Lecture 43 Section 14.1 – 14.3 Mon, Nov 28, 2005 Test of Goodness of Fit Lecture 43 Section 14.1 – 14.3 Mon, Nov 28, 2005

Count Data Count data – Data that counts the number of observations that fall into each of several categories. The data may be univariate or bivariate. Univariate example – Observe a student’s final grade: A – F. Bivariate example – Observe a student’s final grade and year in college: A – F and freshman – senior.

Univariate Example Observe students’ final grades in statistics: A, B, C, D, or F. A B C D F 5 12 8 4 2

Bivariate Example Observe students’ final grade in statistics and year in college. A B C D F Fresh 3 6 2 1 Soph 4 Junior Senior

Observed and Expected Counts Observed counts – The counts that were actually observed in the sample. Expected counts – The counts that would be expected if the null hypothesis were true.

The Chi-Square Statistic Denote the observed counts by O and the expected counts by E. Define the chi-square (2) statistic to be Clearly, if the observed counts are close to the expected counts, then 2 will be small. If even a few observed counts are far from the expected counts, then 2 will be large.

Think About It Think About It, p. 923.

Chi-Square Degrees of Freedom The chi-square distribution has an associated degrees of freedom, just like the t distribution. Each chi-square distribution has a slightly different shape, depending on the number of degrees of freedom.

Chi-Square Degrees of Freedom 2(2) 2(5) 2(10)

Properties of 2 The chi-square distribution with df degrees of freedom has the following properties. 2  0. It is unimodal. It is skewed right (not symmetric!) 2 = df. 2 = (2df). If df is large, then 2(df) is approximately N(df, (2df)).

Chi-Square vs. Normal N(30,60) 2(30) 2(32) N(32, 8)

Chi-Square vs. Normal N(128, 16) 2(128)

The Chi-Square Table See page A-11. The left column is degrees of freedom: 1, 2, 3, …, 15, 16, 18, 20, 24, 30, 40, 60, 120. The column headings represent areas of lower tails: 0.005, 0.01, 0.025, 0.05, 0.10, 0.90, 0.95, 0.975, 0.99, 0.995. Of course, the lower tails 0.90, 0.95, 0.975, 0.99, 0.995 are the same as the upper tails 0.10, 0.05, 0.025, 0.01, 0.005.

Example If df = 10, what value of 2 cuts off an lower tail of 0.05? If df = 10, what value of 2 cuts off a upper tail of 0.05?

TI-83 – Chi-Square Probabilities To find a chi-square probability on the TI-83, Press DISTR. Select 2cdf (item #7). Press ENTER. Enter the lower endpoint, the upper endpoint, and the degrees of freedom. The probability appears.

Example If df = 8, what is the probability that 2 will fall between 4 and 12? Compute 2cdf(4, 12, 8). If df = 32, what is the probability that 2 will fall between 24 and 40? Compute 2cdf(24, 40, 32). If df = 128, what is the probability that 2 will fall between 96 and 160? Compute 2cdf(96, 160, 128).

Tests of Goodness of Fit The goodness-of-fit test applies only to univariate data. The null hypothesis specifies a discrete distribution for the population. We want to determine whether a sample from that population supports this hypothesis.

Examples If we rolled a die 60 times, we expect 10 of each number. If we got frequencies 8, 10, 14, 12, 9, 7, does that indicate that the die is not fair? If we toss a fair coin, we should get two heads ¼ of the time, two tails ¼ of the time, and one of each ½ of the time. Suppose we toss a coin 100 times and get two heads 16 times, two tails 36 times, and one of each 48 times. Is the coin fair?

Examples If we selected 20 people from a group that was 60% male and 40% female, we would expect to get 12 males and 8 females. If we got 15 males and 5 females, would that indicate that our selection procedure was not random (i.e., discriminatory)?

Null Hypothesis The null hypothesis specifies the probability (or proportion) for each category. Each probability is the probability that a random observation would fall into that category.

Null Hypothesis To test a die for fairness, the null hypothesis would be H0: p1 = 1/6, p2 = 1/6, …, p6 = 1/6. The alternative hypothesis will always be a simple negation of H0: H1: At least one of the probabilities is not 1/6. or more simply, H1: H0 is false.

Expected Counts To find the expected counts, we apply the hypothetical probabilities to the sample size. For example, if the hypothetical probability is 1/6 and the sample size is 60, then the expected count is (1/6)  60 = 10.

Example We will use the sample data given for 60 rolls of a die to calculate the 2 statistic. Make a chart showing both the observed and expected counts (in parentheses). 1 2 3 4 5 6 8 (10) 10 14 12 9 7

Example Now calculate 2.

Computing the p-value The number of degrees of freedom is 1 less than the number of categories in the table. In this example, df = 5. To find the p-value, use the TI-83 to calculate the probability that 2(5) would be at least as large as 3.4. p-value = 2cdf(3.4, E99, 5) = 0.6386. Therefore, p-value = 0.6386 (accept H0).

The Effect of the Sample Size What if the previous sample distribution persisted in a much larger sample, say n = 6000? Would it be significant? 1 2 3 4 5 6 800 (1000) 1000 1400 1200 900 700

TI-83 – Goodness of Fit Test The TI-83 will not automatically do a goodness-of-fit test. The following procedure will compute 2. Enter the observed counts into list L1. Enter the expected counts into list L2. Evaluate the expression (L1 – L2)2/L2. Select LIST > MATH > sum and apply the sum function to the previous result, i.e., sum(Ans). The result is the value of 2.

Example To test whether the coin is fair, the null hypothesis would be H0: pHH = 1/4, pTT = 1/4, pHT = 1/2. The alternative hypothesis would be H1: H0 is false. Let  = 0.05.

Expected Counts To find the expected counts, we apply the hypothetical probabilities to the sample size. Expected HH = (1/4) 100 = 25. Expected TT = (1/4)  100 = 25. Expected HT = (1/2)  100 = 50.

Example We will use the sample data given for 60 rolls of a die to calculate the 2 statistic. Make a chart showing both the observed and expected counts (in parentheses). HH TT HT 16 (25) 36 48 (50)

Example Now calculate 2.

Compute the p-value In this example, df = 2. To find the p-value, use the TI-83 to calculate the probability that 2(2) would be at least as large as 8.16. 2cdf(8.16, E99, 2) = 0.0169. Therefore, p-value = 0.0169 (reject H0).