Question
5.28 parts A&D. When asked for evidence, I am supposed to perform a hypothesis test. There is one additional question that is...
If the R&D department used alpha= .10 in place of .01, would the conclusion about whether the additive reduced the mean absorption change from the conclusion using alpha = .01
b. what is the probability of a Type II error if the actual mean concentration is sm c. Do the data appear to have a normal distribution? d. Based on your answer in (c). is the sample size large enough for the test pr dures be valid? Explain. The Level of significance of a Statistical Test 5.28 The R&D; department of a paint company has developed an additive that it hopes will in crease the ability of the company's stain for outdoor decks to resist water absorption. The 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 336 and s 9.2 a. Is there substantial evidence (a i01 that the additive reduces the mean ab- sorption from its current value? b. What is the level of significance (p.value)of your test results? c. What is the probability of a Type II error if the stain with the additive in fact has a mean absorption rate of 30? d. Estimate the mean absorption using a 99% confidence interval. Is the confidence interval consistent with your conclusions from the test of hypotheses?
Explanation / Answer
a)
H0>=35
H1<35
tSTAT=X-µ/S/n
=33.6-35/9.2/50
=-9.90/9.2
=-1.076
tCRIT=-2.4049
Since, tCRIT<tSTAT, we cannot reject the null hypothesis. With 0.10, tCRIT is -3.5- cannot reject the null hypothesis.
d)
LCL=33.6-(2.4049)(9.2)/V50=33.6-3.13=30.47
UCL=33.6+(2.4049)(9.2)/V50=33.6+3.13=36.73
Since this interval consists the hypothesised mean, we cannot reject the null hypothesis.
