Statistics for Engineering - Quiz, Solutions Homework # 3 | STAT 4706, Quizzes of Statistics

Download Statistics for Engineering - Quiz, Solutions Homework # 3 | STAT 4706 and more Quizzes Statistics in PDF only on Docsity! STAT-4706 Solutions Homework # 3 Fall 2012 You must show all of your work for each problem in order to receive full credit. HW is Due 09/27/2012. 1. (16)A manufacturer of MP3 players conducts a set of comprehensive tests on the electrical functions of its products. All MP3 players must pass all tests prior to being sold. Of a random sample of 500MP3 players, 15 failed one or more tests. Find a 90% CI for the proportion of MP3 players from the population that passes all tests. n= 500 p̂= 485 500 =0.97 p̂= (1− p̂ )=0.03 Z α 2 =Z0.05=1.645  PE ± CV*SE =  0.97 ± 1.645√ 0.97∗0.03500 =0.97±0.013  A 90% CI for the population proportion of MP3 player that passes the test is 0.957 < p < 0.983 2. (14) A study is to be made to estimate the percentage of citizens in a town who favor having their water fluoridated. How large a sample is need it if one wishes to be at least 95% confident that the estimate is within 1% of the true percentage? n= Zα /2 2 p̂ q̂ E2 = (1.96)2 (4 ) (0.01 ) 2=9604 (round up) NOTE: Remember if p̂is not given then use p̂=0.5 (most conservative value) 3. (18) A survey of 1000 students found that 274 chose professional baseball team A as their favorite team. In a similar survey involving 760 students, 240 of them chose team A as their favorite. Compute a 95% CI for the difference between the proportions of students favoring team A in the two surveys. Is there a significant difference? nS1=1000 nS2=760 p̂S1= 274 1000 =0.274 p̂S 2= 240 760 =0.3158 Z α 2 =Z0.025=1. 96  PE ± CV*SE =  (0.274−0.3158 )± (1.96 ) √ (0.274 ) (0.726 ) 1000 + (0.3158 ) (0.684 ) 760 =−0.0418±0.0431  The 95% CI for the difference in favorite team is -0.0849< pS1- pS2 < 0.0013. Since the interval contains zero the significance cannot be shown. 1 STAT-4706 Solutions Homework # 3 Fall 2012 4. (16) An electrical firm manufactures light bulbs that have a lifetime that is approximately normally distributed with a mean of 800 hrs. and a standard deviation of 40 hrs. Test the hypothesis that µ = 800 hours against the alternative, µ ≠ 800, if a random sample of 30 bulbs has an average life of 788 hrs. Use a P-value in your answer. Hypotheses: H0: µ = 800 vs. H1: µ ≠ 800 Test: Zobs = X́−μ σ √n = 788−800 40 √30 =−1.64 p-value = 2*P(Z< -1.64) = 2*(0.0505) = 0.1010. Conclusion: Since the p-value is NOT less than α = 0.05 , we conclude that the mean is not significantly different from 800 hours. 5. (16) Test the hypothesis that the average content of containers of a particular lubricant is 10 liters if the contents of a random sample of 10 containers are 10.2, 9.7, 10.1, 10.3, 10.1, 9.8, 9.9, 10.4, 10.3, and 9.8, liters. Use a 0.01 level of significance and assume that the distribution of contents is normal. From the sample: X́=10.06 S = 0.24 Hypotheses: H0: µ = 10 vs. H1: µ ≠ 10 α = 0.01 df = n-1 = 9 Critical Region: t0.005(9)<-3.25 or t0.005(9) > 3.25. Test: Tobs = X́−μ S √n = 10.6−10 0.244 √10 =0.77 Conclusion: Since Tobs < t0.005(9)=3.25, we fail to reject H0 There is not enough evidence to indicate that the average content of lubricant containers is different than 10 liters. 6. (20) A UCLA researcher claims that the average life span of mice can be extendend by as much 8 months when the calories in their diet are reduced by approximately 40% from the time they are weaned. The restricted diets are enriched to normal levels by vitamins and proteins. Suppose that a random sample of size 10 mice is fed a normal diet and has an average life span of 32.1 months, with a standard 2