Assignment 10 Solutions

Assignment 10 Solutions - CE93 Engineering Data Analysis...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CE93 - Engineering Data Analysis Prof. Joan Walker, Spring 2011 Solutions to Assignment 10 1. Average speed of vehicles on a highway a. Observations, n = 50. confidence interval for the mean speed, . . Determine two-sided 99% Because n is large, the distribution of For 99% confidence interval, Confidence interval for b. What value of n gives a 99% confidence interval such that we know within ±1. That is, what n gives a confidence interval (65 – 1 , 65 + 1) = (64, 66). However, n must be integer, so we take the first integer value larger than the number found above. Therefore, n = 240. The additional vehicles observed is n – 50 = 240 – 50 = 190. c. For n=n1=n2=100, the respective sample means are normally distributed: . The difference in sample means are distributed normal as well: CE93 - Engineering Data Analysis Prof. Joan Walker, Spring 2011 Where 2. 30 observations (n = 30), mean = 22 min ( , sample variance = 4.2 2 min ( We want a two-sided 99% confidence interval for the mean, For a sample with large n, the sample mean is normally distributed. When the population variance is unknown, the following statistic is follows a t-distribution with n-1 degrees of freedom: Confidence interval for : CE93 - Engineering Data Analysis Prof. Joan Walker, Spring 2011 3. Bulb lifetime – population is normally distributed. a. An estimate for the population variance is the sample variance, b. Two-sided, 99% confidence interval for For 99% confidence interval, : . The following statistic follows a chi-square distribution with n-1 degrees of freedom: Two-sided confidence interval for : Confidence interval for c. For 90% confidence that 4. Binomial trails with successes 6, 5, 9, and 7. Let’s denote the number of experiments as N = 4. The set of independent trails represents a binomial experiment with trials n and success probability p. CE93 - Engineering Data Analysis Prof. Joan Walker, Spring 2011 a. For n = 12, determine the maximum likelihood for the success probability, p. For a binomial random variable with n trails, and success probability p, its probability mass function (for x successes) is given by: The likelihood function is simply the product of the pmf for each separate set of trials. Taking the logarithm gives us the log likelihood: The log likelihood is differentiable with respect to p, so let’s do that and set it equal to zero. Solving for p: For this problem, For n = 12, we get the result that To ensure that this is a maximum, we can check the second order condition. Taking another derivative with respect to p (Not required to get full credit): CE93 - Engineering Data Analysis Plugging in , we get Prof. Joan Walker, Spring 2011 , so our estimate for is a maximum. b. In part (a) we found the maximum likelihood estimate to be . However, our choice of probability is now constrained to be either 0.5 or 0.75. To determine which probability is more likely, we can calculate the likelihood (or log likelihood) of each one, and see which is larger. The probability that produces a larger likelihood is the maximum likelihood estimate. For p = 0.5, For p = 0.75, The likelihood for p = 0.6 is larger, so our maximum likelihood estimate is c. Now we know that the true probability of success is p = 0.8. We need to determine a maximum likelihood estimate for the number of trials, n. The likelihood function is the same as before This likelihood is not differentiable with respect to n, however. The combinations term includes factorials which are not easily differentiable. But we know certain constraints for n that will help us find a solution. First of all, n must be a positive integer, because it is representing the number of discrete trials. Secondly, n must be at least as large as the largest number of successes found. In this case, the largest number of successes is 9. CE93 - Engineering Data Analysis Prof. Joan Walker, Spring 2011 We will start out with n = 9, and increase n until we find a maximum value of the likelihood function. For n = 9: For n = 10: For n = 11: The likelihood is decreasing as we increase n. The largest value is found when n = 9, so our maximum likelihood estimate for n is 5. Part a: Upper 1-sided 95% confidence interval: Lower 1-sided 95% confidence interval: CE93 - Engineering Data Analysis Prof. Joan Walker, Spring 2011 Part b: 2-sided 95% confidence interval: 6. Two-sided hypothesis test: H0:µ = 8.2 vs. H1:µ ≠ 8.2 -3.3204 (use the normal distribution because variance is known) p-value = 2 [1 − φ(3.3204)] = .0010 This is greater than both α=0.1 and α=0.05, so we can reject H0 at both levels of significance. 7. Two-sided hypothesis test: H0:µ = 30 vs. H1:µ ≠ 30 1.847 (use the t-distribution because variance is unknown) p-value = P{T15 < −1.847} = .04 So we cannot reject H0 at a 95% level of significance. Note: We do not necessarily have to perform a two-sided test. If we conducted one-sided t-test at a 95% level of significance, we would reject H0. In this case, we only care about the sheets not meeting the stress resistance specification, but we don’t really care if they exceed the standard. In addition, , so it we have reason to suspect that the sheets do not meet the specification. In this case, it would be justified to use a one-sided test, which has the stricter rejection criterion. CE93 - Engineering Data Analysis Prof. Joan Walker, Spring 2011 8. Two-sided hypothesis test: Denote X: pH level of Solution A; Y: pH level of Solution B. X and Y n=10, m=10, and Two-sample problems with known variances 1- two-sided confidence interval for is 5% level of significance: 95% confidence interval is = Define a random variable With , and 95% two-sided confidence interval for is The hypothesis test is set up as vs which can also be treated as vs (a) Method 1: is in the 95% confidence interval of Method 2: , we do not reject H0. CE93 - Engineering Data Analysis Prof. Joan Walker, Spring 2011 So we fail to reject H0, and we conclude that the data support the claim. (b) p_value = 2*p(Z<TS)=2* 0.4209 (p_value is the smallest level that we can reject H0. Since p_value is larger than the significance level 0.05, we fail to reject H0.) ...
View Full Document

This note was uploaded on 06/02/2011 for the course CIV ENG 93 taught by Professor Staff during the Spring '08 term at Berkeley.

Ask a homework question - tutors are online