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Sagot :
### 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].
#### 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].
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