CHAPTER 8 Test of Hypotheses Based on a Single Sample Hypothesis testing is the method that decide which of two contradictory claims about the parameter is correct. Here the parameters of interest are population mean and proportion. Hypotheses and Test Procedures In any hypothesis testing problem, there are: Null Hypothesis, denoted by H 0 which is the initially assumption about population parameter. H 0 is an equality claim and the form of null hypothesis is H 0 : θ = θ 0 where θ is parameter of interest. Alternative Hypothesis denoted by H a is the contradictory to H 0 and looks like one of the following cases H a : θ > θ 0 H a : θ < θ 0 H a : θ θ 0 Test of Hypotheses is a method for using sample data to decide that H 0 should be reject or not. Test Procedures is a rule, based on sample data, for deciding whether to reject H 0 and contains: Test Statistic, which is a function of sample data for making decision Reject Region, which is the set of all test statistic values for which H 0 will be rejected. Errors in Hypotheses Testing There are two different types of error Type I error consists of rejecting H 0 when it is true denoted by α = P (type I error) Type II error consists of accepting H 0 when it is false denoted by β = P (type II error) Example: The calibration of a scale is to be checked by weighing a 10-kg test specimen 25 times. Suppose that the results of different weighing are independent of one another and 1
that the weight on each trial is normally distributed with σ = 0.2 kg. Let µ denote the true average weight reading on the scale. a. What hypotheses should be tested? b. Suppose the scale is to be recalibrated if either x 10.1032 or x 9.8968. What is the probability that recalibration is carried out when it is actually unnecessary? c. What is the probability that recalibration is judge unnecessary when in fact µ = 10.1 When µ = 9.83? d. Let z = ( x 10)/(σ/ n). For what value c is the rejection region of part (b) equivalent to the two tailed region either z c or z c? e. If the sample size were only 10 rather than 25, how should the procedure of part (d) be altered so that α = 0.05? f. Using the test part (e), what would you conclude from the following sample data: 9.981 10.006 9.857 10.107 9.888 9.728 10.439 10.214 10.190 9.793 g. Reexpress the test procedure of part (b) in terms of the standardized test statistic Z = ( X 10)/(σ/ n) Test About a Population Mean For this test, consider three different cases: 1. A Normal Population with Known σ Let X 1, X 2, X n be a random sample of size n from a normal population with variance of σ (known), then X N(µ, σ 2 /n) Suppose the null value for population mean is µ 0, when H 0 is true µ x = µ 0, and Z = X µ 0 σ/ n has a standard normal distribution which is the test statistic for H 0 : µ = µ 0. First consider H a : µ > µ 0, the rejection region calculate based on type I error (level of significant denoted by α), if α = 0.05 by using this fact that the distribution of Z is standard normal, the cut of point c is 1.645 and H 0 is rejected if z > 1.645. We have same argument for other kinds of alternatives. In general Null hypothesis: H 0 : µ = µ 0 Test statistic value: z = x µ 0 σ/ n 2
Alternative hypothesis H a : µ > µ 0 z z α (upper-tailed test) H a : µ < µ 0 z z α (lower-tailed test) H a : µ µ 0 z z α/2 or z z α/2 (two-tailed test) - Example: The melting point of 16 samples of a certain brand of hydrogenated vegetable oil was determined, resulting in x = 94.32. Assume that the distribution of melting point is normal with σ = 1.20. Test H 0 : µ = 95 versus H a : µ 95 using a two-tailed level 0.01 test. - 2. Large-Sample Tests When the population is not normal and also σ is unknown, again based on CLT Z = X µ has approximately a standard normal distribution and everything is same as previous case. 3. A Normal Population Distribution As seen before, if X 1, X 2,, X n is a random sample from a normal distribution, then T = X µ 0 has t-distribution with n 1 degrees of freedom, therefore instead of finding critical region from normal distribution, find critical region by using t-distribution. Then in general Null hypothesis: H 0 : µ = µ 0 Test statistic value: t = x µ 0 Alternative hypothesis H a : µ > µ 0 H a : µ < µ 0 H a : µ µ 0 t t α,n 1 (upper-tailed test) t t α,n 1 (lower-tailed test) t t α/2,n 1 or t t α/2,n 1 (two-tailed test) 3
Example: The amount of shaft wear after a fixed mileage was determined for each of n = 8 internal combustion engines having copper lead as a bearing material, resulting in x = 3.72 and s = 1.254. Assuming that the distribution of shaft wear is normal with mean µ, use the t test at level 0.05 to test H 0 : µ = 3.5 versus H a : µ > 3.5. Test Concerning a Population Proportion Based on CLT the estimator ˆp = X/n has a normal distribution with mean p and variance p(1 p)/n, then when n is large and H 0 : p = p 0 is true Z = ˆp p 0 p0 (1 p 0 )/n has approximately a standard normal distribution, and same as before critical value will calculated based on normal distribution. In general Null hypothesis: H 0 : p = p 0 Test statistic value: z = ˆp p 0 p0 (1 p 0 )/n Alternative hypothesis H a : p > p 0 z z α (upper-tailed test) H a : p < p 0 z z α (lower-tailed test) H a : p p 0 z z α/2 or z z α/2 (two-tailed test) These test valid if np 0 10 and n(1 p 0 ) 10 - Example: A random sample of 150 recent donations at a certain blood bank reveals that 92 were type A blood. Does it suggest that the actual percentage of type A donations differ from 40%, the percentage of the population having type A blood Carry out a test of the appropriate hypotheses using a significance level of 0.01. Would your conclusion have been different if a significance level of 0.05 had been used? - P-Value is the smallest level at which H 0 would be rejected. Once P-value has been determined, the conclusion at any particular level α results from comparing the P-value to α: P-value α reject H 0 at level α P-value > α do not reject H 0 at level α 4
Suggested Exercises for Chapter 8: 3, 7, 15, 17, 19, 23, 27, 31, 35, 37, 39, 47, 49, 53, 57, 59, 61, 65 5