Goodness of FIt Test Using TI-84: Videos & Practice Problems

Performing a goodness-of-fit test using a TI-84 involves comparing observed frequencies to expected frequencies based on a claim distribution. For equally distributed categories, expected values are calculated as $e=\frac{n}{k}$, where $n$ is sample size and $k$ is number of categories. The chi-square GOF test on the TI-84 calculates the test statistic and p-value, which is compared to the significance level $\alpha$. A p-value greater than $\alpha$ indicates failure to reject the null hypothesis, supporting the claim distribution's fit to the data.

Goodness of Fit Test Using a TI-84

A goodness of fit test is used to determine whether observed categorical data matches an expected distribution. When testing if customer preferences are equally distributed among categories, such as flavors of a drink, the null hypothesis states that the observed frequencies match the expected equal distribution, while the alternative hypothesis suggests they do not.

To perform this test, first calculate the expected values using the formula $\backslash (\; E\; =\; \backslash frac\{n\}\{k\}\; \backslash )$, where $\backslash (\; n\; \backslash )$ is the total sample size and $\backslash (\; k\; \backslash )$ is the number of categories. For example, if 200 customers are surveyed across 4 flavors, each expected frequency is $\backslash (\; \backslash frac\{200\}\{4\}\; =\; 50\; \backslash )$. The degrees of freedom for the test is calculated as $\backslash (\; k\; -\; 1\; \backslash )$, which in this case is 3.

Using a TI-84 calculator simplifies the computation of the chi-square goodness of fit test. Input the observed frequencies into list L1 and the expected frequencies into list L2. Access the test by pressing the STAT button, navigating to the TESTS menu, and selecting the χ² GOF Test (option D). Ensure the observed list is set to L1, the expected list to L2, and update the degrees of freedom to match the problem (e.g., 3).

After running the test, the calculator provides the chi-square statistic, degrees of freedom, p-value, and contribution values for each category. The p-value is critical for decision-making: if it is greater than the significance level (commonly 0.05), we fail to reject the null hypothesis, indicating insufficient evidence to conclude that the observed distribution differs from the expected. Conversely, a p-value less than the significance level suggests the observed data does not fit the expected distribution.

In the example with a p-value of approximately 0.44, which is greater than 0.05, the conclusion is that customer flavor preferences are evenly distributed, supporting the claim of equal preference among flavors. This process highlights the importance of understanding hypothesis formulation, expected value calculation, degrees of freedom, and interpreting p-values in chi-square goodness of fit tests.

A student performs Goodness of Fit Test using technology to see if the proportion of each candy flavor in a bag matches the expected distribution. They get the following results: ${\chi }^2=18.99$ & $p=0.0019$. What can they conclude about the claimed distribution of candy flavors?

• A. Fail to reject the $H_0$ since there is not enough evidence to suggest $H_{a}$ is true. The claimed distribution is a good fit.
• B. Fail to reject the $H_0$ since there is not enough evidence to suggest $H_{a}$ is true. The claimed distribution is a bad fit.
• C. Reject the $H_0$ since there is enough evidence to suggest $H_{a}$ is true. The claimed distribution is a good fit.
• D. Reject the $H_0$ since there is enough evidence to suggest $H_a$ is true. The claimed distribution is a bad fit.

Answer: (D)

Goodness of Fit Test Using a TI-84 Example 1

In testing whether a call center receives an equal number of calls for each of its four branches (A, B, C, and D), a chi-square goodness of fit test is used to evaluate the claim. The null hypothesis assumes that the observed call frequencies match the expected distribution, which in this case is an equal probability for each branch. The alternative hypothesis states that the observed frequencies do not match this equal distribution.

Given a sample size n of 160 calls and k categories (branches) equal to 4, the expected frequency for each branch is calculated using the formula:

\[E = \frac{n}{k} = \frac{160}{4} = 40\]

This means each branch is expected to receive 40 calls if the claim holds true. The degrees of freedom for the chi-square test is determined by subtracting one from the number of categories:

\[df = k - 1 = 4 - 1 = 3\]

Using a chi-square goodness of fit test with a significance level (α) of 0.01, the observed frequencies are compared against the expected frequencies. The test statistic is calculated and the corresponding p-value is obtained. In this scenario, the p-value is approximately 0.002, which is less than the alpha level of 0.01.

Since the p-value is smaller than the significance level, the null hypothesis is rejected. This indicates there is sufficient evidence to conclude that the observed call frequencies do not fit the claim of equal distribution among the four branches. Therefore, the claim that the call center receives the same number of calls for each branch is not supported by the data.

This process highlights the importance of the chi-square goodness of fit test in assessing how well observed categorical data align with expected distributions, especially when testing claims about equal probabilities across categories.

Goodness of Fit Test Using a TI-84 Example 2

When evaluating whether a store's promotional email coupon distribution matches the claimed percentages, a chi-square goodness of fit test is an effective statistical method. The test begins by establishing hypotheses: the null hypothesis assumes that the observed coupon frequencies align with the claimed distribution, while the alternative hypothesis suggests they do not match.

