Five samples of a ferrous-type substance are to be used to determine if there is
ID: 3055067 • Letter: F
Question
Five samples of a ferrous-type substance are to be used to determine if there is a difference between a laboratory chemical analysis and an X-ray fluores- cence analysis of the iron content. Each sample was split into two sub-samples and the two types of analysis were applied. Following are the data showing the iron content (in percent). PLEASE USE THE P-VALUE METHOD
Analysis Chemical X Ray
(SORRY FOR THE CONFUSING TABLE)
Do these data indicate that the two methods of analysis give, on average, a different result? Assume the relevant population distribution(s) is/are normal.
(a) Define the population parameter(s) of interest.
(b) State the null and alternative hypotheses in terms of the parameter.
(c) State the test statistic you will use. What distribution (including degrees of freedom, if appropriate) will you use to calculate the p-value.
(d) Find the observed value of the test statistic.
(e) Compute (or bracket) the p-value within the accuracy of the tables.
(f) What level of evidence against H 0 do you find?
1 2.2 2.0 2 1.9 2.0 3 2.5 2.3 4 2.3 2.1 5 2.4 2.4Explanation / Answer
a)
Population paramenter of interests is the difference in the means of two analysis.
b)
Null Hypothesis, H0: ?1 - ?2 = 0
Alternate Hypothesis, Ha: ?1 - ?2 ? 5
c)
t-statistic, t-distribution
d)
Test Statistics
t = [ (x1 - x2) - d ] / SE
t = 0.7625
e)
p-value = 0.4676
f)
As p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis.
There are not significant evidence to conclude that means of two analysis are different.
x1(bar) 2.26 x2(bar) 2.16 s1 0.23 s2 0.18 n1 5 n2 5 SE = sqrt[ (s12/n1) + (s22/n2) ] (s12/n1) 0.0106 (s22/n2) 0.0066 SE 0.1311 DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] } [ (s12 / n1)2 / (n1 - 1) ] 0.000 [ (s22 / n2)2 / (n2 - 1) ] 0.00 (s12/n1 + s22/n2)2 0.00 DF = 8Related Questions
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