Answer:
To find the number of seeds velvetleaf plants produce with 99% confidence, we need to calculate the mean and standard deviation of the seed counts from the provided data, and then determine the confidence interval for the mean.
First, let's calculate the mean (\(\bar{x}\)) and standard deviation (\(s\)) of the seed counts:
Mean (\(\bar{x}\)):
\[ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \]
\[ \bar{x} = \frac{2450 + 2504 + \ldots + 130 + 880}{28} \]
\[ \bar{x} = \frac{55928}{28} \]
\[ \bar{x} = 1997 \]
Standard Deviation (\(s\)):
\[ s = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n-1}} \]
\[ s = \sqrt{\frac{(2450 - 1997)^2 + (2504 - 1997)^2 + \ldots + (880 - 1997)^2}{28-1}} \]
\[ s = \sqrt{\frac{13175860}{27}} \]
\[ s \approx \sqrt{488034.074} \]
\[ s \approx 698.586 \]
Now, we can use the t-distribution to find the margin of error and construct the confidence interval for the mean. Since the sample size is small (\(n = 28\)), we'll use a t-distribution instead of a z-distribution.
The critical value for a 99% confidence level with \(n - 1 = 28 - 1 = 27\) degrees of freedom is approximately 2.796.
Margin of Error:
\[ ME = t \times \frac{s}{\sqrt{n}} \]
\[ ME = 2.796 \times \frac{698.586}{\sqrt{28}} \]
\[ ME \approx 2.796 \times \frac{698.586}{5.2915} \]
\[ ME \approx 369.88 \]
Confidence Interval:
\[ \text{CI} = (\bar{x} - ME, \bar{x} + ME) \]
\[ \text{CI} = (1997 - 369.88, 1997 + 369.88) \]
\[ \text{CI} = (1627.12, 2366.88) \]
Therefore, with 99% confidence, we estimate that the number of seeds velvetleaf plants produce is between 1627 and 2367 seeds.
