The bureau of alcohol, tobacco, and firearms (batf) has been concerned about lead levels in california wines. in a previous testing of wine specimens, lead levels ranging from 50 to 700 parts per billion were recorded. how many wine specimens should be tested if the batf wishes to estimate the true mean lead level for california wines to within 10 parts per billion with 95% confidence?

Using the uniform and the z-distribution, it is found that 1353 specimens should be tested.

For an uniform distribution of bounds a and b, the standard deviation is given by:

In this problem, , thus, the estimate is:

The margin of error of a z-confidence interval is given by:

In which:

The first step is finding the critical value, which is z with a p-value of , in which is the confidence level.

In this problem, , thus, z with a p-value of , which means that it is z = 1.96.

Estimate of the standard deviation is of 187.64, thus, .

We want the sample for a margin of error of 10, thus, we have to solve for n when . Then:

Simplifying by 10:

Rounding up:

1353 specimens should be tested.

A similar problem is given at link

 1015

Step-by-step explanation:

Given :In a previous testing of wine specimens, lead levels ranging from 50 to 700 parts per billion were recorded.

Here , Range = 700-50=650    [∵ Range = Max-Min]

According to the thumb rule , the range is approximately 4 times the standard deviation.

Let be the standard deviation, then

Thus,

For confidence interval of 95% , the critical z value =

Formula to find the sample size :

Now, for the margin of error of E=, we have

Hence, the required sample size = 1015

x> 6

step-by-step explanation:

8x+8> 7x+14

8x-7x> 14-8

x> 6

ou would have 12 of each