Probability & Statistics Chapter 11 Chi-Square and F Distributions

Transcription

1 Part I: Hypothesis Tests Using the Chi-Square Distribution Overview of the Chi-Square Distribution Chi ( ) is a Greek letter denoted by the symbol, so chi-square is denoted by the symbol. The graph of is not symmetrical, and lie the Student s t distribution, it depends on the number of degrees of freedom. When the degrees of freedom increase, the graph of the chi-square distribution becomes more bell-like and begins to look more symmetric. We use Table 8 of Appendix II to find critical values of chi-square distributions for which a designated area falls to the right of the critical value.

2 11.1 Chi Square: Tests of Independence Contingency table: 3 x 3, there are 3 rows of cells and 3 columns Row Total Column Total Example 1 Give the size of the contingency tables. Also count the number of cells in each table. (Remember, each cell is a shaded box.) a) Row Total Column Total b) Row Total Column Total

3 Formula for Expected Frequency E Note: If the expected value is not a whole number, do not round it to the nearest whole number. Example 2 The table below contains the observed frequencies O and expected frequencies E for the contingency table giving keyboard arrangement and number of hours it takes a student to learn to type at 20 words per minute. Fill in the missing expected frequencies. Keyboard h h h Row Total A #1 O = 25 E = 24 #2 O = 30 E = 40 #3 O = 25 E = 80 B Standard #4 O = 30 E = 36 #7 O = 35 E = #5 O = 71 E = #8 O = 49 E = 50 #6 O = 19 E = #1 O = 16 E = 20 Column Total Sample size

4 Computing the sample statistic The value is a measure of the sum of differences between observed frequency O and expected frequency E in each cell. When we divide the quantity by E, we take the size of the difference with respect to the size of the expected value., where the sum is over all cells in the contingency table. Example 3 a) Complete the Table from the data Table 11-4 P. 657 Cell O E O E (O E) 2 (O E) 2 /E b) Compute

5 Notice that when the observed frequency and the expected frequency are very close, the quantity is close to zero, and so the statistic is near zero. As the difference increases, the statistic also increases. In order to determine how large the statistic can be before we must reject the null hypothesis of independence, we find a critical value in Table 8 of Appendix II for the specified level of significance and the number of degrees of freedom in the sample. Degrees of Freedom for Test of Independence Degrees of freedom = (number of rows 1) (number of columns 1) or where R = number of cell rows C = number of cell columns Example 4 Determine the number of degrees of freedom in the example of keyboard arrangements (see Table for 1A). Recall that the contingency table had three rows and three columns.

6 Summary: Step 1: Step 2: Set up the hypothesis: H 0 : The variables are independent. H 1 : The variables are not independent. Compute the expected frequency for each cell in the contingency table by use of the formula. Step 3: Compute the statistic for the sample: where O is the observed frequency, E is the expected frequency Step 4: Find the critical value in Table 8 of Appendix II. Use the level of significance and the number of degrees of freedom d.f., to find the critical value. where R is the number of rows, and C is the number of columns of cells in the contingency table. The critical region consists of all values of to the right of the critical value. Step 5: Compare the sample statistic of step 3 with the critical value of step 4. If the sample statistic is larger, reject the null hypothesis of Independence. Otherwise, do not reject the null hypothesis.

7 Example 5 Super Vending Machines Company is to install soda pop machines in elementary schools and high schools. The market analysts wish to know if flavor preference and school level are independent. A random sample of 200 students was taken. Their school level and soda pop preference are given in the table. Is independence indicated at the level of significance? Step 1: State the null and alternate hypothesis. Step 2: Complete the contingency table. Soda Pop High School Elementary School Row Total #1 #2 Kula Kola O = 33 E = 36 O = 57 E = Mountain Mist Jungle Grape O = 30 E = 20 O = 5 E = #3 #5 O = 20 E = 30 O = 35 E = #7 #8 Diet Pop O = 12 E = O = 8 E = 20 Column Total Sample size Step 3: Fill in the table below and use the table to find the sample statistic. Cell O E O E (O E) 2 (O E) 2 /E #4 #

8 Step 4: What is the size of the contingency table? Use the number of rows and columns to determine the number of degrees of freedom. For, use Table 8 of Appendix II to find the critical value. Step 5: Do we reject or fail to reject the null hypothesis that school level and soda pop flavor preference are independent? Calculator Note P. 662

