Question

1. Using confidence intervals to determine how many samples you’ll need to collect so the average voltage will be within the desired accuracy. This will take some assumptions and some work. So, here goes:

The confidence interval (CI) equation is:

Assume that the voltage you are measuring has a true mean (of course we don’t know the actual mean or true value, but that’s OK) of approximately 1 volt. Next, suppose you desire to know your average () vs “true” value (μ) within 0.01 (1/100^th) of a volt.

$3MuA8ZfptsUSrP4mtMAAAAASUVORK5CYII=$ .

Next, you must choose a confidence probability. Let’s assume you want to be 90% confident, so = 0.05 (or 0.95—your choice). Do reverse lookup in a Z table _______ or use Excel function NORM.INV(0.95,0,1) = ________.

Also, you must have a value for σ. For this question, assume it is 0.028. In your lab you’ll have to estimate it—see Notes below).

Note 1: Standard deviation in CI equation is for the entire population. We are going to calculate it for the sample and assume it is “good enough.” After you’ve calculated the minimum sample size you need to collect, just collect a few extra data points to make sure you will have collected enough. You can also rerun your calculations on the larger data set to confirm you are collecting enough data points.

Note 2: Use 10 data points to calculate an estimate of σ. Be careful as Excel does not perform the calculation for standard deviation correctly for small sample sizes of less than 30 data points (that is, it always divides by n, not n-1). It doesn’t make a lot of difference as in the simulation I ran. If you are using the standard deviation function in Excel, you can calculate the corrected standard deviation formula for a small sample size n using the equation = SQRT(STDEV.P(range_of_cells)^2*n/(n-1))

Next, rearrange the CI equation to solve for N = __________

Plug numbers in: N = ____________ (estimated number of samples to collect). Then, round up to next big number (if you get 27, round to 30)

Answer 1

the question Marqun of amor (m) = (M )=0.01 L= 1-0.90 - 0,10 0=0.028 We know that Margin of error (ME) = 2,2 olin (where is the sample size) 0.01 = 1.6449 * 0.028 un > Vña 1.6449 x 0.028 0.01 = p 1.6449x0.028) 0.01 = 21,21 The required sample size is 22.