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

Performing a chi-square goodness of fit test involves comparing observed frequencies to expected values based on a claim distribution. For equal distribution among categories, expected values are calculated using $e=\frac{n}{k}$, where $n$ is sample size and $k$ is number of categories. Degrees of freedom are $k-1$. Using a TI-84 calculator, input observed and expected data, select the chi-square GOF test, and compare the p-value to the alpha level to decide on rejecting the null hypothesis, assessing if observed data fits the expected distribution.

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 fit the claimed distribution, while the alternative hypothesis suggests they do not. For example, if a company surveys 200 customers about their preferred flavor among lemon, berry, mango, and peach, the expected frequency for each flavor can be calculated 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. Here, each expected count is $\backslash (\; \backslash frac\{200\}\{4\}\; =\; 50\; \backslash )$.

The degrees of freedom for the chi-square goodness of fit test is found by subtracting one from the number of categories, $\backslash (df\; =\; k\; -\; 1\backslash )$, which in this case is 3. Using a TI-84 calculator, you can input the observed frequencies into list L1 and the expected frequencies into list L2. Then, by navigating to the chi-square GOF test function under the STAT menu, you specify these lists and the degrees of freedom before calculating the test statistic and p-value.

The chi-square statistic measures the discrepancy between observed and expected frequencies, while the p-value indicates the probability of observing such a discrepancy if 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, suggesting there is insufficient evidence to conclude that preferences are unevenly distributed. In the example, a p-value of approximately 0.44 exceeds 0.05, supporting the conclusion that customer flavor preferences are evenly distributed.

Understanding how to perform and interpret a chi-square goodness of fit test, including calculating expected values, degrees of freedom, and using technology like the TI-84, is essential for analyzing categorical data and making informed decisions based on statistical evidence.

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

A camping retailer wants to verify if they are selling equal amounts of four tent brands: A, B, C, and D. To test this claim, a sample of 160 tents sold over the past year was analyzed using a chi-square goodness of fit test with a significance level of α = 0.01. The null hypothesis assumes that the observed sales frequencies match the expected distribution, meaning each brand sells equally. Conversely, the alternative hypothesis states that the observed frequencies do not match this equal distribution.

Since the claim is that each of the four brands sells equally, the expected frequency for each brand is calculated by dividing the total sample size by the number of categories:
$$e=\frac{n}{k}=\frac{160}{4}=40$$where n is the sample size and k is the number of categories (brands).

The degrees of freedom for the chi-square test is the number of categories minus one, which is 4 - 1 = 3. Using statistical software or a calculator, the observed and expected frequencies are inputted, and the chi-square goodness of fit test is performed. The resulting p-value is approximately 0.00021, which is less than the significance level of 0.01.

Because the p-value is smaller than α, the null hypothesis is rejected. This indicates there is sufficient evidence to conclude that the sales distribution does not fit the claim of equal sales across all four tent brands. In practical terms, the retailer is not selling the same amount of each brand, suggesting that stocking decisions should be adjusted to better reflect actual consumer preferences and sales patterns.

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 proportion 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 serve as a benchmark to compare against the observed frequencies collected from the sample.

The degrees of freedom for this test are calculated as the number of coupon categories minus one. Since there are five coupon types, the degrees of freedom equal:

\[df = 5 - 1 = 4\]

Using statistical software or a calculator, input the observed and expected frequencies into separate lists. Then, perform the chi-square goodness of fit test, specifying the degrees of freedom. The test computes a p-value, which quantifies the probability of observing the data assuming the null hypothesis is true.

In this scenario, 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, indicating strong evidence that the observed coupon distribution does not match the claimed distribution.

This result implies that the store's promotional ads do not accurately represent the actual distribution of emailed coupons. Consequently, the store should revise their advertisements to reflect the true coupon distribution, ensuring transparency and accuracy in their marketing efforts.

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 they do not match.

To perform this test, the expected frequencies are calculated using the formula expected = n × p, where n is the sample size and p is the proportion from the claim distribution. For a sample size of 100 students, the expected values correspond directly to the percentages from last year’s data. For example, if 63% of students used a particular transportation method last year, the expected count for that method in the sample is 63.

The degrees of freedom for the chi-square goodness of fit test is determined by subtracting one from the number of categories (transportation methods), so with five categories, the degrees of freedom is 4. Using statistical software or a calculator, the observed and expected frequencies are inputted, and the chi-square test is conducted to obtain the p-value.

In this case, the p-value was approximately 0.652, which is greater than the significance level α = 0.05. This means there is insufficient evidence to reject the null hypothesis, indicating that the observed commuting habits do not significantly differ from the previous year's distribution. Therefore, the claim that the commuting habits remain consistent is supported.

Consequently, the school can reasonably assume that the commuting patterns of the incoming freshmen will mirror those of last year, and further research on this topic is not necessary at this time.

Here’s what students ask on this topic:

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

To perform a chi-square goodness of fit test on a TI-84, first enter your observed frequencies into list L1 and your expected frequencies into list L2. You calculate expected values using the formula $e=\frac{n}{k}$, where $n$ is the total sample size and $k$ is the number of categories. Then, 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 equal category probabilities?

When the claim is that categories are equally likely, the expected value for each category is calculated using the formula $e=\frac{n}{k}$, where $n$ is the total number of observations and $k$ is the number of categories. This means you divide the total sample size evenly among all categories. For example, if 200 people are surveyed and there are 4 categories, each expected value is $\frac{200}{4}=50$. These expected values are then compared to observed counts in the chi-square test.

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. This formula accounts for the constraint that the sum of observed frequencies must equal the total sample size. For example, if you have 4 categories, the degrees of freedom will be $4-1=3$. This value is used in the chi-square distribution to find critical values and p-values.

What does the p-value indicate in a chi-square goodness of fit test on the TI-84?

The p-value in a chi-square goodness of fit test represents the probability of observing a chi-square statistic as extreme as, or more extreme than, the one calculated from your data, assuming the null hypothesis is true. On the TI-84, after running the test, the p-value is displayed along with the chi-square statistic and degrees of freedom. If the p-value is less than your significance level (commonly 0.05), you reject the null hypothesis, indicating the observed data does not fit the expected distribution. If the p-value is greater, you fail to reject the null hypothesis, suggesting the observed frequencies fit the claimed distribution well.

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

To input data on 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, which you calculate beforehand. For example, if expected values are equal, each entry in L2 will be the same. After entering both lists, you can proceed to the chi-square GOF test by selecting it from the TESTS menu and specifying L1 as observed and L2 as expected.