Contents. 22S39: Class Notes / October 25, 2000 back to start 1

Transcription

1 Contents Determining sample size Testing about the population proportion Comparing population proportions Comparing population means based on two independent samples Comparing population means based on two dependent samples two-sample t tests 22S39: Class Notes / October 25, 2000 back to start 1

2 Determining sample size Given a random sample from N(µ, σ 2 = 4) we want to test H 0 : µ = µ 0, say, 0 vs the alternative that H 1 : µ > µ 0. What sample size and cut-off value should we use to ensure that the test has 5% significance level and it have 90% power of detecting that the mean is shifted upwards by 0.1? The rule (critical region) should be Z = 5% significance level. X 2/ n > z(0.05) = 1.96 to maintain a The sample size is then determined by the power at µ = µ S39: Class Notes / October 25, 2000 back to start 2

3 0.90 = power(µ = 0.1) = P (Z = X µ 0 2/ n 1.96 µ = 0.1) = X 0.1 µ / n 1.96 µ = µ 0 ) = X 0.1 µ 0 2/ n n µ = µ ) Hence, n = z(0.1) = 1.282, implying that n = ( )/(0.05) and so n = Take n = 4205! 22S39: Class Notes / October 25, 2000 back to start 3

4 Testing about the population proportion Eample 1. A certain farm tool is known from past data to have about 5% accident rate. The manufacturer adopted a new design aimed to make the tool less accident-prone. A survey of 400 farmers who used the new tool reports 16 accidents. Does the new design make improvement? First, set up the hypotheses. Let p be the unknown accident rate of the new design. H 0 : p = 0.05 (we probably want to use H 0 : p 0.05 but this reduces to the preceding simple null hypothesis) vs H 1 : p < Use the sample proportion ˆp as the test statistic, which equals 16/400 = 0.04 from data. Under H 0, E(ˆp) = p 0 = 0.15 and σˆp = p 0 (1 p 0 )/n = /400 = By Central Limit Theorem (CLT), under H 0, Z = ˆp N(0, 1). 22S39: Class Notes / October 25, 2000 back to start 4

5 Reject H 0 in favor of H 1 if and only if Z z(0.05) = 1.645, at 5% significance level. From the data, Z = reject H 0. = which is larger than , so do not The data suggest that the new design has not made a statistically significant improvement. An alternative approach: compute p-value= P (Z p = 0.05) = 0.29 > 0.05, so reject H 0, sam econclusion as before. Remark 1: if H 0 = 0.05 and H 1 : p > 0.05, the rule will become reject H 0 if and only Z z(α), at the α 100 significance level. 22S39: Class Notes / October 25, 2000 back to start 5

6 Remark 2: if H 0 = 0.05 and H 1 : p 0.05, the rule will become reject H 0 if and only Z z(α/2), at the α 100 significance level. For eample if α = 0.05, z(α/2) = 1.96 and the rule becomes: reject H 0 if and only Z S39: Class Notes / October 25, 2000 back to start 6

7 Comparing population proportions Eample 2. A company wants to evaluate whether or not a new manufacturing process produce better quality chalk. A sample of 20 chalks are selected from the original process and 16 of them are judged to be of good quality. An independent sample of 40 chalks from the new process yields 38 good-quality chalks. Test whether or not the new process is better than the original process. Let p 1 be the true rate of good quality chalk from the original process, and p 2 that of the new process. H 0 : p 1 = p 2 = p and H 1 : p 2 > p 1. 22S39: Class Notes / October 25, 2000 back to start 7

8 Alternatively, H 0 : p 2 p1 = 0 and H 1 : p 2 p 1 > 0. Test statistic: ˆp 2 ˆp 1. Under H 0, E( ˆp 1 ˆp 2 ) = 0 but its variance equals p(1 p)/n 1 + p(1 p)/n 2 where n 1 = 20 and n 2 = 40 are the sample sizes. Since p is unknown under H 0, estimate it by ( )/( ) = 0.9 so that ˆσ ˆp2 ˆp 1 = / /40 = Use CLT to see that under H 0, Z = ˆp 2 ˆp N(0, 1). Therefore reject H 0 if and only if Z z(0.05) = From the data, ˆp 2 ˆp 1 = 38/40 16/20 = 0.15 and Z = 0.15/ = 1.825, which is greater than We can also compute the p-value of the data: p-value= P (Z 1.825) = S39: Class Notes / October 25, 2000 back to start 8

9 We reject H 0 and conclude that there is evidence suggesting that the new process is an improvement over the old process. How would you modify the rejection rule for the following two sets of hypotheis testing: H 0 : p 2 p1 = 0 vs H 1 : p 2 p 1 < 0: H 0 : p 2 p1 = 0 vs H 1 : p 2 p S39: Class Notes / October 25, 2000 back to start 9

10 Comparing population means based on two independent samples Eample 3: A company comparing two manufacturing process in terms of the thickness of the product. A sample from manufacturing process A yields: X = 12, S 2 1 = 2, n 1 = 10 An independent sample from manufacturing process B yields: Ȳ = 12, S 2 2 = 1, n 2 = 10 Let the thickness of a random product from process A be distributed as N(µ 1, σ 2 ), and that from process B N(µ 2, σ 2 ). Note that we assume common variance. Set up the hypothesis: As the company just wants to compare the two process. It is reasonable to set H 0 : µ 1 µ 2 = 0 vs H 1 : µ 1 µ 2 0. The test statistic is based on X Ȳ. 22S39: Class Notes / October 25, 2000 back to start 10

