Question

A semiconductor manufacturer produces printed circuit boards that are sampled to determine the thickness of their copper plating. The following statements create a data set named Trans, which contains the plating thicknesses (Thick) of 50 boards:

3.412

3.45

3.551

3.451

3.60

3.462

3.586

3.645

3.252

3.62

3.606

3.634

3.852

3.56

3.342

3.341

3.444

3.774

3.632

3.199

3.71

3.654

3.723

3.981

3.934

3.708

3.934

3.315

3.762

3.223

3.469

3.481

3.515

3.535

3.46

3.575

3.488

3.515

3.484

3.482

3.517

3.483

3.467

3.467

3.502

3.471

3.516

3.474

3.5

3.466

a. Using Excel/R find the mean, median, mode, range, variance and standard deviation of the data. Attach your output from Excel/R. Based on the results in part a, construct the intervals and for the data set. Be sure to show your interval below. What percentage of the measurements for the data set falls in each interval? Compare the intervals , and . Explain why the results are different.

Thank you!

Answer 1

a) Using Excel, we calculate the mean, median, mode, range, variance and standard deviation of the given data. The output is as follows:

From part (a), we found the range of the data as 0.782. The minimum value in the data is 3.199 and the maximum value is 3.981.

Now, using these two values, we construct intervals of width = 0.16.

Hence the number of intervals constructed is 6.

To calculate the percentage of measurements for the data set in each interval, we need to calculate the relative frequencies for each interval.

The frequency distribution of the data along with the relative frequencies and the percentage of measurements in each interval is given in the following table:

From the above table, we see that most of the values are in the interval (3.465-3.625), while the first interval (3.145-3.305) and the last interval (3.945-4.105) contain less number of values. This is explained by the results obtained in part (a), namely the mean and median, as these terms are defined as measures of central tendency. Hence the values tend to cluster around them. The first and the last intervals contain very less number of values since we can see the value of variance in part (a) is very less, thus there is less dispersion of data values.