Question

What is the proportion of orders with errors for each of the 20 days?

Dividing the number of orders in the sample by the # of orders with errors gives p_i. The values are provided below.

Day	p(i)
1	0.02
2	0.00
3	0.03
4	0.00
5	0.00
6	0.05
7	0.00
8	0.03
9	0.03
10	0.02
11	0.04
12	0.00
13	0.05
14	0.01
15	0.01
16	0.03
17	0.02
18	0.04
19	0.01
20	0.03

Construct and interpret a p-chart for the GearHedz packaging department. Use a risk of Type I error of 0.03 for your chart.

Risk of Type I error = 0.03 means we’ll have an area of 0.015 in each tail of the distribution. Therefore, we will use z=2.17.

Therefore:

UCL_p=0.021+2.17(0.014338) = 0.052

Center line_p=0.021

LCL_p=0.021-2.17(0.014338) = -.01011

We’ll make the lower limit 0 (because you can’t have a negative proportion).

Therefore:

UCL_p=0.021+2.17(0.014338) = 0.052

Center line_p=0.021

LCL_p=0.021-2.17(0.014338) = -.01011

We’ll make the lower limit 0 (because you can’t have a negative proportion).

How did they calculate Z=2.17????

Answer 1

Result:

Construct and interpret a p-chart for the GearHedz packaging department. Use a risk of Type I error of 0.03 for your chart.

Risk of Type I error = 0.03 means we’ll have an area of 0.015 in each tail of the distribution. Therefore, we will use z=2.17.

Therefore:

UCL_p=0.021+2.17(0.014338) = 0.052

Center line_p=0.021

LCL_p=0.021-2.17(0.014338) = -.01011

We’ll make the lower limit 0 (because you can’t have a negative proportion).

Therefore:

UCL_p=0.021+2.17(0.014338) = 0.052

Center line_p=0.021

LCL_p=0.021-2.17(0.014338) = -.01011

We’ll make the lower limit 0 (because you can’t have a negative proportion).

How did they calculate Z=2.17????

We have to use standard normal distribution.

Risk of Type I error = 0.03 means we’ll have an area of 0.015 in each tail of the distribution

You can use excel function

=NORM.S.INV(0.015)