An engineer is running a plant test on Line 2 and claims that mean moisture cont
ID: 3328017 • Letter: A
Question
An engineer is running a plant test on Line 2 and claims that mean moisture content is out of specification. The specification requires that mean moisture content to be equal to 10.5%. The engineer decided to collect moisture analyses on 5 samples during this testing period. These data resulted in a mean value of 10.7% and a standard deviation of 2%. Using a test level of 0.05 and answer the following questions.
a. Does this set of data supports the engineer’s claim at the specified test level?
b. If your conclusion in Part “a.” is wrong, what type of error did you make?
c. Give a 99% confidence interval for the mean moisture content.
d. With the conditions as given, what is the power if the true mean moisture content is 10.1%?
e. With the conditions as given, if the true mean moisture content is 10.8%, with a type 2 error of 0.05, how many samples should be taken? Use a s = 5%.
Explanation / Answer
a. The sample size is too small and population standard deviation is unknown. Therefore, use, Student's t model to compute the t test statistic for 1-population mean.
t=(xbar-mu)/(s/sqrt n), where, xbar is sample mean, mu is population mean, s is sample standard deviation, and n is sample size.
=(10.7-10.5)/(2/sqrt 5)
=0.22
The p value at 4 df [df=n-1=5-1] and alpha=0.05 (one-tailed) is 0.4183. Per rejection rule based on p value, reject null hypothesis if p value is less than 0.05. Here, p value is not less than 0.05, therefore, fail to reject null hypothesis. There is insufficient sample evidence to support the claim.
b. If conclusion in part a is wrong, this implies the reseracher has incorrectly retained a false null hypothesis. Thus, Type II error is made [Type II error means failure to reject a false null hypothesis].
c. The 99% c.i=xbar+-talpha/2, df=n-1 (s/sqrt n), where, xbar is sample mean, t is t critical at alpha/2 [alpha=0.01, alpha/2=0.005] and n-1 df, s is sample standard deviation, and n is sample size.
=10.7+-4.604(2/sqrt 5)
=(6.582,14.818)
d. This is a right-tailed test, and therefore, one would fail to reject the null (commit Type II error) if one get a t test statistic greater than 2.78.
Now, 2.78=tcritical=(Xcritical-mu)/(2/sqrt 5)
2.78=(Xcritical-10.5)/0.8944
Xcritical=12.99
Thus, P(Type II) error=P[xbar<12.99|mu=10.1]= P(t<[12.99-10.1)/0.8944]=P(t<3.23)=0.0159