ST 314 - Introduction to Statistics for Engineers Solutions to Sample Midterm for 2002 Problem 1. (15 points) The weight of a human joint replacement part is normally distributed with a mean of 2.00 ounces and a standard deviation of 0.05 ounces. (a) What is the probability that a randomly selected part weighs more than 2.10 ounces? Let Y = weight of a randomly selected part. Then P(Y > 2.10) = P( (Y-µ)/σ > (2.10-2.00)/.05)) = P(Z > 2) = 1 -.9772 =.0228. (b) Assuming that the standard deviation remains at 0.05 ounce, what must the mean weight be for the company to state that 99.9% of its parts are less than 2.10 ounces? We want P(Y < 2.10) =.999. But P(Y < 2.10) = P( (Y-µ)/σ < (2.10-µ)/.05)) = P(Z < (2.10-µ)/.05)) so from Table 1 we find: (2.10-µ)/.05) = 3.08 and solving for µ gives µ = 2.10 3.08(.05) = 1.946. (c) Now assuming that the mean weight remains at 2.00 ounces, what must the standard deviation be for the company to state that 95% of its parts are within 0.05 ounces of the mean? We want P(1.95 < Y < 2.05) =.95. But P(1.95 < Y < 2.05) = P((1.95-2.00)/σ < (Y-µ)/σ < (2.05-2.00)/σ) = P( -.05/σ < Z <.05/σ) =.95 so from Table 1 we find.05/σ = 1.96 and thus σ =.05/1.96 =.0255. Problem 2. (30 points) A random sample of 50 suspension helmets worn by motorcycle riders and automobile racing drivers was subjected to an impact test, and on 18 of these helmets some damage was observed. The manufacturer of the helmets claims that 30% of the helmets would show damage. Suppose you suspect that the percentage of damaged helmets is not this low. - 1 -
(a) Set up and conduct the appropriate hypothesis test of the manufacturer s claim. Use a significance level of 0.01, and clearly state all parts of your test (null and alternative hypotheses, test statistic, critical region, and conclusion). From the problem statement we get: n = 50, Y = 18, p^ =.36, q^ =.64, p 0 =.30, q 0 =.70 and α =.01 (1) Hypotheses are: H 0 : p =.30 H a : p >.30 (it is not as low as claimed) (2) Test statistic is: Z = (p^ - p 0 )/[p 0 q 0 /n] 1/2 (3) Upper one-tailed test, so we reject H 0 if Z > Z α = Z.01 = 2.326 (4) Calculating gives: Z = (.36 -.30)/[(.30)(.70)/50] 1/2 =.06/.0648 =.9258 (5) Since Z =.9258 is not greater than Z α = 2.326, we do not reject H 0. There is insufficient evidence to conclude that the proportion of defectives is higher than the claim, at the α =.01 significance level. (b) Find a 99% confidence interval for the true proportion of helmets of this type that would show damage from this impact test. Note that α =.01 so α/2 =.005 and Z α/2 = Z.005 = 2.576. CI is: p^ ± Z α/2 [p^ q^/n] 1/2 =.36 ± (2.576)[(.36)(.64)/50] =.36 ±.175 or [.185,.535 ] (c) Using the point estimate obtained from the preliminary sample of 50 helmets, how many helmets must be tested to be 99% confident that the error in estimating the true proportion of helmets of this type that would show damage from this impact test is less than 0.02? Using the point estimate p^ in place of p 0 in the formula gives: n p^ q^ [Z α/2 /B] 2 = (.36)(.64)[2.576/.02] 2 = 3822.30 or 3823. Problem 3. (20 points) Two catalysts are being analyzed to determine how they affect the mean yield of a chemical process. Specifically, catalyst 1 is currently in use, but catalyst 2 is acceptable. Since catalyst 2 is cheaper, it - 2 -
