CHAPTER 6 Statistical Inference & Hypothesis Testing 6.1 - One Sample Mean μ, Variance σ 2, Proportion π 6.2 - Two Samples Means, Variances, Proportions μ1 vs. μ2 σ12 vs. σ22 π1 vs. π2 6.3 - Multiple Samples Means, Variances, Proportions μ1, …, μk σ12, …, σk2 π1, …, πk

CHAPTER 6 Statistical Inference & Hypothesis Testing 6.1 - One Sample Mean μ, Variance σ 2, Proportion π 6.2 - Two Samples Means, Variances, Proportions μ1 vs. μ2 σ12 vs. σ22 π1 vs. π2 6.3 - Multiple Samples Means, Variances, Proportions μ1, …, μk σ12, …, σk2 π1, …, πk

CHAPTER 6 Statistical Inference & Hypothesis Testing 6.1 - One Sample Mean μ, Variance σ 2, Proportion π 6.2 - Two Samples Means, Variances, Proportions μ1 vs. μ2 σ12 vs. σ22 π1 vs. π2 6.3 - Multiple Samples Means, Variances, Proportions μ1, …, μk σ12, …, σk2 π1, …, πk

“Parameter Estimation” Women in U.S. who have given birth Example: One Mean POPULATION “Random Variable” X = Age (years) Objective 1: “Parameter Estimation” Improve this point estimate of μ to an “interval estimate” of μ, via the… Present: Assume that X follows a “normal distribution” in the population, with std dev σ = 1.5 yrs, but unknown mean μ = ? That is, X ~ N(μ, 1.5). Estimate the parameter value μ. “Sampling Distribution of ” standard deviation σ = 1.5 This is referred to as a “point estimate” of μ from the sample. mean μ = ??? Random Sample size n = 400 mean {x1, x2, x3, x4, … , x400} FORMULA

Population Distribution of X Sampling Distribution of for any sample size n. If X ~ N(μ, σ), then… X = Age of women in U.S. who have given birth μ “standard error” X μ standard deviation σ = 1.5 yrs

Sampling Distribution of μ | μ To achieve Objective 1 — obtain an “interval estimate” of μ — we first ask the following general question: “standard error” Suppose is any random sample mean. Find a “margin of error” (d) so that there is a 95% probability that the interval contains μ.

standard normal distribution d = (z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs Sampling Distribution of μ | μ “standard error” standard normal distribution N(0, 1) 0.95 0.025 0.025 Z -z.025 +z.025

IMPORTANT DEF’NS and FACTS standard normal distribution d = (z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs d is called the “95% margin of error” and is equal to the product of the “.025 critical value” (i.e., z.025 = 1.96) times the “standard error” (i.e., ). | μ The “confidence level” is 95%. The “significance level” is 5%. For any random sample mean the “95% confidence interval” is It contains μ with probability 95%. standard normal distribution N(0, 1) In this example, the 95% CI is 0.95 For instance, if a particular sample yields the 95% CI is (25.6 – 0.147, 25.6 + 0.147) = (25.543, 25.747) yrs. It contains μ with 95% “confidence.” 0.025 0.025 Z -z.025 +z.025

IMPORTANT DEF’NS and FACTS standard normal distribution d = (z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs d = (zα/2)(s.e.) d is called the “95% margin of error” and is equal to the product of the “.025 critical value” (i.e., z.025 = 1.96) times the “standard error” (i.e., ). “100(1 – α)% margin of error” | μ “α/2 zα/2) 1 – α The “confidence level” is 95%. 1 – α. The “significance level” is 5%. α. For any random sample mean the “95% confidence interval” is It contains μ with probability 95%. “100(1 – α)% “confidence interval” 1 – α. standard normal distribution N(0, 1) In this example, the 95% CI is 1 – α 0.95 For instance, if a particular sample yields the 95% CI is (25.6 – 0.147, 25.6 + 0.147) = (25.543, 25.747) yrs. It contains μ with 95% “confidence.” α/2 0.025 α/2 0.025 Z -z.025 -zα/2 +z.025 +zα/2

IMPORTANT DEF’NS and FACTS standard normal distribution d = (zα/2)(s.e.) d is called the “95% margin of error” and is equal to the product of the “.025 critical value” (i.e., z.025 = 1.96) times the “standard error” (i.e., ). “100(1 – α)% margin of error” | μ “α/2 zα/2) 1 – α The “confidence level” is 95%. 1 – α. The “significance level” is 5%. α. For any random sample mean the “95% confidence interval” is It contains μ with probability 95%. “100(1 – α)% “confidence interval” 1 – α. standard normal distribution N(0, 1) What happens if we change α? Example: α = .05, 1 – α = .95 Example: α = .10, 1 – α = .90 Example: α = .01, 1 – α = .99 1 – α 0.95 +2.575 -2.575 -1.96 +1.96 +1.645 -1.645 | α/2 0.025 α/2 0.025 Z Why not ask for α = 0, i.e., 1 – α = 1? Because then the critical values → ± ∞. -z.025 -zα/2 +z.025 +zα/2

