atrigM. Let denote the true average retectometer readng for a new type of paint
ID: 3356603 • Letter: A
Question
atrigM. Let denote the true average retectometer readng for a new type of paint under consideration. Atest of mc -20venus H. ample of sine n from a normal population distribution. what conclusion is 0.05 d appropriate in each of the following situatiens? (Round your P-values to three decimal places) (a) n.19,t-3.3, State the cenclusion in the problem context Do not reject the nul hypohesis. There is "m iere evidence to condude that the new paint has a renectometer O Reject the null hypotheis. There is not sufficient a Reject the nul hypothesis. There is suficient evidence to conclude that the new paint has a reflectometer Do not reject the nul hypothesis. There is not sumcient evidence to condude that the new part has a renectorneter evidence to conclude that the new paint has a reflectometer reading higher than 20. reading higher than 20 (b) n-9, P-value s 1.6, . : 0.01 state the conclusion in th problem centert. O Reject the nall hypothesis. There is suficient evidence to conclude that the new paint has a refectometer reading higher than 20 0 Reject the nul hypothesis. There is not sufficient evidence to cenclude that the new paint has a refleameter reading higher than 20 Do not reject the null hypothesis. There is sufficlent evidence to conclude that the new paint has a reflectometer reading higher than 20. Donot reject the nu. hypothe."Ther-inot iumdent evidence te conclud·th, the new pare han a renectoneter eeaing Ngher than 20. (c n-21-0 valueExplanation / Answer
From the statistical software output of probabilities corresponding to t statistic:
a) Degrees of freedom = n - 1 = 19 - 1 = 18
This is a right tailed test since the alternative hypothesis contains > sign. Hence,
p - value = P(t > 3.3) = 0.002
b) p - value = P(t > 1.6) = 0.074
Do not reject; not sufficient evidence; Option D
c) p - Value = 0.754
Do not reject; not sufficient evidence
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