Inference and Regression

Transcription

1 Inference and Regression Assignment 4 - Solutions Department of IOMS Professor William Greene Phone: Office: KMC 7-90 Home page: wgreene@stern.nyu.edu Course web page: 1. The following 16 numbers came from normal random number generator on a computer: a. What would you guess the mean and variance (µ and σ 2 ) of the generating normal distribution were? b. Give 90%, 95%, and 99% confidence intervals for µ. c. How much larger a sample do you think you would need to halve the length of the confidence interval for µ? a. The sample mean is The sample variance is b. Confidence intervals (based on the t distribution with 15 degrees of freedom). The standard error of the mean is /sqr() where =16. The 90% confidence interval is / ( /4) = The 95% confidence interval is / ( /4) = The 90% confidence interval is / ( /4) = c. Confidence intervals for σ. The confidence intervals for σ 2 are from (-1)s 2 /χ 2 < σ 2 < (-1)s 2 /χ 2 Where the terminal chi squareds are for (1-α/2) and α/2. =16, s 2 = The critical chi squared values are For 1%, (4.60,32.80), 5%, (6.26,27.49), 10%(7.26,25.00) The intervals for σ 2 are as follows: s = The 99% confidence interval 15(s 2 )/32.80 to 15(s 2 )/4.60 = The 95% confidence interval 15(s 2 )/27.49 to 15(s 2 )/6.26 = The 90% confidence interval 15(s 2 )/25.00 to 15(s 2 )/7.26 = Use the square roots for the interval for σ. ( ), ( ), ( ). d. The length of the confidence interval is 2(t*)s/4 where s is the sample standard deviation and 4 is the square root of the sample size. To halve the length of the interval, that 4 must become an 8, so must go from 16 to 64 up by a factor of Suppose that X 1, X 2,..., X n are i.i.d. (µ, σ 2 ), where µ and σ are unknown. How should the constant c be chosen so that the interval (, X + c) is a 95% confidence interval for µ; that is, c should be chosen so that P( < μ X + c) =.95.

2 c should be chosen so that the probability to the left of c is.95. That would be standard deviations above x-bar. So that the probability to the right of c is A coin is thrown independently 10 times to test the hypothesis that the probability of heads is 1/2 versus the alternative that the probability is not 1/2. The test rejects if either 0 or 10 heads are observed. a. What is the significance level of the test? b. If in fact the probability of heads is.1, what is the power of the test? a. The probability of a type 1 error is the Prob(X=0 or 10 p=.5) = 2*1/1024 = 1/512. b. The power of the test is the probability it will reject p=.5 given that p really is.1. This is the probability of 0 p=.1 + probability of 10 p=.1 = = True or false, and state why: a. The significance level of a statistical test is equal to the probability that the null hypothesis is true. b. If the significance level of a test is decreased, the power would be expected to increase. c. If a hypothesis is rejected at the significance level α, the probability that the null hypothesis is true equals α. d. The probability that the null hypothesis is falsely rejected is equal to the power of the test. e. A type I error occurs when the test statistic falls in the rejection region of the test. f. A type II error is more serious than a type I error. g. The power of a test is determined by the null distribution of the test statistic. h. The likelihood ratio is a random variable. a. False. The significance level is the probability that the test will reject the null hypothesis when it is true. b. If the significance level of a test is decreased, the power would be expected to increase. The significance level, α is the probability that the null hypothesis will be rejected when it is true. The power of the test will be the probability that the hypothesis is rejected when it is false. The power of the test is a function of the alternative hypothesis. In general, decreasing the significance level of a test makes it harder to reject the hypothesis. This should operate similarly on the power. Take an example. Suppose the hypothesis is that the mean of a normal distribution is zero. If the significance level is.05, the test is rejected if the sample mean is less than standard errors or greater than standard errors. Suppose the specific alternative is that the mean is.1. Then, the power of the test is the probability that the mean will be greater than 1.86 standard errors from.1 or less than standard errors from.1. The power is ow, reduce the significance level to.01. The probability of the type I error is now.01. The probability that the null will be rejected when the alternative, µ =.1, is true becomes Reducing the significance level reduces the power of the test as well. What does increase when the significance is reduced is the probability of a type II error, not rejecting the null when it is false. This is one minus the power. The probabiliy of a type II error increases when the probability of a type I error decreases. The power of the test is one minus the probability of a type II error. So, the answer to the question is FALSE. c. False. The probability is 1 - α, not α. d. False. The probability that the null is falsely rejected is 1 - α e. True, but it must be qualified that the null hypothesis must also be true. f. ot necessarily. It depends on the application. g. False. The power is determined by the alternative. h. True. It is a function of a statistic computed from the data. 5. Let X 1,..., X 25 be a sample from a normal distribution having a variance of 100. Find the rejection region for a test at level α =.10 of H 0 : μ = 0 versus H A : μ = 1.5. What is the power of the test? Repeat for α =.01.

