Final answer:
To determine a 95% confidence interval for the mean weight of all bags of potatoes, calculate the sample mean and standard deviation, use the t-distribution to find the corresponding t-value, and then compute the margin of error to construct the interval.
Step-by-step explanation:
To find a 95% confidence interval for the mean weight of all bags of potatoes, we can use the following steps:
Calculate the sample mean by adding the weights and dividing by the number of observations.
Find the sample standard deviation.
Use the t-distribution because the sample size is less than 30 and we don't know the population standard deviation.
Determine the t-value for a 95% confidence level and degrees of freedom equal to n - 1.
Calculate the margin of error using the formula: Margin of Error = t-value * (sample standard deviation / sqrt(n)).
Construct the confidence interval by subtracting and adding the margin of error from the sample mean.
The calculation would look like this:
Sample mean (\(\bar{x}\)) = (20.9 + 21.5 + 20.9 + 21.2) / 4
Sample standard deviation (s) = sqrt[\(\sum(x_i - \bar{x})^2 / (n-1)\)]
Degrees of freedom (df) = n - 1
Margin of Error = t-value * (s / sqrt(n))
Confidence Interval = \(\bar{x}\) ± Margin of Error
To answer the specific values, you need to perform the calculations using the sample data provided.