9 11.2 Chi Square: Goodness of Fit Hypotheses H 0 : The population fits the given distribution. H 1 : The population has a different distribution. We use the chi-square distribution to test goodness-of-fit hypotheses. Just as with tests of independence, we compute the sample statistic: with degrees of freedom = n 1 Where E = expected frequency O = observed frequency is summed for each item in the distribution n = number of items in the distribution Then, compare it with an appropriate critical value from Table 8. In the case of a goodnessof-fit test, we use the null hypothesis to compute the expected values. Degrees of Freedom for Goodness-of-Fit Test d.f. = (number of E entries) 1

10 Example 6 According to genetics theory, red-green colorblindness in humans is a recessive sexlinked characteristic. In this case, the gene is carried on the X chromosome only. We will denote an X chromosome with the gene by X c and one without the gene by X n. Women have two X chromosomes, and they will be red-green colorblind only if both chromosomes have the gene, designated X c X c. A woman can have normal vision but still carry the colorblind gene if only one of the chromosomes has the gene, designated X c X n. A man carries an X and Y chromosome; if the X chromosome carries the colorblind gene (X c Y), the man is colorblind. According to genetics theory, if a man with normal vision (X n Y) and a woman carrier (X c X n ) have a child, the probabilities that the child will have red-green colorblindness, have normal vision and not carry the gene, or have normal vision and carry the gene are given by the equally likely events in the table. Mother Father X n Father Y X c X c X n X c Y X n X n X n X n y P(child has normal vision and is not a carrier) = P(X n Y) + P(X n X n ) = ½ P(child has normal vision and is a carrier) = P(X c X n ) = ¼ P(child is red-green colorblind) = P(X c Y) = ¼ To test this genetics theory, Genetics Labs took a random sample of 200 children whose mothers were carriers of the colorblind gene and whose fathers had normal vision. The results are in the table below. We wish to test the hypothesis that the population follows the distribution predicted by the genetics theory in the previous table. (a) State the null and alternate hypotheses.

11 (b) Fill in the rest of the table. Event O E (O E) 2 (O E) 2 /E Red-green colorblind Normal vision, 105 noncarrier Normal vision, carrier 60 (c) There are three expected frequencies listed in this table. Use this information to compute the degrees of freedom. (d) Find the critical value for a 0.01 level of significance. Do we reject the hypothesis that the population follows the distribution predicted by genetics theory or not?

12 11.3 Testing and Estimating a Single Variance or Standard Deviation Two kinds of problems: 1) Test hypotheses about the variance (or standard deviation) of a population. 2) Find confidence intervals for the variance (or standard deviation) of a population. (Remember: standard deviation is just the square root of the variance.) Theorem 11.1 If we have a normal population with variance and a random sample of n measurements is taken from this population with sample variance s 2, then Has a chi-square distribution with degrees of freedom. Example 7 (a) Find the value so that 1% of the area under the curve is to the right of when d.f. = 20. (b) Find the value so that 1% of the area under the curve is to the left of when d.f. = 20 (i) First find the corresponding area to the right of the desired value. (ii) Then look up the value with d.f. = 20 and.

13 Example 8 Certain industrial machines require overhaul when wear on their parts introduces too much variability to pass inspection. A government official is visiting a dentist s office to inspect the operation of an x-ray machine. If the machine emits too little radiation, clear photographs cannot be obtained. However, too much radiation can be harmful to the patient. Government regulations specify an average emission of 60 millirads with standard deviation of 12 millirads, and the machine has been set for these readings. After examining the machine, the inspector is satisfied that the average emission is still 60 millirads. However, there is wear on certain mechanical parts. To test variability, the inspector takes a random sample of 30 x-ray emissions and finds the sample standard deviation to be s = 15 millirads. Does this support the claim that the variance is too high (i.e., the machine should be overhauled)? Use a 1% level of significance. Let be the (population) standard deviation of emissions (in millirads) of the machine in its present condition. (a) Which of the following shall we use for the null hypothesis? Explain. (b) Which of the following shall we use for the alternate hypothesis? Explain. (c) What is the observed value of? (d) What are the degrees of freedom? Are we using a left-, right-, or two-tailed test? Use Table 8 in Appendix Ii to find the chi-square critical value.

