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Two-Sample Inference Procedures with Means

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## Two-Sample Inference Procedures with Means

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**Remember:**We will be interested in the difference of means, so we will use this to find standard error.**Suppose we have a population of adult men with a**mean height of 71 inches and standard deviation of 2.6 inches. We also have a population of adult women with a mean height of 65 inches and standard deviation of 2.3 inches. Assume heights are normally distributed. Describe the distribution of the difference in heights between males and females (male-female). Normal distribution with mx-y =6 inches & sx-y =3.471 inches**Female**Male 65 71 Difference = male - female 6 s = 3.471**What is the probability that the height of a randomly**selected man is at most 5 inches taller than the height of a randomly selected woman? b) What is the 70th percentile for the difference (male-female) in heights of a randomly selected man & woman? P((xM-xF) < 5) = normalcdf(-∞,5,6,3.471) = .3866 (xM-xF) = invNorm(.7,6,3.471) = 7.82**P((xm – xw)< 5) = .0573**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? 6.332 inches**Two-Sample Procedures with means**When we compare, what are we interested in? • The goal of these inference procedures is to compare the responses to two treatments or to compare the characteristics of two populations. • We have INDEPENDENT samples from each treatment or population**Assumptions:**• Have two SRS’s from the populations or two randomly assigned treatment groups • Samples are independent • Both distributions are approximately normally • Have large sample sizes • Graph BOTH sets of data • s’s unknown**H0: m1 - m2 = 0**Ha: m1 - m2 < 0 Ha: m1 - m2 > 0 Ha: m1 - m2 ≠ 0 Hypothesis Statements: H0: m1 = m2 Be sure to define BOTHm1 and m2! Ha: m1< m2 Ha: m1> m2 Ha: m1 ≠ m2**Formulas**Since in real-life, we will NOT know both s’s, we will do t-procedures.**Hypothesis Test:**Since we usually assume H0 is true, then this equals 0 – so we can usually leave it out**Degrees of Freedom**Option 1: use the smaller of the two values n1 – 1 and n2 – 1 This will produce conservative results – higher p-values & lower confidence. Option 2: approximation used by technology Calculator does this automatically!**VS**If there were such a thing as a personality test, do you think that the guys’ personalities would be different from the girls? Dr. Phil’s Survey**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. Assume the absorption time is normally distributed. Twelve people were randomly selected and given an oral 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: mean SD n Brand A 20.1 8.7 12 Brand B 18.9 7.5 12 Is there sufficient evidence that these drugs differ in the speed at which they enter the blood stream?**H0: mA= mB**Ha:mA= mB Have 2 independent randomly assigned treatments Given the absorption rate is normally distributed s’s unknown State assumptions! Hypotheses & define variables! Where mA is the true mean absorption time for Brand A & mB is the true mean absorption time for Brand B Formula & calculations Conclusion in context Since p-value > a, I fail to reject H0. There is not sufficient evidence to suggest that these drugs differ in the speed at which they enter the blood stream.**Suppose that the sample mean of Brand B is 16.5, then is**Brand B faster? No, I would still fail to reject the null hypothesis.**A modification has been made to the process for producing a**certain type of time-zero film (film that begins to develop as soon as the picture is taken). Because the modification involves extra cost, it will be incorporated only if sample data indicate that the modification decreases true average development time by more than 1 second. Should the company incorporate the modification? Original 8.6 5.1 4.5 5.4 6.3 6.6 5.7 8.5 Modified 5.5 4.0 3.8 6.0 5.8 4.9 7.0 5.7**Assume we have 2 independent SRS of film**Both distributions are approximately normal due to approximately symmetrical boxplots s’s unknown H0: mO- mM = 1 Ha:mO- mM > 1 Where mO is the true mean developing time for original film & mM is the true mean developing time for modified film Since p-value > a, I fail to reject H0. There is not sufficient evidence to suggest that the company incorporate the modification.**Called standard error**Confidence intervals:**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. Assume the absorption time is normally distributed. Twelve people were randomly selected and given an oral 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: mean SD n Brand A 20.1 8.7 12 Brand B 18.9 7.5 12 Find a 95% confidence interval difference in mean lengths of time required for bodily absorption of each brand.**Assumptions:**Have 2 independent randomly assigned treatments Given the absorption rate is normally distributed s’s unknown State assumptions! Think “Price is Right”! Closest without going over Formula & calculations From calculator df = 21.53, use t* for df = 21 & 95% confidence level Conclusion in context We are 95% confident that the true difference in mean lengths of time required for bodily absorption of each brand is between –5.685 minutes and 8.085 minutes.**In an attempt to determine if two competing brands of cold**medicine contain, on the average, the same amount of acetaminophen, twelve different tablets from each of the two competing brands were randomly selected and tested for the amount of acetaminophen each contains. The results (in milligrams) follow. Brand ABrand B 517, 495, 503, 491 493, 508, 513, 521 503, 493, 505, 495 541, 533, 500, 515 498, 481, 499, 494 536, 498, 515, 515 Compute a 95% confidence interval for the mean difference in amount of acetaminophen in Brand A and Brand B.**Input Brand A data into L1 and Brand B data into L2.**I am computing a 2-sample T interval for means at a 95% confidence level. Confidence level: (-28.48, -7.189) with 17 df I am 95% confident that the true difference in the mean amount of acetaminophen in Brand A is between 28.5 and 7.2 mg lower than in Brand B.**Note: confidence interval statements**• Matched pairs – refer to “mean difference” • Two-Sample – refer to “difference of means”**Used for two populations with the same variance**When you pool, you average the two-sample variances to estimate the common population variance. DO NOT use on AP Exam!!!!! Pooled procedures: We do NOT know the variances of the population, so ALWAYS tell the calculator NO for pooling!**Robustness:**• Two-sample procedures are more robust than one-sample procedures • BEST to have equal sample sizes! (but not necessary)