10 23 98 252z9851 4 A consumer advocacy group compares the proportions of automobiles produced by two different manufacturers that need major repairs in the first three years of ownership The results are as below Manufacturer 1 Manufacturer 2 x1 45 x2 27 n1 1000 n2 1000 a Do a 2 sided 99 confidence interval for the proportion produced by manufacturer 1 that need major repairs 2 b Do a 2 sided 91 confidence interval for the proportion produced by manufacturer 1 that need major repairs 2 c An automobile columnist states that over 5 of automobiles produced by the first manufacturer need repairs in the first three years State the null and alternative hypotheses to be tested in this problem 1 d Do a test of the hypothesis in c at a 95 confidence level 3 e Repeat d at the 91 confidence level 2 f If the true proportion that need repairs is 6 what is the power of the test 3 g If I claim that the proportion needing repairs is 6 on how large a sample would this have to be based to be accurate within 0 5 2 Solution From page 10 of the Syllabus Supplement Interval for Confidence Hypotheses Interval Proportion p p z 2 s p pq n q 1 p sp a p Test Ratio H 0 p p0 H 1 p p0 z Critical Value p p0 p p cv p 0 z 2 p p x1 45 045 and q 1 p 955 n1 1000 sp pq n 045 955 1000 p0 q 0 n 00655553 p p z 005 s p 045 2 576 00655 045 017 b From page 1 z 045 1 70 So p p z 005 s p 045 1 70 00655 045 011 c H 0 p 05 H1 p 05 p0 q0 05 95 00689 z z 05 1 645 n 1000 p p 0 045 05 0 726 This is below 1 645 so accept H 0 i Test Ratio z p 00689 d p or ii Critical Value p cv p 0 z p 05 1 645 00689 06134 p 045 is below this so accept H 0 or iii Confidence Interval p p z 05 s p 045 1 645 00655 0342 This does not contradict H 0 e We want a point z 09 so that P z z 09 09 From the diagram P 0 z z 09 41 The closest we can come is P 0 z 1 34 4099 So z 09 134 Redo one of the three ways of doing d with 1 645 replaced by 1 34 10 23 98 252z9851 f The power of a test is the probability of rejecting the null hypothesis when it is false It is the opposite of the probability of wrongly accepting the null hypothesis According to d above if 05 we will accept H 0 if p is less than or equal to 06134 So p 06 06134 06 cv power 1 P z 1 P z 1 P z 0 18 1 5 0714 06 94 p 1000 1 5714 4286 g From the outline if we assume a 95 confidence level so that n pqz 2 e2 06 94 1 96 2 005 2 z 2 1 960 8667 5 A consumer advocacy group compares the proportions of automobiles producer by two different manufacturers that need major repairs in the first three years of ownership The results are as below Manufacturer 1 Manufacturer 2 x1 45 x2 27 n1 1000 n2 1000 a Test the hypothesis that the proportions are equal at the 90 level 5 b Do a confidence interval for the difference at the 95 level 4 c Test the hypothesis that the proportion that need repairs is greater for the first manufacturer than the second at the 90 level 2 d Find p values for your results in a and c 2 Those who do not learn from the past are doomed to repeat it 2 10 23 98 252z9851 6 A city sets a requirement that a sewer pipe must have a mean breaking strength exceeding 2500 pounds per linear foot A test of 200 sections of sewer pipe yields an average strength of 2530 pounds per linear foot and a standard deviation of 200 Assume a confidence level of 95 a State null and alternative hypotheses for this problem and a critical value for the sample mean 2 b Given the sample results above do you accept or reject the null hypotheses 1 c What is the power of the test if the true mean is 2535 pounds per square inch 3 d Do a power curve for this test 4 e If we claim that a population has a Poisson distribution with a mean of 9 and the actual value found is 15 do a hypothesis test at the 95 level 3 f Find the lowest value above 9 and the highest value below 9 that would lead to a rejection of the null hypothesis in f Using these find the power of the test if the mean is actually 12 5 5 Solution Interval for Mean known a Confidence Interval Hypotheses x z 2 x H 0 0 H1 0 H 0 2500 H1 2500 05 z 05 Test Ratio z Critical Value x 0 x x cv 0 z 2 x 1 645 n 200 x 2530 and s 200 Since the above can be used for large samples replace x with sx s n 200 200 14 142 If we use the critical value method xcv 0 z s x 2500 1 645 14 142 2523 26 b We accept H 0 if x 2523 26 so reject H 0 We accept H 0 if x 2523 26 so we need the probability that x is less than 2523 26 when the true mean is 2535 2523 2535 P x 2523 2535 P z P z 0 84 5 2995 2005 This 14 142 is the probability of a type II error so that the power is 1 2005 7995 d Try the following points At 1 2500 the power is 5 the significance level c 2523 2511 At 1 2511 the power is 1 P x 2523 2511 1 P z 14 142 1 P z 0 84 1 5 2995 2005 At 1 2523 the critical value the power is 50 At 1 2535 the power is 7995 2523 2546 At 1 2546 the power is 1 P x 2523 2546 1 P z 14 142 1 P z 1 63 1 5 4484 9484 3 10 23 98 252z9851 e f H 0 m 9 H 1 m 9 From the Poisson table for a mean of 9 p value 2 P x 15 2 1 P x 14 2 1 95853 08824 Since this is above 05 accept H 0 Since the number in e is pretty close to 05 try 16 Look for probabilities above 97 5 Also try numbers close to zero Look for probabilities below 2 5 For 16 p value 2 P x 16 2 1 P …
View Full Document
Unlocking...