To construct a 90% confidence interval for the average length of charred portions of fabric, we will use the formula for the confidence interval of the mean when the population variance is unknown and the sample size is sufficiently large (which it is, since \( n = 64 \)).
The formula for the confidence interval is:
\[ \bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} \]
Where:
- \( \bar{x} \) is the sample mean
- \( t_{\alpha/2} \) is the t-value that corresponds to the desired confidence level and degrees of freedom (df = n-1)
- \( s \) is the sample standard deviation
- \( n \) is the sample size
Given:
- Sample mean \( \bar{x} = 3 , \text{cm} \)
- Sample variance \( s^2 = 0.16 , \text{cm}^2 \) ⇒ Sample standard deviation \( s = \sqrt{0.16} = 0.4 , \text{cm} \)
- Sample size \( n = 64 \)
Degrees of freedom:
\[ df = n - 1 = 64 - 1 = 63 \]
For a 90% confidence interval, the significance level \( \alpha = 1 - 0.90 = 0.10 \). Hence, \( \alpha/2 = 0.05 \).
Using a t-table or calculator, we can find \( t_{0.05, 63} \). For \( df=63 \), the t-value approximately equals 1.669.
Now, we can calculate the margin of error (ME):
\[ ME = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} = 1.669 \cdot \frac{0.4}{\sqrt{64}} = 1.669 \cdot \frac{0.4}{8} = 1.669 \cdot 0.05 = 0.08345 \]
Now, we can construct the confidence interval:
\[ \text{Lower limit} = \bar{x} - ME = 3 - 0.08345 \approx 2.91655 , \text{cm} \]
\[ \text{Upper limit} = \bar{x} + ME = 3 + 0.08345 \approx 3.08345 , \text{cm} \]
Thus, the 90% confidence interval for the average length of the charred portions of this fabric is approximately:
\[ (2.9166 , \text{cm}, 3.0834 , \text{cm}) \]
Rounded to four decimal places, the final answer is:
(2.9166 cm, 3.0834 cm).