IMPORTANT DEF’NS and FACTS 95% margin of error (z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs In this example, the 95% CI is For instance, if a particular sample yields the 95% CI is (25.6 – 0.147, 25.6 + 0.147) = (25.543, 25.747) yrs. It contains μ with 95% “confidence.” ? μ | standard normal distribution N(0, 1) 0.95 0.025 +1.96 -1.96 Z In principle, over the long run, the probability that a random interval contains μ will approach 95%. … etc… BUT….

IMPORTANT DEF’NS and FACTS 95% margin of error (z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs In this example, the 95% CI is For instance, if a particular sample yields the 95% CI is (25.6 – 0.147, 25.6 + 0.147) = (25.543, 25.747) yrs. It contains μ with 95% “confidence.” μ | standard normal distribution N(0, 1) 0.95 0.025 +1.96 -1.96 Z In practice, only a single, fixed interval is generated from a single random sample, so technically, “probability” does not apply. In principle, over the long run, the probability that a random interval contains μ will approach 95%. NOW, let us introduce and test a specific hypothesis… BUT….

Random Sample Statistical Inference and Hypothesis Testing Women in U.S. who have given birth Study Question: Has “age at first birth” of women in the U.S. changed over time? Statistical Inference and Hypothesis Testing POPULATION “Random Variable” X = Age at first birth “Null Hypothesis” Year 2010: Suppose we know that X follows a “normal distribution” (a.k.a. “bell curve”) in the population. Present: Is μ = 25.4 still true? H0: public education, awareness programs socioeconomic conditions, etc. Or, is the “alternative hypothesis” HA: μ ≠ 25.4 true? That is, X ~ N(25.4, 1.5). i.e., either or ? (2-sided) μ < 25.4 μ > 25.4 standard deviation σ = 1.5 μ < 25.4 μ > 25.4 Does the sample statistic tend to support H0, or refute H0 in favor of HA? mean μ = 25.4 Random Sample {x1, x2, x3, x4, … , x400} FORMULA mean

Objective 2: Hypothesis Testing… via Confidence Interval We have now seen: 95% CONFIDENCE INTERVAL FOR µ 25.543 25.747 “point estimate” for μ BASED ON OUR SAMPLE DATA, the true value of μ today is between 25.543 and 25.747, with 95% “confidence.” FORMAL CONCLUSIONS: The 95% confidence interval corresponding to our sample mean does not contain the “null value” of the population mean, μ = 25.4. Based on our sample data, we may reject the null hypothesis H0: μ = 25.4 in favor of the two-sided alternative hypothesis HA: μ ≠ 25.4, at the α = .05 significance level. INTERPRETATION: According to the results of this study, there exists a statistically significant difference between the mean ages at first birth in 2010 (25.4 yrs) and today, at the 5% significance level. Moreover, the evidence from the sample data suggests that the population mean age today is older than in 2010, rather than younger. NOTE THAT THE CONFIDENCE INTERVAL ONLY DEPENDS ON THE SAMPLE, NOT A SPECIFIC NULL HYPOTHESIS!!!

Objective 2: Hypothesis Testing… via Confidence Interval What if…? We have now seen: 95% CONFIDENCE INTERVAL FOR µ 25.347 25.053 95% CONFIDENCE INTERVAL FOR µ 25.747 25.543 “point estimate” for μ “point estimate” for μ BASED ON OUR SAMPLE DATA, the true value of μ today is between 25.543 and 25.747, with 95% “confidence.” BASED ON OUR SAMPLE DATA, the true value of μ today is between 25.053 and 25.347, with 95% “confidence.” ? FORMAL CONCLUSIONS: The 95% confidence interval corresponding to our sample mean does not contain the “null value” of the population mean, μ = 25.4. Based on our sample data, we may reject the null hypothesis H0: μ = 25.4 in favor of the two-sided alternative hypothesis HA: μ ≠ 25.4, at the α = .05 significance level. INTERPRETATION: According to the results of this study, there exists a statistically significant difference between the mean ages at first birth in 2010 (25.4 yrs) and today, at the 5% significance level. Moreover, the evidence from the sample data suggests that the population mean age today is older than in 2010, rather than younger. NOTE THAT THE CONFIDENCE INTERVAL ONLY DEPENDS ON THE SAMPLE, NOT A SPECIFIC NULL HYPOTHESIS!!! younger than in 2010, rather than older.