11 Under H 0 : E( X Ȳ ) = 0 with its standard error (estimate of the standard deviation of X Ȳ obtained as follows. S p = (S 1 2 (n 1 1) + S2 2 (n 2 1))/(n 1 + n 2 2) = Hence, ˆσ X Ȳ = Sp/n Sp/n 2 2 = S p 1/n1 + 1/n 2 = Under H 0, T = X Ȳ t(n 1 + n 2 2). Reject H 0 if and only if T t(0.025, n 1 + n 2 2) at 5% significance level. In the eample, t(0.025, 18) = From the data T = = As is larger than 2.101, we reject H 0 and conclude that the two processes are different. What is the p-value for this eample? How would you modify the rejection rule for the following two sets of hypothesis testing: H 0 : µ 1 µ 2 = 0 vs H 1 : µ 1 µ 2 < 0; H 0 : µ 1 µ 2 = 0 vs H 1 : µ 1 µ 2 > 0. 22S39: Class Notes / October 25, 2000 back to start 11

12 22S39: Class Notes / October 25, 2000 back to start 12

13 Comparing population means based on two dependent shamples two-sample t tests When we compare two treatments, e.g., the effectiveness of two sampoos in reducing dandruff, the subjects receiving the two treatments should be identical ecept that they receive different treatments; otherwise, any observed difference may be due to differences in other conditions rather than difference in the treatments. On the other hand, we also want to compare the two treatments for a variety of subjects to ensure that the result holds for a wide population. The conflicting requirement can be resolved by the so-called matched comparison where we find pairs of subjects so that within each pair the subjects are broadly similar but there is diversity between pairs. 22S39: Class Notes / October 25, 2000 back to start 13

14 Net, within each pair, we flip a coin to decide which one receives treatment A and which receives treatment B. For the shampoo eperiment, the two halves of the head of a subject serves as a good pair. We can then include in the eperiment people of different head conditions - dry, greasy, etc., to ensure that the test result holds for a wide spectrum of people. Which half of a head is to recieve shampoo A is decided by a coin flip. For eample, heads means shampoo A on the left half and shampoo B on the right half. A tail will lead to the opposite arrangement. The randomization is to be done for each pair. About 50% of left heads will receive shampoo A. Why? The purpose of the randomization is to equalize the group of subjects receiving shampoo A (treatment A) with that of shampoo B (treatment B). This means that any hidden factor will be equalized when we compare the treatments. 22S39: Class Notes / October 25, 2000 back to start 14

15 For eample, suppose that shampoo A works much better with females than males. Randomization should ensure that the two groups have similar se ratio, thereby eliminating any se bias when comparing the two treatments. The point is that we don t need to know which hidden factors are important, and randomization will mostly take care of them. 22S39: Class Notes / October 25, 2000 back to start 15

16 Eample 4: An eample of the shampoo data with the reponse variable being a measure of dandruff - larger value means more dandruff. X Y D=X-Y We have n = 10 subjects, the X s are the responses from the heads treated with shampoo A, and Y s those of shampoo B. Each row corresponds to the data from a subject. 22S39: Class Notes / October 25, 2000 back to start 16

17 The X and Y are dependent as high (low) X values tend to be followed by high (low) Y values. 22S39: Class Notes / October 25, 2000 back to start 17

18 y rnorm(10) rnorm(10) Figure 1: Upper panel plots the shampoo data and the lower panel displays the X and Y pairs from two independent sample. 22S39: Class Notes / October 25, 2000 back to start 18

19 It is then inappropriate to do the test assuming that the shampoo data are two independent samples. For dependent samples, we may analyse the difference in the response in each pair, i.e., the third column Note that E(D) = E(X Y ) = µ 1 µ 2. Hence, the hypotheses H 0 : µ 1 µ 2 = 0 vs H 1 : µ 1 µ 2 0 becomes H 0 : µ D = 0 vs H 1 : µ D 0 We then apply a t-test to the D s. Reject H 0 if and only if D T = t(0.025, n 1), at 5% significance level. S 2 D /n Now, n = 10, D = , SD 2 = , S d/ n = , so that D T = = , a highly significant result as t(0.025, 9) = S 2 D /n 22S39: Class Notes / October 25, 2000 back to start 19

20 We reject H 0 and conclude that the two shampoos are not of the same efficiency in removing dandruff. Which shampoo is better? If we adopt the wrong approach of treating the two samples as independent, we have the following results. The pooled variance equals and the T-statistic X Ȳ equals T = = / = way below the (1/10+1/10) cut-off point t(0.025, 18) = 2.101, in which case we ll wrongly conclude that the two shampoos are eqaually effective! In other words, the two-sample t-test has enhanced power to detect differences in a variable environment. The key is that the much of the large background variability are removed by taking the difference within the pair. 22S39: Class Notes / October 25, 2000 back to start 20