Two-Sample Inference Procedures with Means

Two-Sample Procedures with Means two differentGoal: Compare two different populations/treatments INDEPENDENT samples from each population/treatment

Remember: When combining two random variables X and Y, This formula only works if X and Y are independent

Suppose we have a population of adult men with a mean height of 71 inches and standard deviation 2.6 inches. We also have a population of adult women with a mean height of 65 inches and standard deviation 2.3 inches. Heights are normally distributed. Describe the distribution of the difference in heights between males and females (male – female). Normal distribution μ M-F = 6 inches & σ M-F = inches

7165 Female Male 6 Difference (male – female) σ = 3.471

a)What is the probability that a randomly selected man is at most 5 inches taller than a randomly selected woman? b) What is the 70th percentile for the difference (male – female) in heights of a randomly selected man and woman? P(x M – x F < 5) = normalcdf(-10^99, 5, 6, 3.471) (x M – x F ) = invNorm(.7, 6, 3.471) = 7.82 =.3866

Calculator Simulation! a) What is the probability that the mean height of 30 men is at most 5 inches taller than the mean height of 30 women? b) What is the 70th percentile for the difference (male – female) in mean heights of 30 men and 30 women? inches P(x M – x W < 5) =.057

Conditions for Two Means Two independent SRS's (or two randomly assigned treatments) Both samp. dist. are approx. normal –Both populations normal –Both n's > 30 –Both graphs are linear σ's known/unknown

Although… impossible In real-life, it's virtually impossible know both σ’s  Two-mean inference always uses t-procedures

Degrees of Freedom Option 1: Use the smaller df: n 1 – 1 or n 2 – 1  Using the larger one overestimates the collective sample sizes Option 2: Calculator's approximation Calculator does this automatically!

Confidence Interval for the Difference of Two Means Standard Error/ Deviation

Two competing headache remedies claim to give fast-acting relief. An experiment was performed to compare the mean lengths of time required for bodily absorption of brand A and brand B. Absorption time is normally distributed. Twelve people were randomly selected and given a dosage of brand A. Another 12 were randomly selected and given an equal dosage of brand B. The length of time in minutes for the drugs to reach a specified level in the blood was recorded. The results follow: meanSDn Brand A Brand B

a) Describe the sampling distribution of the differences in the mean speed of absorption (A – B). b) Construct a 95% confidence interval for the difference in mean lengths of time (A – B) required for bodily absorption of each brand. Normal; s = Conditions: 2 independent randomly assigned treatments Populations are normal  ’s unknown  t

We are 95% confident that the true difference in mean absorption time (A minus B) is between minutes and minutes. From calculator Think “Price is Right”: Closest df without going over If we made lots of intervals this way, 95% of them would contain the true difference in means.

A Subtle Distinction “mean difference”Matched pairs: “mean difference” “difference in means”Two-sample inference: “difference in means”

Hypothesis Statements H 0 : μ 1 – μ 2 = 0 H a : μ 1 – μ 2 < 0 H a : μ 1 – μ 2 > 0 H a : μ 1 – μ 2 ≠ 0 H 0 : μ 1 = μ 2 H a : μ 1 < μ 2 H a : μ 1 > μ 2 H a : μ 1 ≠ μ 2 BOTH Be sure to define BOTH μ 1 and μ 2 !

Test Statistic Since we assume H 0 is true, this part equals 0 – so we can leave it out

c) Is there sufficient evidence that the two brands differ in the speed at which they enter the bloodstream? Conditions: 2 independent randomly assigned treatments Populations are normal σ’s unknown  t

Since p-value > α, we fail to reject H 0. There is not sufficient evidence to suggest these drugs differ in their absorption time. H 0 :  A =  B H a :  A ≠  B Where μ A and μ B are the true mean absorption times p-value =.7210 α =.05

Pooling sameUsed for two populations with the same variance (σ 2 ) Pooling = Averaging the two s 2 to estimate σ 2 We almost never pool for means, since we don't know σ

Robustness more robustTwo-sample procedures: more robust than one-sample procedures Most robustMost robust with equal sample sizes (but not necessary!)

A modification has been made to the process for producing a certain type of film. Since the modification costs extra, it will be incorporated only if sample data indicate that the modification decreases the true average development time. At a significance level of 10%, should the company incorporate the modification? Original Modified Conditions: 2 independent SRS's of film Normal prob. plots linear  approx. normal sampling dist.’s σ's unknown  t

H 0 : μ O = μ M H a : μ O > μ M Where μ O and μ M are the true mean developing times for original and modified film Since p-value < α, we reject H 0. There is sufficient evidence to suggest the company should incorporate the modification. p-value =.076 α =.1