Question

In: Statistics and Probability

** That should be done in Mini-tab. Thanks A metal refining company recently implemented a new...

** 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

Solutions

Expert Solution

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


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