Answered

From science to arts, IDNLearn.com has the answers to all your questions. Our experts are available to provide in-depth and trustworthy answers to any questions you may have.

A restaurant manager wants to know if one of the three specials on the menu (baked fish, roast beef, or chicken salad) is more popular than the others. He selects a random sample of special orders and counts the number of orders in each category. The results are compiled in the following table.

\begin{tabular}{ccc}
\hline
Fish & Beef & Chicken \\
\hline
48 & 46 & 53 \\
\hline
\end{tabular}

1. Select the appropriate null hypothesis for this experiment:
A. One meal is less popular than the other two.
B. One meal is more popular than the other two.
C. There is no difference in popularity among the three meals.
D. Customers prefer the baked fish to the roast beef.

2. Complete the following table of expected frequencies:

\begin{tabular}{ccc}
Fish & Beef & Chicken \\
\hline
[tex]$\square$[/tex] & [tex]$\square$[/tex] & [tex]$\square$[/tex] \\
\end{tabular}

3. The chi-square statistic for these data is [tex]$x^2=$[/tex] [tex]$\square$[/tex].

Sagot :

GinnyAnsweridn
### Step-by-Step Solution

#### Null Hypothesis
First, let's identify the appropriate null hypothesis for this experiment:

Null Hypothesis (H0): There is no difference in popularity among the three meals.

This means that under the null hypothesis, we assume that each type of meal is equally popular.

#### Calculate the Total Number of Orders
From the given data, we have:
- Fish orders: 48
- Beef orders: 46
- Chicken orders: 53

The total number of orders is the sum of these three quantities:
[tex]\[ \text{Total orders} = 48 + 46 + 53 = 147 \][/tex]

#### Expected Frequencies
Under the null hypothesis (equal popularity for each meal), the expected frequency for each category can be calculated by dividing the total number of orders by the number of meal categories (3):

[tex]\[ \text{Expected frequency for each category} = \frac{\text{Total orders}}{3} = \frac{147}{3} = 49 \][/tex]

Thus, the expected frequencies are:
[tex]\[ \begin{array}{ccc} \text{Fish} & \text{Beef} & \text{Chicken} \\ \hline 49 & 49 & 49 \\ \end{array} \][/tex]

#### Chi-Square Statistic
The chi-square statistic is calculated using the formula:
[tex]\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \][/tex]
where [tex]\( O_i \)[/tex] is the observed frequency and [tex]\( E_i \)[/tex] is the expected frequency for each category.

Plugging in the observed and expected frequencies:

- For Fish:
[tex]\[ \frac{(48 - 49)^2}{49} = \frac{(-1)^2}{49} = \frac{1}{49} \][/tex]

- For Beef:
[tex]\[ \frac{(46 - 49)^2}{49} = \frac{(-3)^2}{49} = \frac{9}{49} \][/tex]

- For Chicken:
[tex]\[ \frac{(53 - 49)^2}{49} = \frac{4^2}{49} = \frac{16}{49} \][/tex]

Adding these values together gives:
[tex]\[ \chi^2 = \frac{1}{49} + \frac{9}{49} + \frac{16}{49} = \frac{1 + 9 + 16}{49} = \frac{26}{49} \approx 0.5306 \][/tex]

#### Summary
- Null Hypothesis: There is no difference in popularity among the three meals.
- Expected Frequencies:
[tex]\[ \begin{array}{ccc} \text{Fish} & \text{Beef} & \text{Chicken} \\ \hline 49 & 49 & 49 \\ \end{array} \][/tex]
- Chi-Square Statistic:
[tex]\[ \chi^2 \approx 0.5306 \][/tex]

Therefore:
- The appropriate null hypothesis is: There is no difference in popularity among the three meals.
- The completed expected frequencies table is:
[tex]\[ \begin{array}{ccc} \text{Fish} & \text{Beef} & \text{Chicken} \\ \hline 49 & 49 & 49 \\ \end{array} \][/tex]
- The chi-square statistic is [tex]\( \chi^2 \approx 0.5306 \)[/tex].
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Discover the answers you need at IDNLearn.com. Thank you for visiting, and we hope to see you again for more solutions.