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Two-Sample Inference Procedures with Means
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Two-Sample Procedures with means
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
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Remember: We will be interested in the difference of means, so we will use this to find standard error.
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mx-y =6 inches & sx-y =3.471 inches
Suppose we have a population of adult men with a mean height of 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
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71 65 Female Male 6 Difference = male - female s = 3.471
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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
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Do 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? P((xm – xw)< 5) = .0573 6.332 inches
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Assumptions: Have two SRS’s from the populations or two randomly assigned treatment groups Samples are independent Populations > 10(n1+n2) Both distributions are approximately normally Have large sample sizes Graph BOTH sets of data
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Formulas Since in real-life, we will NOT know both s’s, we will do t-procedures.
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Calculator does this automatically!
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!
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Confidence intervals:
Called standard error
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Pooled procedures: 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!!!!! We do NOT know the variances of the population, so ALWAYS tell the calculator NO for pooling!
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Two competing headache remedies claim to give fast-acting relief
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 Brand B Describe the shape & standard error for sampling distribution of the differences in the mean speed of absorption. (answer on next screen)
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Describe the sampling distribution of the differences in the mean speed of absorption.
Find a 95% confidence interval difference in mean lengths of time required for bodily absorption of each brand. (answer on next screen) Normal distribution with S.E. = 3.316
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Closest without going over
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 minutes.
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Note: confidence interval statements
Matched pairs – refer to “mean difference” Two-Sample – refer to “difference of means”
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Dr. Phil’s Survey VS
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Hypothesis Statements:
H0: m1 = m2 H0: m1 - m2 = 0 Ha: m1 - m2 < 0 Ha: m1 - m2 > 0 Ha: m1 - m2 ≠ 0 Be sure to define BOTH m1 and m2! Ha: m1< m2 Ha: m1> m2 Ha: m1 ≠ m2
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Hypothesis Test: Since we usually assume H0 is true, then this equals 0 – so we can usually leave it out
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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 Brand B Is there sufficient evidence that these drugs differ in the speed at which they enter the blood stream?
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State assumptions! Hypotheses & define variables! 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! H0: mA= mB Ha:mA= mB 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.
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Suppose that the sample mean of Brand B is 16
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.
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Robustness: Two-sample procedures are more robust than one-sample procedures BEST to have equal sample sizes! (but not necessary)
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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 Modified
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H0: mO- mM = 1 Ha:mO- mM > 1
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.
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