Answer :
Sure, let's walk through a detailed step-by-step solution to determine the critical t-value, the margin of error, and the 95% confidence interval for the additional growth of plants.
### Step-by-Step Solution
#### Step 1: Understand the Given Data
We are provided with:
- Sample size ([tex]\( n \)[/tex]) = 11
- Sample mean ([tex]\( \bar{x} \)[/tex]) = 8 inches
- Sample standard deviation ([tex]\( s \)[/tex]) = 3 inches
- Significance level ([tex]\( \alpha \)[/tex]) = 0.05
#### Step 2: Calculate the Degrees of Freedom (df)
Degrees of freedom is calculated as:
[tex]\[ df = n - 1 = 11 - 1 = 10 \][/tex]
#### Step 3: Determine the Critical t-Value ([tex]\( t^ \)[/tex])
The critical t-value at a 0.05 significance level for a two-tailed test can be found using the desired confidence level (1 - [tex]\( \alpha \)[/tex]). Since [tex]\( \alpha = 0.05 \)[/tex], this corresponds to a 95% confidence level.
For [tex]\( df = 10 \)[/tex] and a 95% confidence level:
[tex]\[ t^ = 2.228 \][/tex]
#### Step 4: Calculate the Margin of Error
The margin of error (ME) can be calculated using the formula:
[tex]\[ \text{ME} = t^ \times \left( \frac{s}{\sqrt{n}} \right) \][/tex]
Substitute the values:
- [tex]\( t^ = 2.228 \)[/tex]
- [tex]\( s = 3 \)[/tex]
- [tex]\( n = 11 \)[/tex]
[tex]\[ \text{ME} = 2.228 \times \left( \frac{3}{\sqrt{11}} \right) \approx 2.015 \][/tex]
#### Step 5: Calculate the Confidence Interval
The confidence interval (CI) can be calculated using the sample mean ([tex]\( \bar{x} \)[/tex]) and the margin of error (ME):
[tex]\[ \text{CI} = \left( \bar{x} - \text{ME}, \bar{x} + \text{ME} \right) \][/tex]
Substitute the values:
- [tex]\( \bar{x} = 8 \)[/tex]
- [tex]\( \text{ME} = 2.015 \)[/tex]
[tex]\[ \text{CI} = \left( 8 - 2.015, 8 + 2.015 \right) = (5.985, 10.015) \][/tex]
### Summary
- [tex]\( t^* \)[/tex] at the 0.05 significance level = 2.228
- Margin of error = 2.015
- Confidence interval = [5.985, 10.015]
So, the additional growth of plants in one week can be estimated with 95% confidence to lie between 5.985 inches and 10.015 inches.
### Step-by-Step Solution
#### Step 1: Understand the Given Data
We are provided with:
- Sample size ([tex]\( n \)[/tex]) = 11
- Sample mean ([tex]\( \bar{x} \)[/tex]) = 8 inches
- Sample standard deviation ([tex]\( s \)[/tex]) = 3 inches
- Significance level ([tex]\( \alpha \)[/tex]) = 0.05
#### Step 2: Calculate the Degrees of Freedom (df)
Degrees of freedom is calculated as:
[tex]\[ df = n - 1 = 11 - 1 = 10 \][/tex]
#### Step 3: Determine the Critical t-Value ([tex]\( t^ \)[/tex])
The critical t-value at a 0.05 significance level for a two-tailed test can be found using the desired confidence level (1 - [tex]\( \alpha \)[/tex]). Since [tex]\( \alpha = 0.05 \)[/tex], this corresponds to a 95% confidence level.
For [tex]\( df = 10 \)[/tex] and a 95% confidence level:
[tex]\[ t^ = 2.228 \][/tex]
#### Step 4: Calculate the Margin of Error
The margin of error (ME) can be calculated using the formula:
[tex]\[ \text{ME} = t^ \times \left( \frac{s}{\sqrt{n}} \right) \][/tex]
Substitute the values:
- [tex]\( t^ = 2.228 \)[/tex]
- [tex]\( s = 3 \)[/tex]
- [tex]\( n = 11 \)[/tex]
[tex]\[ \text{ME} = 2.228 \times \left( \frac{3}{\sqrt{11}} \right) \approx 2.015 \][/tex]
#### Step 5: Calculate the Confidence Interval
The confidence interval (CI) can be calculated using the sample mean ([tex]\( \bar{x} \)[/tex]) and the margin of error (ME):
[tex]\[ \text{CI} = \left( \bar{x} - \text{ME}, \bar{x} + \text{ME} \right) \][/tex]
Substitute the values:
- [tex]\( \bar{x} = 8 \)[/tex]
- [tex]\( \text{ME} = 2.015 \)[/tex]
[tex]\[ \text{CI} = \left( 8 - 2.015, 8 + 2.015 \right) = (5.985, 10.015) \][/tex]
### Summary
- [tex]\( t^* \)[/tex] at the 0.05 significance level = 2.228
- Margin of error = 2.015
- Confidence interval = [5.985, 10.015]
So, the additional growth of plants in one week can be estimated with 95% confidence to lie between 5.985 inches and 10.015 inches.