14 (e) Sketch the critical region on a chi-square curve. Locate the observed chi0-square value on the sketch. (f) Do we reject overhauled? or not? Should the inspector recommend that the machine be Theorem 11.2 Let a random sample of size n be taken from a normal population with population standard deviation, and let c be a chosen confidence level. Then, ( ) and ( ) where n = sample size is the sample standard deviation chi-square value from Table 8 of Appendix II using and chi-square value from Table 8 of Appendix II using and

15 Example 9 A few miles off the Kona coast of the island of Hawaii a research vessel lies anchored. This ship makes electrical energy from the solar temperature differential of (warm) surface water versus (cool) deep water. The basic idea is that the warm water is flushed over coils to vaporize a special fluid. The vapor is under pressure and drives electrical turbines. Then some electricity is used to pump up cold water to cool the vapor back to a liquid, and the process is repeated. Even though some electricity is used to pump up the cold water, there is plenty left to supply a moderate-sized Hawaiian town. The subtropic sun always warms up surface water to a reliable temperature, but ocean currents can change the temperature of the deep, cooler water. If the deep-water temperature is too variable, the power plant cannot operate efficiently or possibly not operate at all. To estimate the variability of deep ocean water temperatures, a random sample of 25 near-bottom readings gave a sample standard deviation of. Find a 99% confidence interval for the variance and standard deviation of deep-water temperatures. (a) Determine the following values: c = n = d.f. = s = (b) What is the value of? of? (c) Find a 99% confidence interval for. (d) Find a 99% confidence interval for.

16 Part II: Hypothesis Tests Using the F Distribution 11.4 Testing Two Variances Use independent random samples from two populations to test the claim that the population variances are equal. F Probability Distribution Discovered by the English statistician Sir Ronald Fisher ( ). What we need to assume for a test of two population variances: 1) The two populations are independent of each other. (No relation whatsoever between specific values of the two distributions.) 2) The two populations each have a normal probability distribution.

17 How to Set Up the Test: Step 1: Get two independent random samples, one from each population We use the following notation: Population I (larger s 2 ) Population II (smaller s 2 ) n1 = sample size n2 = sample size sample variance sample variance population variance population variance Step 2: Set up the Hypotheses The null hypothesis will be that we have equal population variance. Alternate hypothesis will be either: or Step 3: Compute the sample test statistic For two normally distributed populations with equal variances (, the sampling distribution we will use is the F distribution (see Table 9 of Appendix II). The F distribution depends on two degrees of freedom. Properties of the F Distribution: The F distribution is not symmetrical. It is skewed to the right. Values of the F distribution are always greater than or equal to zero. A specific F distribution is determined from two degrees of freedom. These are called degrees of freedom for the numerator d.f. N and degrees of freedom for the denominator d.f. D. For our tests of two variances, it can be shown that d.f. N = n 1-1 d.f. D = n 2 1 Step 4: Use table 9 in Appendix II Find a critical F value.

18 Example 10 A large variance in blood chemistry components can result in health problems as the body attempts to return to equilibrium. J.B. O Sullivan and C.M. Mahan conducted a study reported in the American Journal of Clinical Nutrition (vol. 19, pp ) that concerned the glucose (blood sugar) levels of pregnant and nonpregnant women at Boston City Hospital. For both groups, a fasting (12-hour) fast blood glucose test was done. The following data are in units of milligrams of glucose per 100 milliliters of blood. Glucose Test: Nonpregnant Women Glucose Test: Pregnant Women Medical researchers question if the variance of the glucose test results for nonpregnant women is different (either way) compared with the variance for pregnant women. Let s conduct a test using a 5% level of significance. (a) What assumptions must be made about the two populations and the samples? (b) Use a calculator to compute the sample variance for each data group. Population I Population II n 1 = n 2 = = = (c) What is the null hypothesis? (d) What is the alternate hypothesis?

19 (e) Compute the sample F statistic d.f. N = n 1-1 and d.f. D = n 2 1. (f) Use Table 9 from Appendix II with level of significance value for the two-tailed test. to find a critical (g) What about the left tail? Do we need a critical value for the left tail? (h) Sketch the critical region on the F distribution and show the location of the sample F statistic on the diagram. Use the sketch to conclude the test. (i) What does the conclusion mean to a person doing medical research? See Calculator Note P

20 11.5 One Way ANOVA: Comparing Several Sample Means A lot of statistical applications in psychology, social science, business administration, and natural science involve many means and many data groups. Questions commonly asked are: o Which of several alternative methods yields the best results in a particular setting> o Which of several treatments leads to highest incidence of patient recovery? o Which of several investment schemes leads to greatest economic gain? Using the previous methods we have learned of comparing only two means will require many tests of significance to answer the preceding questions. Statisticians developed a method called analysis of variance (abbreviated ANOVA). We will study single-factor analysis of variance (one-way ANOVA). An ANOVA test is used to determine the existence (or nonexistence) of a statistically significant difference among the group means. 3 Basic Assumptions for ANOVA: 1) We assume that each of our k groups of measurements is obtained from a population with a normal distribution. 2) Each group is randomly selected and is independent of all other groups. In particular, this means that we will not use the same subjects in more than one group and that the scores of one subject will not have an effect on the scores of another subject. 3) We assume that the variables from each group come from distributions with approximately the same standard deviation.

