Question
A cross between white and yellow summer squash gave progeny of the following colours:
Colour White Yellow Green
Number of progeny 150 35 10
Expected frequencies A B C
A certain genetic model predicted the ratio of these colours to be 6:3:2. Perform an appropriate hypothesis test to test whether the observed results are consistent with what was expected based on the genetic model. Test at 2.5% significance level. What is the value of C in the table?
35.4
11.5
21.4
8.2
Colour White Yellow Green
Number of progeny 150 35 10
Expected frequencies A B C
A certain genetic model predicted the ratio of these colours to be 6:3:2. Perform an appropriate hypothesis test to test whether the observed results are consistent with what was expected based on the genetic model. Test at 2.5% significance level. What is the value of C in the table?
35.4
11.5
21.4
8.2
Answers
GPT-4o mini
To perform a hypothesis test to evaluate whether the observed results (progeny colors) are consistent with the expected genetic model ratio of 6:3:2, we can use the Chi-square test.
### Step 1: Calculate the Total Number of Progeny
Total number of progeny = 150 (White) + 35 (Yellow) + 10 (Green) = 195
### Step 2: Determine the Expected Frequencies
According to the genetic model, the expected ratio is 6:3:2. To calculate the expected counts for each color, we first find the total parts in the ratio:
Total parts = 6 + 3 + 2 = 11
Now, we can calculate the expected frequencies (A for White, B for Yellow, and C for Green):
- **A (White)** = (6/11) * 195 = 106.36
- **B (Yellow)** = (3/11) * 195 = 53.18
- **C (Green)** = (2/11) * 195 = 35.45
### Step 3: Calculate Expected Frequencies
Based on the calculations:
- \( A \approx 106.36 \)
- \( B \approx 53.18 \)
- \( C \approx 35.45 \)
### Step 4: Perform Chi-square Test
The Chi-square test statistic is calculated using the formula:
\[
\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}
\]
where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency.
Let's calculate \( \chi^2 \):
- For White:
\[
\frac{(150 - 106.36)^2}{106.36} = \frac{43.64^2}{106.36} \approx 18.41
\]
- For Yellow:
\[
\frac{(35 - 53.18)^2}{53.18} = \frac{(-18.18)^2}{53.18} \approx 6.44
\]
- For Green:
\[
\frac{(10 - 35.45)^2}{35.45} = \frac{(-25.45)^2}{35.45} \approx 18.03
\]
### Step 5: Calculate Total \(\chi^2\)
\[
\chi^2 = 18.41 + 6.44 + 18.03 \approx 42.88
\]
### Step 6: Determine the Critical Value
Our degrees of freedom for this test can be calculated as:
\[
\text{df} = k - 1 = 3 - 1 = 2 \quad (\text{k: number of categories})
\]
At the 2.5% significance level, using a Chi-square table, the critical value for \( \chi^2 \) with 2 degrees of freedom is approximately 9.21 (values might vary slightly based on table source).
### Step 7: Decision
Since our calculated \(\chi^2\) value (42.88) is much greater than the critical value (9.21), we reject the null hypothesis.
### Conclusion
The observed results are not consistent with the expected ratios based on the genetic model of 6:3:2.
### Answer to the Question Regarding Value of C
The value of C in the table is approximately **35.45**, rounding to your nearest expected C values leads to **35.4**, making it the best answer choice based on calculations.
### Step 1: Calculate the Total Number of Progeny
Total number of progeny = 150 (White) + 35 (Yellow) + 10 (Green) = 195
### Step 2: Determine the Expected Frequencies
According to the genetic model, the expected ratio is 6:3:2. To calculate the expected counts for each color, we first find the total parts in the ratio:
Total parts = 6 + 3 + 2 = 11
Now, we can calculate the expected frequencies (A for White, B for Yellow, and C for Green):
- **A (White)** = (6/11) * 195 = 106.36
- **B (Yellow)** = (3/11) * 195 = 53.18
- **C (Green)** = (2/11) * 195 = 35.45
### Step 3: Calculate Expected Frequencies
Based on the calculations:
- \( A \approx 106.36 \)
- \( B \approx 53.18 \)
- \( C \approx 35.45 \)
### Step 4: Perform Chi-square Test
The Chi-square test statistic is calculated using the formula:
\[
\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}
\]
where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency.
Let's calculate \( \chi^2 \):
- For White:
\[
\frac{(150 - 106.36)^2}{106.36} = \frac{43.64^2}{106.36} \approx 18.41
\]
- For Yellow:
\[
\frac{(35 - 53.18)^2}{53.18} = \frac{(-18.18)^2}{53.18} \approx 6.44
\]
- For Green:
\[
\frac{(10 - 35.45)^2}{35.45} = \frac{(-25.45)^2}{35.45} \approx 18.03
\]
### Step 5: Calculate Total \(\chi^2\)
\[
\chi^2 = 18.41 + 6.44 + 18.03 \approx 42.88
\]
### Step 6: Determine the Critical Value
Our degrees of freedom for this test can be calculated as:
\[
\text{df} = k - 1 = 3 - 1 = 2 \quad (\text{k: number of categories})
\]
At the 2.5% significance level, using a Chi-square table, the critical value for \( \chi^2 \) with 2 degrees of freedom is approximately 9.21 (values might vary slightly based on table source).
### Step 7: Decision
Since our calculated \(\chi^2\) value (42.88) is much greater than the critical value (9.21), we reject the null hypothesis.
### Conclusion
The observed results are not consistent with the expected ratios based on the genetic model of 6:3:2.
### Answer to the Question Regarding Value of C
The value of C in the table is approximately **35.45**, rounding to your nearest expected C values leads to **35.4**, making it the best answer choice based on calculations.