Textbook Notes (369,140)
BIOL499A (11)
Chapter 24

# Chapter 24final.pdf

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Department
Biology (Biological Sciences)
Course Code
BIOL499A
Professor
Blaine Mullins

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Chapter 24: Comparing Two Population Means – Two Independent Samples In this chapter, we will talk about:  The difference between independent and dependent samples,  Two-sample hypothesis tests of means when 1 2 is unknown,  Confidence intervals fo 1  2 when 1 2 is unknown  Two-sample hypothesis tests of means when 1 and 2are unknown and unequal,  Confidence intervals fo 1  2when  1and 2 are unknown and unequal, For comparing two treatments, the two basic types of design are:  Independent Samples,  Matched pairs samples Goal for Independent samples:  Case 1:  1 2 is unknown: Use t-test with pooling  Case 2:  1and  2are unknown and unequal: Use t-test without pooling or conservative t-test. Suppose 10 male students and 20 femalestudents were selected from last year stat 151 classes and we found: nys0, 75, 12 1 1 1 and n2 2 2, 70, 9 Do we have enough evidence to conclude that the average of male student marks is higher than the average of female student marks? Interested parameter:   1 2 yy Estimate: 1 2 To test the hypotheH 0 1 2 0, the test statistic is: Estimate Null Hypothesis y1 2 0    ~t distribution S.E.(Estimate) S.E.(y1 2 ) ) 1 () Var Y1 2 Va1  Y2 Y = (1)Var Y 1 2 (-1)Var Y ) = Var Y1 2 Var Y ) 1 2 =  n1 2 If 1 2 ,then 2 2 2 2 1 2  211 =a1 2   n1 2 1 2 12n  Therefore, SD Y   11 1 2 nn 1 2 S.    ˆ 11 s1 1 2 nn nn 1 2 1 2 Question: Recall 1 1t,n2s12and 20, 9.Should I use s1 22 or 9as an estimate of Inference about  1  2 for Two Independent Samples: We consider following two cases:  Case 1:  1  2s unknown  Case 2:  1 and  2are unknown and unequal Case 1 (Unknown population standard deviations are equal): Suppose that the two normal population distributions have same standard deviation. Call the common-and still unknown-standard deviation of both populations  . Both sample variances S1 and S 2 2 estimates . The best way to combine these two estimates is: 2 2 )(11 1 2S22 2 Spooled p n1 2 2 This is called the pooled estimator of  2 because it combines the information in both samples. To test the hypothesisH 0: 1  2 0 , the test statistic is: Y1 2 0 TS   ~ t distribution, with df  n1 n 2 2 ,ifH 0s true 11 Sp  n1 2 For H : 0  , we reject H ifTS  t * a 1 2 0 * For H a 1 2 , we reject H 0if TS  t For H : 0  , we reject H if TS  t * or TS  t * a 1 2 0 A 100(1)% confidence interval for    is given by: 1 2 11 Y1 2t * p n1 2 Example 24.1: A study was conducted to find out whether the children of aggressive families have a higher beha vior problem checklist (BPC) score. The sample mean and standard deviation of BPC scores for 13 children whose parents were classified as aggressive were 7.92 and 3.45 respectively. For a sample of 12 children whose parents were classified as non-aggressive, the mean and standard deviation of BPC score were 5.80 and 2.87, respectively. Based on the data, a psychiatrist claimed that the children of aggressive families have higher behavior problem checklist (BPC) scores on average. Test the psychiatrist's claim at   0.01 . Solution: Example 24.2: A psychologist was in terested in the effects of two different kinds of drugs on the mean time to complete a task. She randomly chose 10 subjects and assigned 5 of them to each drug A, and B. The following data represent the time in minutes required to complete the task. DruAg 30 24 30 25 26 DruBg 26 25 21 26 27 Assume the time to complete the task for the two drugs are approximately normally distributed and have equal vari ances. Is there enough evidence to conclude that the population mean time to complete the task is smaller for drug B than drug A? Solution: 2 )(n1As A B B s p nA B 2 H0: 1) Ha: 2)   0.05 YYA B 0 3) TS   ~t distribution if H 0is true. S 11  p nA B df  n  n  2  A B 4) We reject H 0if 5) The value of TS is: 6) Since At   0.05,there enough evid ence to conclude that the population mean time to complete the task is smaller for drug B than drug A 7) P-value = Example 24.3: The operations manager at a light bulb factory wants to determine if there is any difference in the average life expectancy of bulbs manufactured on two different types of machines. A random sample of 35 light bulbs obtained from machine I indicates a sample mean of 375 hours, and a sample standard deviation 110 hours. A similar sample of 25 from machine II indicates a sample mean of 362 hours and a sample standard deviation 125 hours. (a) Find a 95% confidence interval for the difference of the average life of bulbs produced by the two types of machines. Solution: LargestS 125 Since   2  SmallestS 110 )((1n s 2 2 s p 1 1 22 n1 2 2  1  0.95   0.05   0.025 2 df  n1 n 2 2 A 95% confidence interval for  1  2is: * 11 y1 2 p  n1 2 Conclusion: It is estimated with 95% confidence that the average life of bulbs produced by first machine is between hours smaller to hours larger than the average life of bulbs produced by second machine Since zero is in the confidence interval, Hence, we conclude that there is no difference in the average life of bulbs produced by the two machines. (b) Using the 0.05 level of significance, is there any evidence of a difference in the average life of bulbs produced by the two types of machines? Solution: H : 0 1) H : a 2)   0.05 Y1 2 0 3) TS   ~ t distribution if H 0 is true. S 11 p n1 2 df  n  n  2  1 2 4) We reject H 0if 5) The value of TS is: 6) Since At   0.05 , there e
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