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Chapter 22 Comparing 2 Proportions © 2006 W.H. Freeman and Company

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Objectives (Chapter 22) Comparing two proportions (comparing ABILITY in two different contexts) Comparing two independent samples Large-sample CI for two proportions Test of statistical significance

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Point Estimator: Two random samples are selected from two populations. The number of successes in each sample is recorded. The sample proportions are computed. 3 Sample 1 Sample size n 1 Number of successes x 1 Sample proportion Sample 1 Sample size n 1 Number of successes x 1 Sample proportion Sample 2 Sample size n 2 Number of successes x 2 Sample proportion Sample 2 Sample size n 2 Number of successes x 2 Sample proportion x n 1 1 ˆ p 1

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Comparing two independent samples We often need to estimate the difference p 1 – p 2 between two unknown population proportions based on independent samples. We can compute the difference between the two sample proportions and compare it to the corresponding, approximately normal sampling distribution model for

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Large-sample CI for two proportions For two independent samples of sizes n 1 and n 2 with sample proportion of successes 1 and 2 respectively, an approximate level C confidence interval for p 1 – p 2 is Use this method when C is the area under the standard normal curve between −z* and z*.

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Icing the Kicker: 95% Confidence Interval Football coaches often employ the “icing the kicker” strategy. To ice the kicker the opposing coach calls for a timeout just before the kicker attempts a field goal, hoping that the delay interrupts the kicker’s concentration and causes him to miss the kick. So the 95% CI is 0.024 ± 0.0672 = ( 0.0432, 0.0912) We are 95% confident that the interval 4.32% to 9.12% captures the true difference in the ABILITY of kickers to make a field goal when iced and their ABILITY to make a field goal when not iced. Because 0 is in the interval, we do not have convincing evidence that there is a significant difference in the ABILITY of kickers to make field goals when iced and when not iced. Standard error of the difference p 1 − p 2 : Made FGn Timeout (icing) p 1 15719779.7% No Timeout p 2 37748877.3% 95% CI

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7 Example: 95% confidence interval for p 1 – p 2 The age at which a woman gives birth to her first child may be an important factor in the risk of later developing breast cancer. An international study conducted by WHO selected women with at least one birth and recorded if they had breast cancer or not and whether they had their first child before their 30 th birthday or after. CancerSample Size Age at First Birth > 30 683322021.2% Age at First Birth <= 30 149810,24514.6% The parameter to be estimated is p1 – p2. p1 = cancer rate when age at 1 st birth >30 p2 = cancer rate when age at 1 st birth <=30 We estimate that the cancer rate when age at first birth > 30 is between.05 and.082 higher than when age <= 30.

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Beware!! Common Mistake !!! A common mistake is to calculate a one-sample confidence interval for p a one-sample confidence interval for p and to then conclude that p and p are equal if the confidence intervals overlap. This is WRONG because the variability in the sampling distribution for from two independent samples is more complex and must take into account variability coming from both samples. Hence the more complex formula for the standard error.

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INCORRECT Two single-sample 95% confidence intervals: The confidence interval for the rightie BA and the confidence interval for the leftie BA overlap, suggesting no significant difference between Ryan Howard’s ABILITY to hit right- handed pitchers and his ABILITY to hit left-handed pitchers. Rightie interval: (0.274, 0.366) HitsABphat(BA) Rightie126394.320 Leftie50222.225 Leftie interval: (0.170, 0.280) 0.095.023.167

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Reason for Contradictory Result 10

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Hypothesis Tests for p 1 p 2 If the null hypothesis is true, then we can rely on the properties of the sampling distribution of to estimate the probability of selecting 2 samples with proportions This test is appropriate when =0

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12 Do NFL Teams With Domes Have an Unfair Advantage? Do NFL teams that play their home games in a dome have an advantage over teams that do not play their home games in a dome? Home Wins Home Games Home field is dome 5079.633 Home field is not dome 103187.551 The parameter to be estimated is p1 – p2. p1 = home win rate for dome teams p2 = home win rate for non-dome teams Do not reject H 0 :p 1 – p 2 = 0. There is no significant difference between the home win rate for dome teams and the home win rate for non-dome teams.

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Gastric Freezing Gastric freezing was once a treatment for ulcers. Patients would swallow a deflated balloon with tubes, and a cold liquid would be pumped for an hour to cool the stomach and reduce acid production, thus relieving ulcer pain. The treatment was shown to be safe, significantly reducing ulcer pain, and was widely used for years. A randomized comparative experiment later compared the outcome of gastric freezing with that of a placebo: 28 of the 82 patients subjected to gastric freezing improved, while 30 of the 78 in the control group improved. Conclusion: The gastric freezing was no better than a placebo (P-value 0.69), and this treatment was abandoned. ALWAYS USE A CONTROL! H 0 : p gf - p placebo = 0 H a : p gf - p placebo > 0 p gf = proportion that receive relief from gastric freezing p placebo = proportion that receive relief using a placebo

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Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 11 Section 2 – Slide 1 of 25 Chapter 11 Section 2 Inference about Two Means: Independent.

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 11 Section 2 – Slide 1 of 25 Chapter 11 Section 2 Inference about Two Means: Independent.

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