesis (H_a) suggests that at least one of the observed frequencies differs from the expected frequencies.

The test statistic for a goodness of fit test is calculated using the chi-squared statistic, represented as:

\[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]

where \(O_i\) represents the observed frequency for each category, and \(E_i\) is the expected frequency. This formula quantifies the discrepancy between what was observed and what was expected. For instance, if the observed frequency of rolling a 1 is 13, the calculation for that category would be:

\[\frac{(13 - 10)^2}{10} = \frac{9}{10} = 0.9\]

After calculating the chi-squared value for all categories, you sum these values to obtain the overall chi-squared statistic. In our example, this might yield a chi-squared value of 11.2.

To interpret the chi-squared statistic, you also need to determine the degrees of freedom, calculated as \(k - 1\), where \(k\) is the number of categories. In this case, with 6 categories, the degrees of freedom would be 5. Using statistical tables or software, you can find the p-value associated with the chi-squared statistic. For a chi-squared value of 11.2 and 5 degrees of freedom, the p-value might be approximately 0.0476.

Finally, you compare the p-value to your significance level (α), which is often set at 0.05. If the p-value is less than α, you reject the null hypothesis. In this case, since 0.0476 is less than 0.05, you would reject the null hypothesis, concluding that the observed frequencies do not match the expected frequencies, indicating that the die is likely not fair.

When conducting a goodness of fit test, ensure that the sample is random, that there are observed frequencies for all categories, and that the expected frequencies are sufficiently large (typically at least 5) to validate the test's assumptions. This method provides a robust framework for assessing the fit of observed data to theoretical distributions.

In this example, we explore a goodness of fit test to evaluate customer satisfaction survey responses across five categories: very poor, poor, neutral, good, and very good. The manager hypothesizes that the responses will not be uniformly distributed across these categories. To test this claim, a random sample of 100 survey responses is collected, and we will analyze the observed frequencies of responses.

To begin, we establish our null and alternative hypotheses. The null hypothesis (H₀) posits that the frequencies for all rating categories are equal, indicating a uniform distribution of responses. Conversely, the alternative hypothesis (H_a) suggests that at least one category's frequency is significantly different from the others. This distinction is crucial, as the null hypothesis represents the default assumption that there is no difference among the categories.

Next, we confirm the conditions for conducting the goodness of fit test. We have a random sample, observed frequencies for each category, and we need to ensure that the expected frequencies are greater than or equal to five. The total sample size (n) is 100, and with five categories (k), we calculate the expected frequency for each category as:

Expected frequency (E) = n / k = 100 / 5 = 20.

Now, we compute the chi-squared test statistic using the formula:

χ² = Σ((O_i - E)² / E),

where O_i represents the observed frequencies. Plugging in the observed values, we calculate:

χ² = (13 - 20)² / 20 + (14 - 20)² / 20 + (26 - 20)² / 20 + (29 - 20)² / 20 + (18 - 20)² / 20.

After performing the calculations, we find that χ² = 10.3. The degrees of freedom (df) for this test is calculated as:

df = k - 1 = 5 - 1 = 4.

Using the chi-squared value and degrees of freedom, we determine the p-value, which is found to be 0.0357. We compare this p-value to our significance level (α = 0.05). Since the p-value is less than α, we reject the null hypothesis.

This rejection indicates that there is sufficient evidence to support the alternative hypothesis, suggesting that the frequencies of at least one of the rating categories differ significantly from the others. In conclusion, the claimed distribution of equal frequencies does not fit the observed data well, indicating a significant difference in customer satisfaction ratings.

In statistical analysis, the goodness of fit test is a crucial method used to determine whether observed data aligns with expected distributions based on a specific claim. This test is particularly useful when dealing with distributions where probabilities are not equal, as is the case with Benford's Law. Benford's Law states that in many real-world datasets, lower digits (like 1, 2, and 3) appear more frequently than higher digits (like 7, 8, and 9).

To conduct a goodness of fit test under these conditions, the calculation of expected frequencies differs from scenarios where probabilities are equal. Instead of simply dividing the total sample size by the number of categories, expected frequencies are calculated by multiplying the total sample size by the probability of each category. For instance, if the sample size (n) is 100 and the probability of a digit appearing is 0.301, the expected frequency (E) for that digit would be:

$$E = n \times P = 100 \times 0.301 = 30.1$$

Once the expected frequencies are determined, the chi-squared statistic can be calculated using the formula:

$$\chi^2 = \sum \frac{(O - E)^2}{E}$$

where O represents the observed frequencies. This formula quantifies the discrepancy between observed and expected frequencies, allowing for a statistical assessment of fit. The larger the difference between observed and expected values, the greater the contribution to the chi-squared statistic.

After calculating the chi-squared values for each category, these values are summed to obtain the overall chi-squared statistic. For example, if the calculated values for each category yield a total of 17.92, this statistic alone does not determine whether to reject or fail to reject the null hypothesis. To make this determination, one must also consider the degrees of freedom and calculate the corresponding p-value.

Understanding how to compute expected frequencies with unequal probabilities is essential for accurately applying the goodness of fit test in various statistical contexts. This knowledge lays the groundwork for further exploration and practice in statistical analysis.

In conducting a goodness of fit test, we aim to determine whether the observed frequencies of a categorical variable align with the expected frequencies based on a claimed distribution. In this scenario, a regional airline categorizes its customers into three ticket types: business, leisure, and last minute, with claimed proportions of 50%, 35%, and 15%, respectively. A random sample of 200 ticket purchases is analyzed to see if these proportions hold after changes to the pricing model.

The first step involves formulating the null and alternative hypotheses. The null hypothesis (H₀) posits that the distribution of ticket types matches the claimed distribution, while the alternative hypothesis (H_a) suggests that at least one of the proportions has changed. This sets the stage for our analysis.

Next, we calculate the expected frequencies for each category using the formula:

e = n * p

where n is the total sample size (200) and p is the claimed proportion for each category. Thus, the expected frequencies are:

• Business: e = 200 * 0.5 = 100
• Leisure: e = 200 * 0.35 = 70
• Last Minute: e = 200 * 0.15 = 30

With observed frequencies of 98 for business, 71 for leisure, and 31 for last minute, we can now compute the chi-squared test statistic using the formula:

χ² = ∑ (O - E)² / E

Calculating each term:

• For business: (98 - 100)² / 100 = 0.04
• For leisure: (71 - 70)² / 70 = 0.0143
• For last minute: (31 - 30)² / 30 = 0.0333

Summing these values gives us a chi-squared statistic of χ² = 0.088.

To determine the p-value, we need the degrees of freedom, calculated as the number of categories minus one (k - 1). Here, we have 3 categories, resulting in 2 degrees of freedom. Using statistical software or a chi-squared distribution table, we find the p-value corresponding to χ² = 0.088 with 2 degrees of freedom, which is approximately 0.9571.

Finally, we compare the p-value to our significance level (α = 0.1). Since 0.9571 is greater than 0.1, we fail to reject the null hypothesis. This indicates that there is insufficient evidence to suggest that the distribution of ticket types has changed after the pricing model adjustments. Therefore, we conclude that the claimed distribution of 50%, 35%, and 15% remains a good fit for the observed data.