In: Statistics and Probability
** That should be done in Mini-tab.
Thanks
A metal refining company recently implemented a new process for
producing iron nuggets directly from raw iron ore and coal. To
assess the stability of the process, a quality engineer sampled and
measured the percentage change in carbon content every 4 hours. The
change in carbon content for a stable process is known to follow a
normal distribution with a mean of 3.5%.
a. Prepare a normal probability plot of the % change in carbon content measurements. At α = 0.05, does the data fit a normal distribution?
b.State the hypothesis and perform the appropriate test to evaluate the claim that the observed process mean does not follow the specified stable mean change in carbon content. (α = 0.05)
% Change |
3.25 |
3.3 |
3.23 |
3.51 |
3.1 |
3.6 |
3.65 |
3.5 |
3.4 |
3.35 |
3.48 |
3.5 |
3.25 |
3.6 |
3.55 |
3.6 |
3.55 |
3.48 |
3.42 |
3.4 |
3.5 |
3.78 |
3.7 |
3.5 |
3.45 |
3.75 |
3.52 |
3.1 |
3.25 |
3.4 |
3.2 |
3.45 |
3.3 |
Here, we have given that
X: The change in carbon content for a stable process
and X follows the normal distribution with a mean of 3.5%
By using MINITAB software we can perform this anlaysis.
(A)
Now, We want to find the probability plot
put the data into MINITAB Software
Graph<<select Probability plot <<< select variable <<OK
we get
Interpretation
Here all the data points are following within this reference line. that is here we conclude that the data follows the normal distribution.
(B)
Now,we want to perform the one sample t-test using MINITAB Software.
Claim: To check weather the observed process mean does not follow the specified stable mean change in carbon content. ( = 0.05)
The Hypothesis is as follows,
v/s
Now,
Put the data in MINITAB software
Stat<<Basic statistics<<< one sample t<< select data<< OK
we get the following output
One-Sample T: % Change
Test of mu = 3.5 vs not = 3.5
Variable N Mean StDev SE Mean T-statistics
P-value
% Change 33 3.44303 0.17202 0.02994 -1.90 0.066
Here we get
The t-statistics is -1.90
and
P-value is 0.066
Decision:
Here P-value > 0.05
we do not reject the Ho Null hypothesis
Conclusion:
we conclude that the observed process mean does not follow the specified stable mean change in carbon content 0.05