Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8SUMMARY: Why are these methods so important?STATISTICS IN A NUTSHELL• Click on image for full .pdf article• Links in article to access datasetsWomen in U.S. who have given birthμ > 25.4 POPULATION“Random Variable” X = Age at first birthmean μ = 25.4H0:“Null Hypothesis”μ < 25.4 That is, X ~ N(25.4, 1.5).μ < 25.4 standard deviation σ = 1.5μ > 25.4 Year 2010: Suppose we know that X follows a “normal distribution” (a.k.a. “bell curve”) in the population. Or, is the “alternative hypothesis” HA: μ ≠ 25.4 true?Random Samplemean Statistical Inference and Hypothesis Testing{x1, x2, x3, x4, … , x400}Study Question:Has “age at first birth” of women in the U.S. changed over time?• public education, awareness programs• socioeconomic conditions, etc.FORMULADoes the sample statistic tend to support H0, or refute H0 in favor of HA? i.e., either or ? (2-sided)Present: Is μ = 25.4 still true? x = 25.6x = 25.6IF H0 is true, then we would expect a random sample mean to lie between 25.253 and 25.547, with 95% probability.95% CONFIDENCE INTERVAL FOR µ25.74725.453In order to answer this question, we must account for the amount of variability of different values, from one random sample of n = 400 individuals to another.BASED ON OUR SAMPLE DATA, the true value of μ today is between 25.453 and 25.747, with 95% “confidence” (…akin to “probability”). We will see three things:25.54725.25395% ACCEPTANCE REGION FOR H0IF H0 is true, then we would expect a random sample mean that is at least 0.2 away from 25.4 (as ours was), to occur with probability .00383 (= 0.383%)… VERY RARELY! ,which is less t “P-VALUE” of our sampleTHEORYTHEORYEXPERIMENTEXPERIMENTxx = 25.6m= 25.4xm= 25.4x = 25.6x25.425.6IF H0 is true, then we would expect a random sample mean that is at least 0.2 away from 25.4 (as ours was), to occur with probability .00383 (= 0.383%)… VERY RARELY! ,which is less t IF H0 is true, then we would expect a random sample mean to lie between 25.253 and 25.547, with 95% probability.95% CONFIDENCE INTERVAL FOR µ25.74725.453In order to answer this question, we must account for the amount of variability of different values, from one random sample of n = 400 individuals to another.BASED ON OUR SAMPLE DATA, the true value of μ today is between 25.453 and 25.747, with 95% “confidence” (…akin to “probability”). We will see three things:25.54725.25395% ACCEPTANCE REGION FOR H0“P-VALUE” of our sampleHOW CAN WE USE ANY OR ALL OF THESE THREE OBJECTS TO TEST THE NULL HYPOTHESIS H0: µ = 30?HOW CAN WE USE ANY OR ALL OF THESE THREE OBJECTS TO TEST THE NULL HYPOTHESIS H0: µ = 30?xxx = 25.6m= 25.4xm= 25.4x = 25.625.425.6IF H0 is true, then we would expect a random sample mean that is at least 0.2 away from 25.4 (as ours was), to occur with probability .00383 (= 0.383%)… VERY RARELY! ,which is less t IF H0 is true, then we would expect a random sample mean to lie between 25.253 and 25.547, with 95% probability.95% CONFIDENCE INTERVAL FOR µ25.74725.453In order to answer this question, we must account for the amount of variability of different values, from one random sample of n = 400 individuals to another.BASED ON OUR SAMPLE DATA, the true value of μ today is between 25.453 and 25.747, with 95% “confidence” (…akin to “probability”). We will see three things:25.54725.25395% ACCEPTANCE REGION FOR H0“P-VALUE” of our sampleOur data value lies in the 5% REJECTION REGION.SIGNIFICANCE LEVEL (α)<Less than .05xxx = 25.6m= 25.4xm= 25.4x = 25.625.425.6IF H0 is true, then we would expect a random sample mean that is at least 0.2 away from 25.4 (as ours was), to occur with probability .00383 (= 0.383%)… VERY RARELY! ,which is less t IF H0 is true, then we would expect a random sample mean to lie between 25.253 and 25.547, with 95% probability.95% CONFIDENCE INTERVAL FOR µ25.74725.453BASED ON OUR SAMPLE DATA, the true value of μ today is between 25.453 and 25.747, with 95% “confidence” (…akin to “probability”). 25.54725.25395% ACCEPTANCE REGION FOR H0“P-VALUE” of our sampleIn order to answer this question, we must account for the amount of variability of different values, from one random sample of n = 400 individuals to another.We will see three things:Our data value lies in the 5% REJECTION REGION.SIGNIFICANCE LEVEL (α)<Less than .05FORMAL CONCLUSIONS: The 95% confidence interval corresponding to our sample mean does not contain the “null value” of the population mean, μ = 25.4. The 95% acceptance region for the null hypothesis does not contain the value of our sample mean, . The p-value of our sample, .00383, is less than the predetermined α = .05 significance level.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 years old) 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, by about 0.2 years.FORMAL CONCLUSIONS: The 95% confidence interval corresponding to our sample mean does not contain the “null value” of the population mean, μ = 25.4. The 95% acceptance region for the null hypothesis does not contain the value of our sample mean, . The p-value of our sample, .00383, is less than the predetermined α = .05 significance level.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 years old) 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, by about