The problem describes a scenario where a pizza company wants to determine if the number of different toppings ordered is the same. The provided data includes the observed number of pepperoni, sausage, and cheese pizzas ordered last week. The task is to complete the frequency table by calculating the total observed orders, calculating the expected orders under the assumption of equal preference for each topping, and then performing a test for a difference. It is implied that we perform a chi-square goodness-of-fit test to check if there is a significant difference between the observed and expected frequencies.

1. Problem Description

The problem describes a scenario where a pizza company wants to determine if the number of different toppings ordered is the same. The provided data includes the observed number of pepperoni, sausage, and cheese pizzas ordered last week. The task is to complete the frequency table by calculating the total observed orders, calculating the expected orders under the assumption of equal preference for each topping, and then performing a test for a difference. It is implied that we perform a chi-square goodness-of-fit test to check if there is a significant difference between the observed and expected frequencies.

2. Solution Steps

First, we need to compute the total number of pizzas ordered.

$Total = Pepperoni + Sausage + Cheese$

$Total = 325 + 275 + 255 = 855$

Now we fill in the "Total" cell for the "Observed" row. Next, we want to calculate the expected number of pizzas for each topping, assuming an equal preference. We divide the total number of pizzas by the number of topping types (which is 3).

$Expected = Total / NumberOfToppings$

$Expected = 855 / 3 = 285$

We fill the "Expected" row with 285 for each topping and for the "Total" we put 855 because the sum of the expected values should be equal to the total.

Now we set up our null and alternative hypotheses:

$H_0$: The number of pepperoni, sausage, and cheese pizzas ordered is the same.

$H_1$: The number of pepperoni, sausage, and cheese pizzas ordered is not the same.

We will perform a chi-square goodness-of-fit test. The test statistic is calculated as:

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

Where $O_i$ is the observed frequency for category i and $E_i$ is the expected frequency for category i.

$\chi^2 = \frac{(325 - 285)^2}{285} + \frac{(275 - 285)^2}{285} + \frac{(255 - 285)^2}{285}$

$\chi^2 = \frac{(40)^2}{285} + \frac{(-10)^2}{285} + \frac{(-30)^2}{285}$

$\chi^2 = \frac{1600}{285} + \frac{100}{285} + \frac{900}{285}$

$\chi^2 = \frac{2600}{285} \approx 9.123$

The degrees of freedom for this test are $df = NumberOfCategories - 1 = 3 - 1 = 2$.

Using a chi-square distribution table or calculator, we can find the p-value associated with $\chi^2 = 9.123$ and $df = 2$. Using a calculator, the p-value is approximately 0.

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Since the problem only says to "test for a difference" and does not state a significance level, we will assume a standard significance level of $\alpha = 0.05$.

Since the p-value (0.0104) is less than the significance level (0.05), we reject the null hypothesis. This means there is a significant difference in the number of pepperoni, sausage, and cheese pizzas ordered.

3. Final Answer

Observed Total: 855

Expected Pepperoni: 285

Expected Sausage: 285

Expected Cheese: 285

Expected Total: 855

$\chi^2$ statistic: 9.123

p-value: 0.0104

Conclusion: There is a significant difference in the number of pepperoni, sausage, and cheese pizzas ordered.