An artist experimenting with clay to create pottery with a special texture has b

Question

An artist experimenting with clay to create pottery with a special texture has been expenencing 50. with these special ieces About 40% breakin the kin dunng firing Hoping to solve this problem, she buys s re more expensive clay from another supplier. She plans to maké and fire 10 pieces and will decide to use the new clay at most one of them breaks. a) Suppose the new, expensive clay really is no better than her usual clay. What's the probability that this test convinces her to use it anyway? (Hint: Use a Binomial model) b) If she decides to switch to the new clay and it is no better, what kind of error did she commit? c) f the ne clay really can reduce breakage to only 20%, what's the bability that her test will not detect the improvement? d) How can she improve the power of her test? Offer at least two suggestions Page 532 and P Inc, or its affiliate(s). All rights reserved. Privacy Policy Terms of Use I SKMBT C28017

Explanation / Answer

Question 50

Here by the current method, there are 40% of the kiln break during firing.

Pr(kiln breaking by previous clay texture) = 0.4

(a) Now, her employed new method and check it on 10 pots. Her will accept ir if only at max 1 pot will break. As we know that the new method is same as earlier method.

so Pr(kiln breaking with clay texture) = 0.4

so he will accept the new clay = Pr(X 1) = BIN (X 1; 10; 0.4) = 10C0 (0.4)0(0.6)10 + 10C1 (0.4)1(0.60)9 = 0.0464

(b) If he switches to new clay despit it has similar probability of kiln breaking. Then he will commit TYPE I error. As he rejected the null hypothesis p = 0.04 despit it being correct.

(c) Now, With the new clay. Let say the probabiliy of kiln failure = 0.2

so Now, we have to find the probability of number of kiln failure that would more than 1.

Pr(X > 1) = BIN (X > 1; 10 ; 0.2) = 1 - BIN (X 1; 10 ; 0.2)

BIN (X 1; 10 ; 0.2) = Pr (X = 0 ; 10 ; 0.2) + Pr( X =1 ; 10 ; 0.2) = 10 C0 (0.2)0(0.8)10 + 10C1 (0.2)(0.8)9

= 0.1074 + 0.2684 = 0.3758

Pr(X > 1) = 1 - 0.3758 = 0.6242

so probability that test will not detect the experiment = 0.6242

Here it is type II error.

(d) She can improve the power of the test.

(i) she should increase the sample size.

(ii) she should develop a new critieria to check the testng method. LIke she should make it atmost till the half of the expected value which would be np/2 where n is the sample size.

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