should be adopted, providing it does not change the process yield. A test is run in the pilot plant using samples of size 8 for each catalyst, and results in the following: Catalyst 1 Catalyst 2 ---------------------------------------------------------- Average 92.255 94.733 Variance 5.712 8.880 (a) You are to answer the question: is there any difference between the mean yields of the two catalysts? To do so, you must assume that the two populations are normally distributed and have equal variances. How comfortable are you in making the assumption that the two population variances are equal? Provide a reason for your answer. Rule of thumb says you should be comfortable with the assumption of equal variances if the ratios of: (larger sample variance)/(smaller sample variance) < 4. Here we have (8.880)/5.712) = 1.555 < 4, so I m comfortable with assuming the two population variances are equal. (b) Conduct the appropriate hypothesis test on the mean yields of the two catalysts. Use a significance level of 0.10, and clearly state all parts of your test (null and alternative hypotheses, test statistic, critical region, and conclusion). This is a test on the means from two independent samples. (1) Hypotheses are: H 0 : µ 1 - µ 2 = 0 H a : µ 1 - µ 2 0 (looking for any difference in the two means). (2) Test statistic is: t = ( y 1 - y 2 )/s p {1/n 1 + 1/n 2 } 1/2 (3) With α =.10 and n 1 + n 2 2 = 8 + 8 2 = 14, we find t 14,.05 = 1.761 so we reject H 0 if t > 1.761 since this is a two-tailed test. (4) Given that: y 1 = 92.255, s 1 2 = 5.712, y 2 = 94.733 and s 2 2 = 8.880. So s 2 p = [7(5.712) + 7(8.880)]/14 = 7.296 and s p = 2.70. Thus, t = [( y 1 - y 2 )]/s p {1/n 1 + 1/n 1 } 1/2 = [(92.255 94.733)]/(2.70){1/8 + 1/8} 1/2 = -2.478/1.35 = -1.836. (5) Since t = -1.836 = 1.836 is greater than t 14,.05 = 1.761, we reject H 0. There is sufficient evidence to conclude that the two mean yields differ at the α =.10 significance level. - 3 -
Problem 4. (35 points) Aircrew escape systems are powered by a solid propellant. The burning rate of this propellant is an important product characteristic. Specifications require that the target mean burning rate must be 50 cm/s. We know that the standard deviation of burning rate is 2 cm/s. An aerospace engineer selects a random sample of 25 propellant packages and measures their burning rates. From this sample an average burning rate of 50.9 cm/s is obtained. The engineer wants to determine, based on this sample, if the true mean burning rate has changed from the target. (a) Conduct a hypothesis test to determine whether the true mean burning rate has changed from the target. Use a 0.05 significance level for your test. Clearly state all parts of your test (null and alternative hypotheses, test statistic, critical region, and conclusion). From the problem statement, we know σ = 2, µ 0 = 50 and n = 25. With y = 50.9 we have the test: (1) Hypotheses are: H 0 : µ = 50 H a : µ 50 (looking for any change in the mean burning rate). (2) Test statistic is: Z = ( y - µ 0 )/(σ/ n). [Use Z statistic since σ is known.] (3) With α = 0.05 we find Z α/2 = Z.025 = 1.96 so we reject H 0 if Z > 1.96. (4) Calculating gives: Z = ( y - µ 0 )/(σ/ n) = (50.9 50)/(2/ 25) = 2.25. (5) Since z = 2.25 = 2.25 is greater than Z α/2 = 1.96, we reject H 0. At the α = 0.05 significance level, there is sufficient evidence to conclude that the true mean burning rate has changed from the target value of 50. (b) What is the p-value associated with your test statistic in part (a)? p-value = 2*P(Z > z 0 ) = 2*P(Z > 2.25) = 2(1 -.9878) = 2(.0122) =.0244. (c) Construct a 95% confidence interval for the true mean burning rate. Explain how this confidence interval can be used to test the hypothesis in part (a). The CI is: [ y ± z α/2 (σ/ n) ] = [ 50.9 ± 1.96 (2/ 25) ] = [ 50.9 ± 0.784 ] or [ 50.116, 51.684] Since the 95% confidence interval for µ does not contain the nominal value of µ 0 = 50, we would reject µ 0 = 50 as a plausible value for µ at the 95% confidence level. - 4 -
(d) Suppose we wanted the error in estimating the true mean burning rate of the rocket propellant to be less than + 0.5 with 95% confidence. What size of sample would be required? We want: z α/2 (σ/ n) B where α =.05, σ = 2, and B = 0.5. This gives:n [z α/2 *σ/b] 2 = [(1.96)(2)/(0.5)] 2 = (7.84) 2 = 61.46 so n = 62. - 5 -