Distribution. Lecture 34 Section Fri, Oct 31, Hampden-Sydney College. Student s t Distribution. Robb T. Koether.

Transcription

1 Lecture 34 Section 10.2 Hampden-Sydney College Fri, Oct 31, 2008

2 Outline

3 Exercise 10.4, page 633. A psychologist is studying the distribution of IQ scores of girls at an alternative high school. She wants to test the hypothesis that the average IQ score is 100 against the alternative that it is higher than 100. She assumes that the normal distribution with standard deviation 15 is a good model for these IQ scores. She takes a random sample of nine girls from this high school and measures their IQ. The average IQ score for these nine girls was 114.

4 Exercise 10.4, page 633. Assuming that the appropriate assumptions are met, compute the test statistic and p-value for this test. State your decision at level α = 0.01 and write up your conclusion in words.

5 Solution Let s just show all seven steps. (1) The hypotheses are H 0 : µ = 100. H 1 : µ > 100. (2) α = (3) The test statistic is z = x µ 0 σ/ n.

6 Solution (4) Compute z = / 9 = 14 5 = 2.8. (5) The p-value is normalcdf(2.8,e99) = (6) Reject H 0. (7) The average IQ score of the girls at the alternative high school is higher than 100.

7 Last time, we made a very important, but unrealistic, assumption: The value of σ was known. What happens if σ is not known? We have to use s as an approximate the value of σ. Does that have any ramifications? Yes! But first, let s do the Z-test on the TI-83.

8 Last time, we made a very important, but unrealistic, assumption: The value of σ was known. What happens if σ is not known? We have to use s as an approximate the value of σ. Does that have any ramifications? Yes! But first, let s do the Z-test on the TI-83.

9 Last time, we made a very important, but unrealistic, assumption: The value of σ was known. What happens if σ is not known? We have to use s as an approximate the value of σ. Does that have any ramifications? Yes! But first, let s do the Z-test on the TI-83.

10 Last time, we made a very important, but unrealistic, assumption: The value of σ was known. What happens if σ is not known? We have to use s as an approximate the value of σ. Does that have any ramifications? Yes! But first, let s do the Z-test on the TI-83.

11 Last time, we made a very important, but unrealistic, assumption: The value of σ was known. What happens if σ is not known? We have to use s as an approximate the value of σ. Does that have any ramifications? Yes! But first, let s do the Z-test on the TI-83.

12 Last time, we made a very important, but unrealistic, assumption: The value of σ was known. What happens if σ is not known? We have to use s as an approximate the value of σ. Does that have any ramifications? Yes! But first, let s do the Z-test on the TI-83.

13 Testing on the TI-83 TI-83 Testing for the Mean Press STAT. Select TESTS. Select Z-Test. Press ENTER. A window appears requesting information. Select Data if you have the sample data entered into a list. Otherwise, select Stats.

14 Testing on the TI-83 The Stats Option TI-83 Testing for the Mean (Stats Option Enter µ 0, the hypothetical mean. Enter σ. (Remember, σ is known.) Enter x. Enter n, the sample size. Select the type of alternative hypothesis. Select Calculate and press ENTER.

15 Testing on the TI-83 The Stats Option TI-83 Testing for the Mean A window appears with the following information. The title Z-Test. The alternative hypothesis. The value of the test statistic Z. The p-value of the test. The sample mean. The sample size.

16 Example The Stats Option Example (TI-83 Testing for the Mean (Stats Option)) A large number of students at a high school take the AP test in Statistics. A random sample of 36 students reveals the following scores: Score Frequency

17 Example The Stats Option Example (TI-83 Testing for the Mean (Stats Option)) Assume that the population standard deviation is σ = 1. Use the TI-83 to test the hypothesis that the average score for the high school is above 3.

18 Testing on the TI-83 The Data Option TI-83 Testing for the Mean (Data Option) Enter the data into list L 1. (Enter the frequencies into list L 2.) Enter the hypothetical mean µ 0. Enter σ. (Why?) Identify the list that contains the data (L 1 ). Skip Freq. (It should be 1 unless you entered the frequencies in list L 2 ). Select the alternative hypothesis. Select Calculate and press ENTER.

19 Example The Data Option Example TI-83 Testing for the Mean (Data Option) Re-work the previous example about AP Statistics scores using the Data option.

20 Now we will assume more realistically that the value of σ is unknown. Instead of the test statistic we will use the statistic x µ 0 σ/ n, x µ 0 s/ n. Does this statistic have a normal distribution? No. So rather than call it z, we will call it t.

21 Now we will assume more realistically that the value of σ is unknown. Instead of the test statistic we will use the statistic x µ 0 σ/ n, x µ 0 s/ n. Does this statistic have a normal distribution? No. So rather than call it z, we will call it t.

22 Now we will assume more realistically that the value of σ is unknown. Instead of the test statistic we will use the statistic x µ 0 σ/ n, x µ 0 s/ n. Does this statistic have a normal distribution? No. So rather than call it z, we will call it t.

23 Definition ( distribution) If the population from which we sample is normal, then the distribution of the statistic is distribution. t = x µ s/ n The t distribution was discovered by W. S. Gosset in See the MathWorld article -

24 The t The shape of the t distribution is very similar to the shape of the standard normal distribution. It is Symmetric Unimodal Centered at 0. But it is wider than the standard normal. That is because of the additional variability introduced by using s instead of σ.

25 The t Furthermore, the shape of the t distribution depends on the sample size. It has a slightly different shape for each sample size. However, it approaches the standard normal z as n gets larger and larger. In fact, if n 30, then the t distribution is almost exactly standard normal.

26 Definition ( freedom) If the sample size is n, then t is said to have n 1 degrees of freedom. We use df to denote degrees of freedom." We will use the notation t 5 to denote the t distribution with 5 degrees of freedom (i.e., sample size 6). (The book uses t(5).)

27 Standard Normal vs. t The distributions t 1, t 2, t 10, t 30, and Z

28 To find probabilities concerning t, use the TI-83 function tcdf. tcdf takes three parameters: The left endpoint (sometimes -E99). The right endpoint (sometimes E99). The number of degrees of freedom.

29 Example () Find P(t > 3) with df = 1. Find P( 3 < t < 3) with df = 1. Find P( 3 < t < 3) with df = 10. Find P( 3 < t < 3) with df = 30.

30 Example ( ) Test the hypotheses concerning the average AP Statistics score, assuming that the value of σ is not known. Assume that the population of test scores is normal.

31 Example ( ) (1) µ = average AP Statistics score. H 0 : µ = 3 H 1 : µ > 3 (2) α = (3) t = x µ 0 s/ n.

32 Example ( ) Use the TI-83 to compute the statistics x and s. We get x = s =

33 Example ( ) (4) t = / 36 = = (5) p-value = tcdf(1.065,e99,35) = (6) Accept H 0. (7) The average AP Statistics score is not greater than 3.

34 Read Section 10.2, pages Let s Do It! 10.3, 10.4, 10.5.