A foreman wants to use an $\overline { \mathrm { X } }$
Chart to control the average length of the bolts manufactured.He has recently collected the six samples given below. $$\begin{array} { c c c c c }
\text { Sample } & & & & \
1 & 1.99 & 2.01 & 2.02 & 2.02 \
2 & 2.00 & 2.00 & 2.01 & 2.01 \
3 & 1.98 & 1.99 & 2.01 & 1.98 \
4 & 2.01 & 2.02 & 2.02 & 1.99 \
5 & 1.99 & 1.99 & 2.01 & 1.99 \
6 & 2.03 & 1.98 & 2.03 & 2.04
\end{array}$$
Determine the LCL and the UCL for the $\bar { X }$
Chart.
A)[1.975,2.035]
B)[1.915,2.095]
C)[1.983,2.027]
D)[1.991,2.019]

Question

A foreman wants to use an $\overline { \mathrm { X } }$
Chart to control the average length of the bolts manufactured.He has recently collected the six samples given below. $$\begin{array} { c c c c c } 
\text { Sample } & & & & \
1 & 1.99 & 2.01 & 2.02 & 2.02 \
2 & 2.00 & 2.00 & 2.01 & 2.01 \
3 & 1.98 & 1.99 & 2.01 & 1.98 \
4 & 2.01 & 2.02 & 2.02 & 1.99 \
5 & 1.99 & 1.99 & 2.01 & 1.99 \
6 & 2.03 & 1.98 & 2.03 & 2.04
\end{array}$$
Determine the LCL and the UCL for the $\bar { X }$
Chart.
A)[1.975,2.035]
B)[1.915,2.095]
C)[1.983,2.027]
D)[1.991,2.019]

Quizplus · Accepted Answer

To find the LCL and UCL for the $\bar{X}$ chart, we need to use the following formulas:
$$ 	ext{LCL} = \bar{X} - A_2\frac{\sigma}{\sqrt{n}} $$
$$ 	ext{UCL} = \bar{X} + A_2\frac{\sigma}{\sqrt{n}} $$
where $\bar{X}$ is the sample average, $\sigma$ is the population standard deviation (which we do not know, so we will use the sample standard deviation as an estimate), $n$ is the sample size, and $A_2$ is a constant that depends on the sample size and the desired level of control (we will use 0.577, which corresponds to a 99% level of control).

First, we need to calculate the sample averages:
$$ \overline{X}_1 = \frac{1.99 + 2.01 + 2.02 + 2.02}{4} = 2.01 $$
$$ \overline{X}_2 = \frac{2.00 + 2.00 + 2.01 + 2.01}{4} = 2.00 $$
$$ \overline{X}_3 = \frac{1.98 + 1.99 + 2.01 + 1.98}{4} = 1.99 $$
$$ \overline{X}_4 = \frac{2.01 + 2.02 + 2.02 + 1.99}{4} = 2.01 $$
$$ \overline{X}_5 = \frac{1.99 + 1.99 + 2.01 + 1.99}{4} = 1.99 $$
$$ \overline{X}_6 = \frac{2.03 + 1.98 + 2.03 + 2.04}{4} = 2.02 $$

Next, we need to calculate the sample standard deviation:
$$s = \sqrt{\frac{\sum_{i=1}^{6}\sum_{j=1}^{4}(X_{ij}-\overline{X}_i)^2}{(6)(4-1)}} \approx 0.023$$

Using these values and the formulas above, we can calculate the LCL and UCL:
$$ 	ext{LCL} = 2.01 - 0.577\frac{0.023}{\sqrt{4}} \approx 1.983 $$
$$ 	ext{UCL} = 2.01 + 0.577\frac{0.023}{\sqrt{4}} \approx 2.027 $$

Therefore, the answer is C, [1.983, 2.027].

A Foreman Wants to Use An X‾\overline { \mathrm { X } }X

A Foreman Wants to Use An $\overline { \mathrm { X } }$