Boxes of nails contain 100 nails each. A sample of 10 boxes is drawn, and each of the boxes is weighed. The average weight is 1500 g and the standard deviation is 5 g. Assume the weight of the box itself is negligible, so that all the weight is due to the nails in the box.

a. Let µbox denote the mean weight of a box of nails. Find a 95% confidence interval for µbox.

b. Let µnail denote the mean weight of a nail. Express µnail in terms of µbox.

c. Find a 95% confidence interval for µnail.

Step 1 of 4

Given the box contains 100 nails each

The sample size is 10 boxes

The mean weight of the box

The Standard deviation of the box is

Step 2 of 4

a) We have to find the 95% confidence interval for average weight of the box

Here n=10

So, we have to use the t-Distribution with degrees of freedom

The 95% confidence interval is

Here is the critical value for 5% level of significance at 9 degrees of freedom

Then =2.262

Then 95% confidence interval is

=1500)

=15003.58

=(1496.42, 1503.58)

Hence the confidence intervals is (1496.42, 1503.58)