Question

The weights of four randomly and independently selected bags of potatoes labeled 20.0 pounds were found to be 20.9​, 21.5​, 20.9​, and 21.2 pounds. assume normality. answer parts​ (a) and​ (b) below. question content area bottom part 1

a. find a​ 95% confidence interval for the mean weight of all bags of potatoes.

Answer 1

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.