IF the null hypothesis H0: μ = 25.4 is indeed true, then… Objective 2: Hypothesis Testing… via Acceptance Region Objective 2: Hypothesis Testing… via Confidence Interval IF the null hypothesis H0: μ = 25.4 is indeed true, then… 95% margin of error (z.025)(s.e.) = (1.96)(.075 yrs) = 0.147 yrs Sampling Distribution of “Null” Distribution of μ “standard error” … and out here… …with 5% probability. 0.95 … we would expect a random sample mean to lie in here, with 95% probability… 0.025 0.025 95% ACCEPTANCE REGION FOR H0 | μ = 25.4 25.253 25.547 25.4

Objective 2: Hypothesis Testing… via Acceptance Region Our data value lies in the 5% REJECTION REGION. We have now seen: 95% ACCEPTANCE REGION FOR H0 25.253 25.547 IF H0 is true, then we would expect a random sample mean to lie between 25.253 and 25.547, with 95% probability. FORMAL CONCLUSIONS: The 95% acceptance region for the null hypothesis does not contain the sample mean of Based on our sample data, we may reject the null hypothesis H0: μ = 25.4 in favor of the two-sided alternative hypothesis HA: μ ≠ 25.4, at the α = .05 significance level. INTERPRETATION: According to the results of this study, there exists a statistically significant difference between the mean ages at first birth in 2010 (25.4 yrs) and today, at the 5% significance level. Moreover, the evidence from the sample data suggests that the population mean age today is older than in 2010, rather than younger. NOTE THAT THE ACCEPTANCE REGION ONLY DEPENDS ON THE NULL HYPOTHESIS, NOT ON THE SAMPLE!!!

IF the null hypothesis H0: μ = 25.4 is indeed true, then… Objective 2: Hypothesis Testing… via Acceptance Region Objective 2: Hypothesis Testing… via “p-value” - measures the strength of the rejection IF the null hypothesis H0: μ = 25.4 is indeed true, then… > 1.96 what is the probability of obtaining a random sample mean that is as, or more, extreme than the one actually obtained? i.e., 0.2 yrs OR MORE away from μ = 25.4, ON EITHER SIDE (since the alternative hypothesis is 2-sided)? < .05 statistically significant | μ = 25.4 25.253 25.547 95% ACCEPTANCE REGION FOR H0 0.025 0.95 0.00383

IF the null hypothesis H0: μ = 25.4 is indeed true, then… Objective 2: Hypothesis Testing… via Acceptance Region Objective 2: Hypothesis Testing… via “p-value” - measures the strength of the rejection IF the null hypothesis H0: μ = 25.4 is indeed true, then… > 1.96 what is the probability of obtaining a random sample mean that is as, or more, extreme than the one actually obtained? i.e., 0.2 yrs OR MORE away from μ = 25.4, ON EITHER SIDE (since the alternative hypothesis is 2-sided)? < .05 α statistically significant | μ = 25.4 25.253 25.547 95% ACCEPTANCE REGION FOR H0 0.025 0.95 1 – α α / 2 100(1 – α)% ACCEPTANCE REGION FOR H0 -zα/2 +zα/2 0.00383

p-value If p-value < , then reject H0; significance!  = .05 If p-value < , then reject H0; significance! ... But interpret it correctly! p-value

~ Summary of Hypothesis Testing for One Mean ~ Assume the population random variable is normally distributed, i.e., X  N(μ, σ). NULL HYPOTHESIS H0: μ = μ0 (“null value”) Test null hypothesis at significance level α. ALTERNATIVE HYPOTHESIS HA: μ  μ0 i.e., either μ < μ0 or μ > μ0 (“two-sided”) CONFIDENCE INTERVAL Compute the sample mean Compute the 100(1 – α)% “margin of error” = (critical value)(standard error) Then the 100(1 – α)% CI = Formal Conclusion: Reject null hypothesis at level α, Statistical significance! Otherwise, retain it. zα/2 if CI does not contain μ0.

