Inference About Two Means: Independent Samples

Transcription

1 Inference About Two Means: Independent Samples MATH 130, Elements of Statistics I J. Robert Buchanan Department of Mathematics Fall 2017

2 Motivation Suppose we wish to study the mean absorption in muscle tissue of two different drugs. A single test subject cannot receive both drugs since the drugs may interfere with each other. A simple random sample of 36 test subjects is given drug A and independently a simple random sample of 45 test subjects is given drug B. The sample mean and sample standard deviation in the absorption of each drug is found.

3 Motivation Suppose we wish to study the mean absorption in muscle tissue of two different drugs. A single test subject cannot receive both drugs since the drugs may interfere with each other. A simple random sample of 36 test subjects is given drug A and independently a simple random sample of 45 test subjects is given drug B. The sample mean and sample standard deviation in the absorption of each drug is found. Question: how do we test a hypothesis in this situation?

4 Welch s t Suppose a simple random sample of size n 1 is taken from a population with unknown mean µ 1 and unknown standard deviation σ 1. Additionally, a simple random sample of size n 2 is taken from a second population with unknown mean µ 2 and unknown standard deviation σ 2. If the two populations are normally distributed or if the sample sizes are sufficiently large (n 1 30 and n 2 30) then t = (x 1 x 2 ) (µ 1 µ 2 ) s 2 1 n 1 + s2 2 n 2 approximately follows Student s t-distribution where the degrees of freedom is the smaller of n 1 1 and n 2 1 and x 1 is the sample mean and s 1 is the sample standard deviation from the first population, and x 2 is the sample mean and s 2 is the sample standard deviation from the second population.

5 Hypothesis Testing Using Independent Samples Steps: 1. Determine the null and alternative hypotheses. Two-tailed Left-tailed Right-tailed H 0 : µ 1 = µ 2 H 0 : µ 1 = µ 2 H 0 : µ 1 = µ 2 H 1 : µ 1 µ 2 H 1 : µ 1 < µ 2 H 1 : µ 1 > µ 2 2. Select a level of significance α. 3. Compute the test statistic: t 0 = (x 1 x 2 ) (µ 1 µ 2 ) s 2 1 n 1 + s2 2 n 2 4. Use the classical or P-value approach to make a decision. 5. State the conclusion.

6 Example (P-Value Approach, 1 of 3) Many cheeses are produced in the shape of a wheel. Due to differences in the amount of water in a cheese, the weights of different types of cheese can vary. A random sample of 16 wheels of Gouda had a mean of 1.20 pounds and a standard deviation of 0.32 pounds. A random sample of 14 wheels of Brie had a mean of 1.05 pounds and a standard deviation of 0.25 pounds. At the α = 0.10 significance level test the claim that mean weights of wheels of Gouda and Brie are the same.

7 Example (P-Value Approach, 2 of 3) H 0 : µ 1 = µ 2 H 1 : µ 1 µ 2 (two-tailed test) α = 0.10, α/2 = 0.05, df = min{16 1, 14 1} = 13, t α/2 = ±1.771 Test statistic: t 0 = (x 1 x 2 ) (µ 1 µ 2 ) s 2 1 n 1 + s2 2 n 2 = ( ) (0) (0.32) (0.25)2 14 =

8 Example (P-Value Approach, 3 of 3) < t 0 = < Decision: do not reject H 0. P-value = 2P(t 0 > 1.771) > 0.10 Conclusion: the sample data do not warrant rejection of the claim that wheels of Gouda and Brie cheese weigh the same.

9 Example (Classical Approach, 1 of 3) The purchasing department for a regional supermarket chain is considering two sources from which to purchase 10-lb bags of potatoes. A random sample taken from each source gives the following results: Idaho Supers Idaho Best Number of bags Mean weight Sample variance At the 0.05 level of significance, is there a difference between the mean weights of the 10-lb bags of potatoes?

10 Example (Classical Approach, 2 of 3) H 0 : µ 1 = µ 2 H 1 : µ 1 µ 2 (two-tailed test) α = 0.05, α/2 = 0.025, df = min{100 1, 100 1} = 99, t α/2 = ±1.987 Test statistic: t 0 = (x 1 x 2 ) (µ 1 µ 2 ) s 2 1 n 1 + s2 2 n 2 = ( ) (0) = 2.561

11 Example (Classical Approach, 3 of 3) t Decision: reject H 0. Conclusion: the sample data warrant rejection of the claim that Idaho Supers and Idaho Best 10-lb bags weigh the same.

12 Example (Classical Approach, 1 of 3) Bausch & Lomb measured lenses from two different groups once by two different instruments and the differences were recorded. The data is as follows: Group A Group B At the α = 0.05 level of significance, test the claim that the mean of Group B is less than the mean of Group A.

13 Example (Classical Approach, 2 of 3) H 0 : µ A = µ B H 1 : µ A > µ B (right-tailed test) α = 0.05, df = min{25 1, 23 1} = 22, t α = Test statistic: t 0 = (x 1 x 2 ) (µ 1 µ 2 ) s 2 1 n 1 + s2 2 n 2 = (0.7 ( 5.6)) (0) (8.3) (12.2)2 23 = 2.074

14 Example (Classical Approach, 3 of 3) t Decision: reject H 0. Conclusion: the sample data support the claim that the mean of Group B is less than the mean of Group A.

15 Confidence Intervals Suppose a simple random sample of size n 1 is taken from a population with unknown mean µ 1 and unknown standard deviation σ 1. Additionally, a simple random sample of size n 2 is taken from a second population with unknown mean µ 2 and unknown standard deviation σ 2. If the two populations are normally distributed or if the sample sizes are sufficiently large (n 1 30 and n 2 30) then a (1 α) 100% confidence interval estimate of µ 1 µ 2 is given by Lower bound: (x 1 x 2 ) t α/2 s 2 1 n 1 + s2 2 n 2 Upper bound: (x 1 x 2 ) + t α/2 s 2 1 n 1 + s2 2 n 2 where t α/2 is computed using the smaller of n 1 1 and n 2 1 as the number of degrees of freedom.

16 Example (1 of 2) According to USA Today the longest average workweeks for non-supervisory employees in private industry are in mining (45.4 hours) and manufacturing (42.3 hours). The shortest average workweeks are in retail trade (29.0 hours) and services (32.4 hours). A study conducted in Missouri found these results: Industry n Mean Hours Std. Dev. Mining Manufacturing Determine the 95% confidence interval estimate for the difference in the average length of the workweek between mining and manufacturing.

17 Example (2 of 2) α = 0.05, α/2 = 0.025, df = min{15 1, 10 1} = 9, t α/2 = Margin of Error: E = t α/2 s 2 1 n 1 + s2 2 n 2 = (5.5) (4.9)2 10 = % Confidence Interval: ([x 1 x 2 ] E, [x 1 x 2 ] + E) = ( , ) = ( 0.8, 8.8)