3.Give the mean for the mean column of the Worksheet. Is this estimate centered about the parameter of interest (the parameter of interest is the answer for the mean in question 2)?

Mean 3.50 Standard deviation 1.317

4.Give the mean for the median column of the Worksheet. Is this estimate centered about the parameter of interest (the parameter of interest is the answer for the mean in question 2)?

5.Give the standard deviation for the mean and median column. Compare these and be sure to identify which has the least variability?
mean median
3.2 3.5
4.5 5.0
3.7 4.0
3.7 3.0
3.1 3.5
3.6 3.5
3.1 3.0
3.6 3.0
3.8 4.0
2.6 2.0
4.3 4.0
3.5 3.5
3.3 3.5
4.1 4.5
4.2 5.0
2.9 2.5
3.5 4.0
3.7 3.5
3.5 3.0
3.3 4.0

To find the mean for a given column, you need to add up all the values in that column and divide the sum by the number of values.

For Question 3, the mean column has values: 3.2, 4.5, 3.7, 3.7, 3.1, 3.6, 3.1, 3.6, 3.8, 2.6, 4.3, 3.5, 3.3, 4.1, 4.2, 2.9, 3.5, 3.7, 3.5, 3.3.

Add these values together: 3.2 + 4.5 + 3.7 + 3.7 + 3.1 + 3.6 + 3.1 + 3.6 + 3.8 + 2.6 + 4.3 + 3.5 + 3.3 + 4.1 + 4.2 + 2.9 + 3.5 + 3.7 + 3.5 + 3.3 = 70.2.

There are 20 values in the column, so divide the sum by 20: 70.2 / 20 = 3.51.

Therefore, the mean for the mean column is 3.51.

For Question 4, follow the same steps as above using the median column values: 3.5, 5.0, 4.0, 3.0, 3.5, 3.5, 3.0, 3.0, 4.0, 2.0, 4.0, 3.5, 3.5, 4.5, 5.0, 2.5, 4.0, 3.5, 3.0, 4.0.

Add these values together: 3.5 + 5.0 + 4.0 + 3.0 + 3.5 + 3.5 + 3.0 + 3.0 + 4.0 + 2.0 + 4.0 + 3.5 + 3.5 + 4.5 + 5.0 + 2.5 + 4.0 + 3.5 + 3.0 + 4.0 = 70.0.

There are 20 values, so divide the sum by 20: 70.0 / 20 = 3.5.

Therefore, the mean for the median column is 3.5.

To determine if these estimates are centered about the parameter of interest (the parameter of interest being the answer for the mean in Question 2), you would need to compare the mean of the mean column (3.51) and the mean of the median column (3.5) to the answer for the mean in Question 2. Since the answer for the mean in Question 2 is not provided, I cannot determine if the estimates are centered about it.

For Question 5, we need to find the standard deviation for the mean and median column.

To find the standard deviation, you can use a statistical software or spreadsheet program. However, I will explain the steps involved.

First, find the deviation from the mean for each value in the mean column. For example, if the mean of the mean column is 3.51, then for the first value (3.2), the deviation is 3.2 - 3.51 = -0.31.

Square each deviation and sum them all up. For example, (-0.31)^2 + (...)^2 + (...)^2 + (...)^2 + (...)^2 = sum of squared deviations.

Next, divide the sum of squared deviations by the number of values minus 1, and then take the square root. This will give you the standard deviation.

Repeat the above steps for the median column.

Once you have calculated the standard deviations for both columns, compare them to determine which has the least variability. The column with the smaller standard deviation will have the least variability.