21 Steps for ANOVA Test: 1) Determine the Null and Alternate Hypotheses. The null hypothesis is simply that all the group population means are the same. The alternate hypothesis is that not all the group population means are equal. 2) Find SS TOT SS TOT represents total variability of the data. Total variability can occur in two ways: 1. Scores may differ from one another because they belong to different groups with different means (recall that the alternate hypothesis says that the means are not all equal). This difference is called betweengroup variability and is denoted by SS BET. 2. Inherent differences unique to each subject and differences due to chance may cause a particular score to be different from the mean of its own group. This difference is called within-group variability and is denoted SS W. 3) Find SS BET Between-group variability measures the variability of group means. = mean of all x values from all groups ( ) Where n i = sample size of group i = sample mean of group i = mean for values from all groups = sum of data in group i. = sum of data from all groups.

22 4) Find SS W 5) Find Variance Estimates (Mean Squares), where k is the number of groups, where N is the total sample size We are looking for the following variance estimates: MS BET, the variance between groups (read mean square between) MS W, the variance within groups (ream mean square within) 6) Find the F Ratio and Complete the ANOVA Test The logic of our ANOVA test rests on the fact that one of the variances, MS BET, can be influenced by population differences among means of the several groups, whereas the other variance, MS W, cannot be so influenced. If the null hypothesis is true, MS BET and MS W should both estimate the same quantity. Therefore, if H 0 is true, the F ratio, should be approximately one, and variations away from 1 should occur only because of sampling errors. The decision to reject or not to reject the null hypothesis is determined by the relative size of the F ratio.

23 Example 11 A psychologist is studying pattern-recognition skills under four laboratory settings. In each setting, a fourth-grade child is given a pattern-recognition test with 10 patterns to identify. In setting I, the child is given praise for each correct answer and no comment about wrong answers. In setting II, the child is given criticism for each wrong answer and no comment about correct. In setting III, the child is given no praise or criticism, but the observer expresses interest in what the child is doing. In setting IV, the observer remains silent in an adjacent room watching the child through a one-way mirror. A random sample of fourth-grade children was used, and each child participated in the test only once. The test scores (number correct) for each group follow. Pattern-Recognition Experiment Group I (Praise) Group II (Criticism) Group III (Interest) Group IV (Silence) n 1 = 5 n 2 = 4 n 3 = 6 n 4 = 5 x 1 x 2 x 3 x N = k = a) Fill in the missing entries of the table. b) What assumptions are we making about the data to apply a single-factor ANOVA test? c) What are the null and alternate hypotheses?

24 d) Find the Value of SS TOT. e) Find SS BET. f) Find SSW and check your calculations using the formula SS TOT = SS BET + SS W g) Find d.f.bet and d.f.w. h) Find the mean squares MS BET and MS W. i) Find the F ratio. j) Find F 0.01, the critical value for an level of significance. Sketch the critical region and your observed F ratio. Does the test indicate we should reject the null hypothesis or not? Explain. k) Make a summary table of this ANOVA test. Source of variation Between groups Within groups Total Sum of Squares Degrees of Freedom Mean Square (variance) F Ratio F Critical Value Test Decision See Calculator Note P. 710

25 11.6 Introduction to Two-Way ANOVA Two-Way ANOVA involves two variables. These variables are called factors. The levels of a factor are different values the factor can assume. Basic assumptions of two-way ANOVA Just as for one-way ANOVA, our application of two-way ANOVA requires some basic assumptions: 1) The measurements in each cell of a two-way ANOVA model are assumed to be drawn from a population with a normal distribution. 2) The measurements in each cell of a two-way ANOVA model are assumed to come from distributions with approximately the same variance. 3) The measurements in each cell come from independent random samples. 4) There are the same number of measurements in each cell. Procedure to conduct a two-way ANOVA Test 1) Establish the Hypotheses Three sets of hypotheses are: a) H 0 : There is no difference in population means among the levels of the row factor. H 1 : At least two population means are different among the levels of the row factor. b) H 0 : There is no difference in population means among the levels of the row factor H1: At least two population means are different among the levels of the column factor.