The actual yielding point of a sample of mild steel subjected to increasing stre

Question

The actual yielding point of a sample of mild steel subjected to increasing stress will usually differ from the theoretical yielding point. Let p denote the true proportion of samples that yield before their theoretical yielding point. If on the basis of a sample it can be concluded that more than 20% of all specimens yield before the theoretical point, the production process will have to be modified. (a) If 12 of 52 specimens yield before the theoretical point, what is the P-value when the appropriate test is used? (Round your answer to four decimal places.) P-value = What would you advise the company to do? Because the P-value is rather small, H_0 will be rejected at any reasonable alpha, so the production process will have to be modified. Because the P-value is rather small, H_0 will be rejected at any reasonable alpha, so no modification appears necessary. Because the P-value is rather large, H_0 would not be rejected at any reasonable alpha, so no modification appears necessary. Because the P-value is rather large, H_0 would not be rejected at any reasonable alpha, so the production process will have to be modified. (b) If the true percentage of "early yields" is actually 50% (so that the theoretical point is the median of the yield distribution) and a level 0.01 test is used, what is the probability that the company concludes a modification of the process is necessary? (Round your answer to four decimal places.)

Explanation / Answer

here p=0.2

std error =(p(1-p)/n)1/2 =0.0555

phat =12/52 =0.2308

hence test stat z=(phat-p)/std error =0.5547

p value =0.2896

becasue the p value is rather large , Ho would not be rejected...., so no modifications appears necessary

b)for true yield p=0.5

std error =0.0693

p value =0.3286

hence probabilty company conclude a modification is necessary =1-0.3286 =0.6714

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