cistG

2021-09-18

The Level of Significance of a Statistical Test The  and $D$ 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. 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 $\stackrel{―}{x}=33.6$ and $s=9.2$
a) Is there substantial evidence $\left(\alpha =0.01\right)$ that the additive reduces the mean absorption 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?

Bentley Leach

Step 1
a) As we are testing here whether the mean has decreased from the mean level of 35, therefore the null and the alternative hypothesis here are given as:
${H}_{0}:\mu =35$
${H}_{a}:\mu <35$
The test statistic here is computed as:
${t}^{\cdot }=\frac{\stackrel{―}{X}-{\mu }_{0}}{\frac{s}{\sqrt{n}}=\frac{33.6-35}{\frac{9.2}{\sqrt{50}}=-1.08}}$
For $n-1=49$ degrees of freedom, as this is a one tailed test, the p-value here is obtained from t distribution tables as :
$p=P\left({t}_{49}<-1.08\right)=0.1436>0.01$ which is the level of significance.
As the p-value here is high, we cannot reject the null hypothesis here and therefore we conclude here that the mean is not less than 35 units here.
b) The p-value above obtained is 0.1436.
c) For 49 degrees of freedom, we have from t distribution tables:
$P\left({t}_{49}<-2.405\right)=0.01$
Therefore the critical mean value here is computed as:
$\stackrel{―}{X}={\mu }_{0}-2.405×\frac{s}{\sqrt{n}}=35-2.405×\frac{9.2}{\sqrt{50}}=31.8709$
The probability of type II error for a true mean of 30 is computed here as:
$P\left(\stackrel{―}{X}>31.8709\right)$
Converting it into a t distribution variable, we have here:
$P\left(t>\frac{31.8709-30}{\frac{9}{\sqrt{50}}}\right)$
$P\left(t>1.44\right)$
Getting it from the t distribution tables, we have here:
$P\left(t>1.44\right)=0.0781$
Therefore 0.0781 is the required probability here.
d) for the critical value obtained in previous part, the 99% confidence interval here is obtained as:
$\stackrel{―}{y}±t×\frac{s}{\sqrt{n}}$
$33.6±-1.299×\frac{9.2}{\sqrt{50}}$
$33.6±1.69$

As the confidence interval here contains 35, therefore the test is not significant here and we cannot conclude here that the mean is less than 35.

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