The R&D; department of a paint company has developed an additive that it hopes w
ID: 3177589 • Letter: T
Question
The R&D; department of a paint company has developed an additive that it hopes will increase the ability of the company's stain for outdoor decks to resist water absorption. I he current formulation of the stain has a mean absorption rate of 35 units. Before changing the stain, a study was designed to evaluate whether the mean absorption rate of the stain with the additive was decreased from the current rate of 35 units. The stain with the additive was applied to 50 pieces of decking material. The resulting data were summarized to y - 33.6 and s = 9.2 Is there substantial evidence (alpha = .01) that the additive reduces the mean absorption from its current value? What is the level of significance (p-value) of your test results? What is the probability of a Type II error if the stain with the additive in fact has a mean absorption rate of 30? Estimate the mean absorption using a 99% confidence interval. Is the confidence interval consistent with your conclusions from the test of hypotheses?Explanation / Answer
We are to test the null hypothesis H0: Mu=35 versus the alternative H1:Mu<35
a) Test statitsic t=(y_bar-Mu)/(s/sqrt(n)) = (33.6-35)/(9.2/sqrt(50)) = -1.08
Degree of freedom =n-1=50-1=49
Left tailed Critical value at 0.01 level is tcritical=-2.4049 using excel fucntion =TINV(0.02,49) Here 0.02 is used because for one tail test excel require to double signifcance level.
As caluclted t=-1.08>-2.4049 , we do not reject the nul hypothesis at 0.01 level.Hence, we conclude that the additive does not reuce the mean absorption rate below 35 units.
b) P-value=0.1427 using excel fucntion =TDIST(1.08,49,1)
c) P(Type-II error )=P(Accept null /Mu=30)
=P(y_bar>=35/Mu=30)
=P(t>(35-30)/(9.2/sqrt(50)) )
=P(t>3.8430)
=0.00018 using excel function =TDIST(A20,49,1)
d) 99% confidence interval Lower Limit = y_bar- t(0.01,49,two tailed) *s/sqrt(n)
=33.6- 2.6800*9.2/sqrt(50) using excel fuction =TINV(0.01,49)
=30.1131
99% confidence interval Upper Limit = y_bar+t(0.01,49,two tailed) *s/sqrt(n)
=33.6+2.6800*9.2/sqrt(50) using excel fuction =TINV(0.01,49)
=37.0869
As the confidence interval includes 35, so mean is not different from 35 and hence confidence interval is consistent with hypothesis test
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