~ Summary of Hypothesis Testing for One Mean ~ Assume the population random variable is normally distributed, i.e., X  N(μ, σ). NULL HYPOTHESIS H0: μ = μ0 (“null value”) Test null hypothesis at significance level α. ALTERNATIVE HYPOTHESIS HA: μ  μ0 i.e., either μ < μ0 or μ > μ0 (“two-sided”) ACCEPTANCE REGION Compute the sample mean Compute the 100(1 – α)% “margin of error” = (critical value)(standard error) Then the 100(1 – α)% AR = Formal Conclusion: Reject null hypothesis at level α, Statistical significance! Otherwise, retain it. zα/2 if AR does not contain

~ Summary of Hypothesis Testing for One Mean ~ Assume the population random variable is normally distributed, i.e., X  N(μ, σ). NULL HYPOTHESIS H0: μ = μ0 (“null value”) Test null hypothesis at significance level α. ALTERNATIVE HYPOTHESIS HA: μ  μ0 i.e., either μ < μ0 or μ > μ0 (“two-sided”) p-value Compute the sample mean Compute the z-score If +, then the p-value = 2 P(Z ≥ z-score ). If –, then the p-value = 2 P(Z ≤ z-score ). Formal Conclusion: Reject null hypothesis Statistical significance! Otherwise, retain it. Remember: “The smaller the p-value, the stronger the rejection, and the more statistically significant the result.” Z ~ N(0, 1) z-score if p < α.

statistically significant Objective 2: Hypothesis Testing… 1-sided tests The alternative hypothesis usually reflects the investigator’s belief! 2-sided test H0: μ = 25.4 HA: μ  25.4 p-value In this case,  = .05 is split evenly between the two tails, left and right. 1-sided tests “Right-tailed” H0: μ  25.4 HA: μ > 25.4 Here, all of  = .05 is in the right tail. < .05 The alternative hypothesis usually reflects the investigator’s belief! statistically significant | μ = 25.4 95% ACCEPTANCE REGION FOR H0 0.95 25.253 25.547 0.025 .00383

? < .05 statistically significant Use 1-sided tests sparingly! Objective 2: Hypothesis Testing… 1-sided tests 2-sided test H0: μ = 25.4 HA: μ  25.4 p-value In this case,  = .05 is split evenly between the two tails, left and right. 1-sided tests “Right-tailed” H0: μ  25.4 HA: μ > 25.4 Here, all of  = .05 is in the right tail. < .05 statistically significant Use 1-sided tests sparingly! 0.95 0.05 0.025 0.025 .00383 .00383 95% ACCEPTANCE REGION FOR H0 95% ACCEPTANCE REGION FOR H0 | μ = 25.4 25.253 25.547 ?

? < .05 >> .05 statistically significant p-value 25.2 25.2 Objective 2: Hypothesis Testing… 1-sided tests 2-sided test H0: μ = 25.4 HA: μ  25.4 p-value 25.2 25.2 In this case,  = .05 is split evenly between the two tails, left and right. 25.2 1-sided tests “Right-tailed” H0: μ  25.4 HA: μ > 25.4 Here, all of  = .05 is in the right tail. “Left-tailed” H0: μ  25.4 HA: μ < 25.4 Here, all of  = .05 is in the left tail. < .05 >> .05 The alternative hypothesis usually reflects the investigator’s belief! strong support of null hypothesis statistically significant 0.95 0.05 95% ACCEPTANCE REGION FOR H0 | μ = 25.4 ?

Subject: basic calculation of p-values for z-test STATBOT 301 Subject: basic calculation of p-values for z-test Calculate… from H0 Test Statistic “z-score” = sign of z-score? 1 – table entry table entry HA: μ ≠ μ0? HA: μ < μ0 HA: μ > μ0 2 × table entry 2 × (1 – table entry) – + Calculate… from H0 Test Statistic “z-score” = HA: μ ≠ μ0? HA: μ < μ0 HA: μ > μ0 1 – table entry table entry sign of z-score? 2 × table entry 2 × (1 – table entry) – +

“Parameter Estimation” Women in U.S. who have given birth Example: One Mean POPULATION “Random Variable” X = Age (years) Objective 1: “Parameter Estimation” Improve this point estimate of μ to an “interval estimate” of μ, via the… Present: Assume that X follows a “normal distribution” in the population, with std dev σ = 1.5 yrs, but unknown mean μ = ? That is, X  N(μ, 1.5). Estimate the parameter value μ. “Sampling Distribution of “ But how do we know that the variance is the same as in 2010? standard deviation σ = 1.5 This is referred to as a “point estimate” of μ from the sample. mean μ = ??? Random Sample size n = 400 {x1, x2, x3, x4, … , x400} FORMULA mean

… which leads us to… (1.5)2 = 2.25 in our example

All have postively-skewed tails. HOWEVER……

In practice,  2 is almost never known, so the sample variance s 2 is used as a substitute in all calculations!