Measurements on the percentage of enrichment of 12 fuel rods
Solution
A)
Formulating the null and alternative hypotheses,
Ho: u = 2.95
Ha: u =/ 2.95
As we can see, this is a two tailed test.
Getting the test statistic, as
X = sample mean = 2.997272727
uo = hypothesized mean = 2.95
n = sample size = 11
s = standard deviation = 0.110008264
Thus, t = (X - uo) * sqrt(n) / s = 1.425219281
Also, the p value is, as df = n - 1 = 11,
p = 0.184550102
As P > 0.01, we FAIL TO REJECT THE NULL HYPOTHESIS.
Thus, there is no significant evidence that the mean percentage of enrichment is not 2.95. [CONCLUSION]
*******************
B)
Note that
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.005
X = sample mean = 2.997777778
t(alpha/2) = critical t for the confidence interval = 3.355387331
s = sample standard deviation = 0.108717268
n = sample size = 9
df = n - 1 = 8
Thus,
Lower bound = 2.876181596
Upper bound = 3.119373959
Thus, the confidence interval is
( 2.876181596 , 3.119373959 )
As 2.95 is within this interval, then yes, it is okay. [ANSWER]
***********************
c)
We can use a normal distribution plot and see if it is linear.

