Inference About Two Means: Independent Samples

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

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.

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.

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.

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.

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 = (1.20 1.05) (0) (0.32) 2 16 + (0.25)2 14 = 1.4391

Example (P-Value Approach, 3 of 3) 1.350 < t 0 = 1.4391 < 1.771 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.

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 100 100 Mean weight 10.2 10.4 Sample variance 0.36 0.25 At the 0.05 level of significance, is there a difference between the mean weights of the 10-lb bags of potatoes?

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 = (10.2 10.4) (0) 0.36 100 + 0.25 100 = 2.561

Example (Classical Approach, 3 of 3) t 0 2.561 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.

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: 4 5 2 3 7 10 11 1 Group A 3 7 5 3 4 5 7 4 1 18 0 17 12 9 4 17 2 13 12 5 11 15 7 33 10 Group B 6 2 16 2 0 19 6 17 4 19 22 4 8 10 6 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.

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 = 1.717 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) 2 25 + (12.2)2 23 = 2.074

Example (Classical Approach, 3 of 3) t 0 2.074 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.

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.

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 15 47.5 5.5 Manufacturing 10 43.5 4.9 Determine the 95% confidence interval estimate for the difference in the average length of the workweek between mining and manufacturing.

Example (2 of 2) α = 0.05, α/2 = 0.025, df = min{15 1, 10 1} = 9, t α/2 = 2.262 Margin of Error: E = t α/2 s 2 1 n 1 + s2 2 n 2 = 2.262 (5.5) 2 15 + (4.9)2 10 = 4.8 95% Confidence Interval: ([x 1 x 2 ] E, [x 1 x 2 ] + E) = (47.5 43.5 4.8, 47.5 43.5 + 4.8) = ( 0.8, 8.8)