STAT 400 Lecture AL1 Examples for 8.1 (Part 1) Spring 2015 Dalpiaz 0. A car manufacturer claims that, when driven at a speed of 50 miles per hour on a highway, the mileage of a certain model follows a normal distribution with mean 0 = 30 miles per gallon and standard deviation = 4 miles per gallon. A consumer advocate thinks that the manufacturer is overestimating average mileage. The advocate decides to test the null hypothesis H 0 : = 30 against the alternative hypothesis H 1 : 30. a) Suppose the actual overall average mileage is indeed 30 miles per gallon. What is the probability that the sample mean is 29.4 miles per gallon or less, for a random sample of n = 25 cars? P( X 29.4 ) = 254304.29ZP = P( Z 0.75 ) = ( 0.75 ) = 0.2266. b) A random sample of 25 cars yields x = 29.4 miles per gallon. Based on the answer for part (a), is there a reason to believe that the actual overall average mileage is not 30 miles per gallon? If = 30, it is not unusual to see the values of the sample mean x at 29.4 miles per gallon or even lower. It does not imply that = 30, but we have no reason to doubt the manufacturer’s claim. c) Suppose the actual overall average mileage is indeed 30 miles per gallon. What is the probability that the sample mean is 28 miles per gallon or less, for a random sample of n = 25 cars? P( X 28 ) = 2543028ZP = P( Z 2.50 ) = ( 2.50 ) = 0.0062. d) A random sample of 25 cars yields x = 28 miles per gallon. Based on the answer for part (c), is there a reason to believe that the actual overall average mileage is not 30 miles per gallon? If = 30, it is very unusual to see the values of the sample mean x at 28 miles per gallon or lower. It does not imply that < 30, but we have a very good reason to doubt the manufacturer’s claim.e) Suppose the consumer advocate tests a sample of n = 25 cars. What is the significance level associated with the rejection region “Reject H 0 if x < 28.6” ? = significance level = P( Type I Error ) = P( Reject H 0 | H 0 true ). Need P( X 28.6 | = 30 ) = ? .n ZX P( X < 28.6 | = 30 ) = 25 4306.28ZP = P( Z < 1.75 ) = ( 1.75 ) = 0.0401. f) Suppose the consumer advocate tests a sample of n = 25 cars. Find the rejection region with the significance level = 0.05. n = 25. = 0.05. Rejection Region: Reject H 0 if Z = n σμ0X < – z . Z = 25 430X < – 1.645. 254645.130X = 28.684. g) Suppose that the sample mean is x = 29 miles per gallon for a sample of n = 25 cars. Find the p-value of the appropriate test. H 0 : 30 vs. H 1 : 30. Left – tailed. Z = 2543029X σμ0n = – 1.25. P( Z 1.25 ) = ( 1.25 ) = 0.1056.h) State your decision ( Accept H 0 or Reject H 0 ) for the significance level = 0.05. P-value > Do NOTReject H 0 P-value < Reject H 0 Since 0.1056 > 0.05, Do NOT Reject H 0 at = 0.05. i) Construct a 95% confidence interval for the overall average miles-per-gallon rating for this model, . = 4 is known. n = 25. The confidence interval : nσz2αX . 95% confidence level, = 0.05, /2 = 0.025, 2αz = 1.96. 25496.129 29 1.568 ( 27.432 ; 30.568 ) j) What is the minimum sample size required if we want to estimate to within 0.5 miles per gallon with 95% confidence? = 0.5, = 4, 95% confidence level, = 0.05, /2 = 0.025, 2αz = 1.96. 222 5.0496.1αεσzn= 245.8624. Round up. n = 246. k) Construct a 95% confidence upper bound for . The confidence upper bound for : nσzX . 95% confidence level, = 0.05, z = 1.645. 254645.129 29 + 1.316 ( 0 ; 30.316 )1. The overall standard deviation of the diameters of the ball bearings is = 0.005 mm. The overall mean diameter of the ball bearings must be 4.300 mm. A sample of 81 ball bearings had a sample mean diameter of 4.299 mm. Is there a reason to believe that the actual overall mean diameter of the ball bearings is not 4.300 mm? a) Perform the appropriate test using a 10% level of significance. Claim: 4.300 H 0 : = 4.300 vs. H 1 : 4.300 Test Statistic: is known Z = 81005030042994X 0μ...n = – 1.80. Rejection Region: 2 – tailed. Reject H 0 if Z < – z /2 or Z > z /2 = 0.10 2 = 0.05. z 0.05 = 1.645. Reject H 0 if Z < – 1.645 or Z > 1.645. Decision: The value of the test statistic does fall into the Rejection Region. Reject H 0 at = 0.10. OR P-value: p-value = P( Z – 1.80 ) + P( Z 1.80 ) = 0.0359 + 0.0359 = 0.0718. Decision: 0.0718 < 0.10. P-value < . Reject H 0 at = 0.10.OR Confidence Interval: is known. The confidence interval : n2zX. 90% conf. level. = 0.10 2 = 0.05. z 0.05 = 1.645. 81005064512994 ... 4.299 0.0009138889 Decision: 90% confidence interval for does not cover 4.300. Reject H 0 at = 0.10. Two-tailed test same Confidence Interval Accept H 0 Covers 0 Reject H 0 Does not cover 0 b) State your decision (Accept H 0 or Reject H 0 ) for the significance level = 0.05. 0.0718 > 0.05. P-value > . Do NOT Reject H 0 ( Accept H 0 ) at = 0.05.2. A trucking firm believes that its mean weekly loss due to damaged shipments is at most $1800. Half a year (26 weeks) of operation shows a sample mean weekly loss of $1921.54 with a sample standard deviation of $249.39. a) Perform the appropriate test. Use the significance level = 0.10. Claim: 1800 H 0 : 1800 vs. H 1 : > 1800 Test Statistic: is unknown T = 26 39.249180054.1921 X sμ0n = 2.485. Rejection Region: Right – tailed. Reject H 0 if T > t = 0.10. n – 1
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