TMGT 361
Assignment VII BInstructions
See the general assignment instructions for information about the quality and quantity expectations and evaluation criteria. SPC & statistics. You can collect and use your own data instead of the data sets provided by the instructor for the unit VII assignments. Note that there are 4 assignments based on unit VII but only one quiz.

VII B. Using data set VIIB, do the following. Show your work (by hand or software output). Calculate appropriate descriptive statistics. Create a histogram and assess normality. What are the raw scores that correspond to +/- 1.96σ and +/-3σ? Calculate a CI for the mean. You pick the interval, e.g., 95%. Test µ = 10.0. You select the Type I error rate.

Calculate appropriate descriptive statistics.

The data follows:
Product …show more content…
So as to calculate the corresponding raw scores, we will have to incorporate the z score formula.
The z score formula follows.

= the mean (usually of a sample, e.g., our sample of data)
X = a raw score (usually of a sample, e.g., our sample of data) = the standard deviation (usually of a sample, e.g., our sample of data)

Calculating the raw scores is similar to calculating the z scores of +/- 1.96 σ and +/-3σ. To start with, we will need the mean and the standard deviation. The standard deviation is found as 5^(-9). The mean in turn is 9.915384615.

This is the same as asking what are the z scores of +/- 1.96 and +/- 3. I am going to round off to one decimal place. The mean of our data was 9.8 and the standard deviation was 5^(-9)

X = (1.96*5^(-9)) + 9.9 = 9.90001

X = (-1.96*5^(-9)) + 9.9 = 9.8999

9.8999 and 9.90001 are the raw values that correspond to minus and plus (which we would show as +/- even though the minus value is before the plus, i.e., we usually do not show the symbol as -/+).
+/- 3z works the same way.

X = (3*5^(-9)) + 9.9 = 9.900000015

X = (-3*5^(-9)) + 9.9 =