IDNLearn.com is your go-to resource for finding expert answers and community support. Explore a wide array of topics and find reliable answers from our experienced community members.
Sagot :
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.
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. IDNLearn.com is committed to providing accurate answers. Thanks for stopping by, and see you next time for more solutions.