3 The rejection region is the values of x-bar that are greater than 1.28 standard errors above zero That is, where the probability that x-bar is above the value is.1. That is 1.28 standard errors. The standard error is σ/sqr(n) = 10/5 = 2. So, the rejection region is values of x-bar above The power of the test is the probability that it will reject H 0 when it is false. That is the probability that x-bar will exceed 2.56 when µ is 1.5. If µ is 1.5, then this is the probability that z is greater than ( )/2 = prob(z >.53) = If α is.01, then the 1.28 becomes 2.33 so the critical value is The probability that z exceeds 4.66 if µ is 1.5 is Prob(z > ( )/2) = Prob(z > 1.58) = Let X 1,..., X n be a random sample from an exponential distribution with the density function f (x θ) = θexp[ θx]. Derive a likelihood ratio test of H 0 : θ = θ 0 versus H A : θ θ 0, and show that the rejection region is of the form { x exp[ θ 0 x ] c}. The likelihood under the null hypothesis is θ exp( θ Σ x ). Under the alternative, 0 0 the likelihood is maximized by setting θ = 1/x. So, the likelihood under the alternative hypothesis is (1/x) exp( 1/ x Σ x ). But, Σ x = x, so the likelihood i i i i under the alternative is (1/x) exp( ). The likelihood ratio is 0 exp( 0x) exp( 0x) 0 = θ 0 = θ0 θ θ θ θ L = ( x) ( x exp( x) ). (1/x) exp( ) exp( 1) exp( 1) ( ) The rejection region is values of x for which x exp( θ x) is small. 7. a. In 1965, a newspaper carried a story about a high school student who reported getting 9207 heads and 8743 tails in 17,950 coin tosses. Is this a significant discrepancy from the null hypothesis H 0 : p = ½? b. Jack Youden, a statistician at the ational Bureau of Standards, contacted the student and asked him exactly how he had performed the experiment (Youden 1974). To save time, the student had tossed groups of five coins at a time, and a younger brother had recorded the results, shown in the following table: umber of Heads Frequency Are the data consistent with the hypothesis that all the coins were fair (p = ½)? a. The mean would be.5(17950) = 8975 and the variance =.5(.5)17950 = The standard deviation is heads is ( )/ = standard deviations from the mean. Using the normal approximation, this seems like a significant difference. b. In tosses of 5 coins, assuming they are fair, the probabilities are 1/32, 5/32, 10/32, 10/32, 5/32, 1/32 = (.03125,.15625,.3125,.3125,.15625,.03125). The expected values in 3590 tosses of 5 coins are (112.18, , , , , ). Computing the chi-squared = Σ [(Observed Expected) 2 /Expected] = with 5 degrees of freedom. The critical value is about Something wrong in here. i 0 i 8. If X ~ (μ x, σ x 2 ) and Y is independent (μ y, σ y 2 ), what is π = P(X < Y ) in terms of μ x, μ y, σ x, and σ y?

4 Prob(X < Y) = Prob(X Y < 0). Z = X Y is normally distributed with mean µ X - µ Y and variance σ X 2 + σ Y 2. So, the desired probability is prob(z < [0 (µ x - µ y )]/sqr(σ X 2 + σ Y 2 )]. 9. Lin, Sutton, and Qurashi (1979) compared microbiological and hydroxylamine methods for the analysis of ampicillin dosages. In one series of experiments, pairs of tablets were analyzed by the two methods. The data in the following table give the percentages of claimed amount of ampicillin found by the two methods in several pairs of tablets. What are E[X Y] and the standard deviation of X Y? If the pairing had been erroneously ignored and it had been assumed that the two samples were independent, what would have been the estimate of the standard deviation of X Y? Analyze the data to determine if there is a systematic difference between the two methods. Microbiological Method Hydroxylamine Method The means are and The difference is The standard deviation of x-bar y-bar is the square root of Var(x-bar)+var(ybar)-2Cov(xbar,ybar) = sqr( ( ))/15 =.308. If it were assumed that the pairings were independent, the covariance term in the result above would be omitted, and the standard deviation of xbar-ybar would be estimated by sqr( )/15 = Are they different? Rice seems to think not. Whichever method is used, the difference in the means is not statistically different from zero. The independence assumption seems dubious given that the correlation of the two sets of values is.98.

