Question

Management at Webster Chemical Company is concerned as to whether caulking tubes are being properly capped. If a significant proportion of the tubes are not being​ sealed, Webster is placing its customers in a messy situation. Tubes are packaged in large boxes of . Several boxes are​ inspected, and the following numbers of leaking tubes are​ found: Sample Tubes Sample Tubes Sample Tubes 1 8 15 2 9 16 3 10 17 4 11 18 5 12 19 6 13 20 7 14 Total Calculate​ p-chart ​-sigma control limits to assess whether the capping process is in statistical control.

Answer 1

Answer:

$UCL = 0.078$

$LCL = 0$

Step-by-step explanation:

Poorly formatted question (see attachment)

From the attachment, we have:

$n = 135$ --- the sample size

$Samples = 20$ --- The number of samples

$\sum np = 87$ --- Number of leaking tubes

First, we calculate the number of observations

$\sum n = n * Samples$

$\sum n= 135 * 20$

$\sum n= 2700$

Using 3 sigma control limit, we have:

$z = 3$

Calculate $\bar p$

$\bar p = \frac{\sum np}{\sum n}$

So, we have:

$\bar p = \frac{87}{2700}$

$\bar p = 0.0322$

Next, calculate the standard deviation

$s_p = \sqrt{\frac{\bar p* (1-\bar p)}{n}}$

So, we have:

$s_p = \sqrt{\frac{0.0322* (1-0.0322)}{135}}$

$s_p = \sqrt{\frac{0.03116}{135}}$

$s_p = \sqrt{0.0002308}$

$s_p = 0.0152$

The control limits is then calculated as:

$UCL = \bar p + z * s_p$ --- upper control limit

$LCL = \bar p - z * s_p$ --- lower control limits

So, we have:

$UCL = \bar p + z * s_p$

$UCL = 0.0322 +3 * 0.0152$

$UCL = 0.078$

$LCL = \bar p - z * s_p$

$LCL = 0.0322 - 3 * 0.0152$

$LCL = -0.013$

Since the calculated LCL is less than 0, we simply set it to 0

$LCL = 0$

So, the p chart control limits are:

$UCL = 0.078$

$LCL = 0$