A random sample of 20 hogs on farmer Brown's hog farm had an average weight of 350 pounds and a SD of 70 pounds. Find 95% confidence interval estimate of the mean weight of all farmer Brown's pigs.

E=CV(SD/square root of n)

E=1.96(70/square root of 20)
E=6.86
mean-e<mean<mean+e
350 - 6.86 < 350 < 350 + 6.86

343.14 < mean < 356.86

I think that is right

To find the 95% confidence interval estimate of the mean weight of all of Farmer Brown's pigs, we can use the formula:

Confidence Interval = Sample Mean ± (Critical Value * Standard Deviation / √n)

Where:
- Sample Mean is the average weight of the sample hogs (350 pounds).
- Critical Value is a value obtained from the t-distribution table for a given confidence level and degrees of freedom. For a 95% confidence level with 20-1 degrees of freedom, the critical value is approximately 2.093.
- Standard Deviation is the standard deviation of the sample hogs (70 pounds).
- n is the sample size (20 hogs).

By substituting these values into the formula, we can calculate the confidence interval:

Confidence Interval = 350 ± (2.093 * 70 / √20)

First, calculate the value inside the parentheses:

2.093 * 70 / √20 ≈ 62.375

Now, plug this value into the confidence interval:

Confidence Interval = 350 ± 62.375

To find the lower and upper bounds of the confidence interval, we subtract and add the value obtained from the sample mean:

Lower bound: 350 - 62.375 ≈ 287.625
Upper bound: 350 + 62.375 ≈ 412.375

Therefore, the 95% confidence interval estimate of the mean weight of all Farmer Brown's pigs is approximately 287.625 to 412.375 pounds.