5 10. A market research team conducted a survey to investigate the relationship of personality to attitude toward small cars. A sample of 250 adults in a metropolitan area were asked to fill out a 16-item selfperception questionnaire, on the bases of which they were classified into three types: cautions conservative, middle of the roader and confident explorer. There were then asked to five their overall opionion of small cars, favorable, neutral, or unfavorable. Is there a relationship between personality type and attitude toward small cars? If so, what is the nature of the relationship? Personality types Attitude Cautious Midroad Explorer Favorable eutral Unfavorable There is an error in the problem. The table contains 299 responses, not 250. Use the chi squared test of independence. The observed frequencies are in the table. I convert these to proportions Cautious Midroad Explorer Total Favorable.264 (.205).194 (.208).164 (.208).622 eutral.033 (.030).027 (.030).030 (.030).090 Unfavorable.033 (.095).114 (.096).141* (.096).288 (* rounded up) Total Expected proportions are the products of the marginals, given in parentheses. The chi squared is 299 times the sum over the 9 cells of [(observed expected) 2 /expected] matrix ; list ; a=[.264,.194,.164/.033,.027,.030/.033,.114,.141] $ A matrix ; list ; cs = 1'a$ matrix ; list ; rs = a*[1/1/1] $ matrix ; list ; f = rs*cs $ F matrix ; list ; d = a-f $ Observed - expected matrix ; list ; dd = dirp(d,d) $ Square element by element matrix ; list ; fi = diri(f)$ Reciprocals of elements matrix ; list ; c = dirp(dd,fi)$ Product, element by element matrix ; list ; c2 = 299* 1'c*[1/1/1]$ Sum all elements * 299 result is This is a chi squared with (3-1)(3-1) = 4 degrees of freedom. This is much larger than the critical chi squared with 4 degrees of freedom of The hypothesis of independence is rejected. The test does not imply the nature of the relationship; it merely suggests that the two opinions are correlated. 11. A random variable X is said to follow a lognormal distribution if Y = ln(x) follows a normal distribution. The lognormal is sometimes used as a model for heavy tailed skewed distributions. The following data are the dorsal lengths in millimeters of taxonomically distinct octopods. a. Calculate the density function of the lognormal distribution. (What are the parameters µ and σ.) b. Examine whether the lognoral distribution roughly fits the data.

6 Hint: According to the problem, the data given come from a lognormal population. Thus, logs (base e) will be normally distributed. Take the logs, then sort the data. Divide the observed data into a set of five ranges. Observe the proportions of observations that fall in those ranges. You can compute the counterparts to these proportions as the predicted probabilities for the ranges from a normal distribution with the mean and variance that occur in the data. A chi squared goodness of fit can then be used. These are the logs of the sorted observations. The mean is The standard deviation is a. If x ~ [µ,σ 2 ] then x µ f( x) = exp σ 2π 2 σ x y = e, x = log y, dx = 1/ y dy log y µ f( y) = exp yσ 2π 2 σ b. (Hint: According to the problem, the data given come from a lognormal population. Thus, logs (base e) will be normally distributed. Take the logs, then sort the data. Divide the observed data into a set of ranges. Observe the proportions of observations that fall in those ranges. You can compute the counterparts to these proportions as the predicted probabilities for the ranges from a normal distribution with the mean and variance that occur in the data. A chi squared goodness of fit can then be used. The sorted data on the logs of the variables are listed below:

7 The mean and standard deviation of the data are and I divided the data into the 5 sets above. The boundaries of 5 regions are [-, ], ( , ], ( , ], ( , ], ( , - ]. Standardized (by taking (value )/.78543), these are (-,-.86), (-.86,-.32), (-.32,+.32), (+.32,+.77),(+.77,+ ) Using the normal table, the probabilities for these 5 intervals are.195,.180,.251,.154,.220. (Rounding error of.001, so I reduced the top cell to make these add to 1.000). Based on my division of the data (yours might be different), the 5 sample proportions are.191,.191,.191,.191,.236. (again, a bit of rounding error in the top cell). The chi squared is Σ [(Observed Expected) 2 /Expected] = The number of degrees if 5-1 = 4. The critical chi squared for 4 degrees of freedom is 9.49 (5% significance), so the normality hypothesis is not rejected.