Given the parameters of the sample:
- Sample size ([tex]\(n\)[/tex]) = 11
- Sample standard deviation ([tex]\(s\)[/tex]) = 3 inches
- Sample mean ([tex]\(\bar{x}\)[/tex]) = 8 inches
- Significance level ([tex]\(\alpha\)[/tex]) = 0.05
We need to determine the following:
1. The critical value ([tex]\(t^\)[/tex]) at a 0.05 significance level.
2. The margin of error.
3. The confidence interval for the mean additional growth of plants.
### Step 1: Determining the Degrees of Freedom
The degrees of freedom ([tex]\(df\)[/tex]) can be calculated as:
[tex]\[ df = n - 1 = 11 - 1 = 10 \][/tex]
### Step 2: Critical Value ([tex]\(t^\)[/tex])
For a two-tailed test at a 0.05 significance level with 10 degrees of freedom, the critical value ([tex]\(t^\)[/tex]) can be found from the t-distribution table. The critical value [tex]\(t^\)[/tex] is approximately:
[tex]\[ t^ \approx 2.228 \][/tex]
### Step 3: Calculating the Margin of Error
The margin of error (ME) is given by:
[tex]\[ ME = t^ \times \left(\frac{s}{\sqrt{n}}\right) \][/tex]
Plugging in the values:
[tex]\[ ME = 2.228 \times \left(\frac{3}{\sqrt{11}}\right) \approx 2.015 \][/tex]
### Step 4: Constructing the Confidence Interval
The confidence interval is calculated as:
[tex]\[ \left( \bar{x} - ME, \bar{x} + ME \right) \][/tex]
Placing the given values into this expression:
[tex]\[ CI = (8 - 2.015, 8 + 2.015) \][/tex]
[tex]\[ CI = (5.985, 10.015) \][/tex]
Thus, the results are:
1. Critical value ([tex]\(t^\)[/tex]) at the 0.05 significance level:
[tex]\[ t^ = 2.228 \][/tex]
2. Margin of error:
[tex]\[ ME = 2.015 \][/tex]
3. Confidence interval:
[tex]\[ CI = [5.985, 10.015] \][/tex]
Therefore, the additional growth of plants in one week, with a 95% confidence interval, is between 5.985 inches and 10.015 inches.