The claim distribution provides the expected proportions for each coupon type. To calculate the expected frequencies, multiply the total sample size n by each claimed probability p, using the formula:

\[E = n \times p\]

For example, with a sample size of 1,000 coupons and claimed percentages of 50%, 20%, 15%, 10%, and 5%, the expected counts are 500, 200, 150, 100, and 50 respectively. These expected values are compared against the observed frequencies collected from the sample.

The degrees of freedom for the chi-square test is determined by subtracting one from the number of categories, so with five coupon types, the degrees of freedom is:

\[df = 5 - 1 = 4\]

Using statistical software or a calculator, input the observed and expected frequencies, specify the degrees of freedom, and perform the chi-square goodness of fit test. The resulting p-value indicates the probability that the observed differences could occur if the null hypothesis were true.

In this case, the p-value was approximately \(9.4 \times 10^{-19}\), which is significantly less than the chosen significance level \(\alpha = 0.1\). Since the p-value is smaller than \(\alpha\), the null hypothesis is rejected, providing strong evidence that the observed coupon distribution does not fit the claimed distribution.

This outcome implies that the store's promotional ads do not accurately represent the actual distribution of emailed coupons. Consequently, the store will need to revise their advertisements to reflect the true coupon distribution, ensuring transparency and compliance with advertising standards.

Goodness of Fit Test Using a TI-84 Example 3

When testing whether the commuting habits of incoming freshmen match the distribution of last year's class, a goodness of fit test is an effective statistical method. The null hypothesis (H₀) assumes that the observed frequencies of commuting methods align with the claimed distribution from the previous year, while the alternative hypothesis (H_a) suggests that the observed frequencies differ from this distribution.

To perform this test, the expected frequencies are calculated using the formula E = n × p, where n is the sample size and p is the proportion from the claimed distribution. For a sample size of 100 students, the expected values correspond directly to the percentage values of each commuting method from last year. For example, if 63% of students used a particular method, the expected count is 63.

The degrees of freedom (df) for the chi-square goodness of fit test is determined by subtracting one from the number of categories:
$$\mathrm{df}=k-\; 1$$where k is the number of commuting methods. In this case, with five methods, df = 5 - 1 = 4.

Using statistical software or a calculator, the observed and expected frequencies are entered into separate lists, ensuring each method's observed and expected values align in the same row. The chi-square goodness of fit test is then conducted, yielding a p-value. If the p-value exceeds the significance level (α = 0.05), the null hypothesis is not rejected, indicating insufficient evidence to conclude that the observed distribution differs from the claimed one.

In this scenario, the p-value was approximately 0.652, which is greater than 0.05. This result means there is no significant difference between the observed commuting habits of the incoming freshmen and last year's distribution. Therefore, the school can reasonably assume that the commuting habits remain consistent and does not need to conduct further research on this matter.

Here’s what students ask on this topic:

How do you perform a goodness of fit test using a TI-84 calculator?

To perform a goodness of fit test on a TI-84, first enter your observed frequencies into list L1 and your expected frequencies into list L2. The expected values are calculated based on the claim distribution, often using the formula $e=\frac{n}{k}$, where $n$ is the total sample size and $k$ is the number of categories. Next, press the STAT button, scroll right to the TESTS menu, and select option D: χ² GOF Test. Set the observed list to L1, the expected list to L2, and input the degrees of freedom as $k-1$. Finally, select Calculate to get the chi-square statistic and p-value. Compare the p-value to your significance level to decide whether to reject the null hypothesis.

What is the formula for calculating expected values in a goodness of fit test with equally distributed categories?

When categories are equally distributed, the expected value for each category is calculated using the formula $e=\frac{n}{k}$, where $n$ is the total number of observations or sample size, and $k$ is the number of categories. This formula assumes that each category is equally likely, so the expected count is the total divided evenly among all categories. For example, if 200 people are surveyed and there are 4 categories, each expected value would be $\frac{200}{4}=50$.

How do you determine the degrees of freedom for a chi-square goodness of fit test?

The degrees of freedom (df) for a chi-square goodness of fit test is calculated as $\mathrm{df}=k-1$, where $k$ is the number of categories or groups in your data. This formula accounts for the fact that the total number of observations is fixed, so only $k-1$ categories can vary independently. For example, if you have 4 flavors in a survey, the degrees of freedom would be $4-1=3$.

What does the p-value from a chi-square goodness of fit test indicate when using a TI-84?

The p-value from a chi-square goodness of fit test on a TI-84 indicates the probability of observing the data, or something more extreme, assuming the null hypothesis is true. If the p-value is greater than the significance level (commonly 0.05), you fail to reject the null hypothesis, meaning there is not enough evidence to say the observed frequencies differ from the expected distribution. Conversely, a p-value less than or equal to the significance level suggests rejecting the null hypothesis, indicating the observed data does not fit the claimed distribution well. The TI-84 displays this p-value after running the test, helping you make this decision.

How do you input observed and expected frequencies into the TI-84 for a goodness of fit test?

To input observed and expected frequencies into the TI-84, press the STAT button and select 1:Edit. Enter your observed frequencies into list L1 by typing each value and pressing ENTER. Then, move to list L2 and enter the expected frequencies similarly. For example, if you have four categories with expected counts of 50 each, enter 50 four times in L2. After entering the data, you can proceed to the STAT > TESTS menu to perform the chi-square goodness of fit test using these lists.