1. Hypothesis Testing

2. Suppose we believe the average systolic blood pressure of healthy adults is normally distributed with mean μ = 120 and variance σ 2 = 50. To test this assumption, we sample the blood pressure of 42 randomly selected adults. Sample statistics are Mean X = 122.4 Variance s 2 = 50.3 Standard Deviation s = √50.3 = 7.09 Standard Error = s / √n = 7.09 / √42 = 1.09

3. Central Limit Theorem The distribution of all sample means of sample size n from a Normal Distribution (μ, σ 2 ) is a normally distributed with Mean = μ Variance = σ 2 / n For our case: Mean μ = 120 Variance σ 2 / n = 50 / 42 = 1.19 Note: Theoretically we can test the hypothesis regarding the mean and the hypothesis regarding the variance; however one usually presumes the sample variances are stable from sample to sample and any one sample variance is an unbiased estimator of the population variance. As such, hypothesis testing is most frequently associated with testing assumptions regarding the population mean.

4. Hypothesis Testing Test the assumption H 0 : μ = 120 vs. H 1 : μ ≠ 120 using a level of significance α = 5% Note: If our sample came from the assumed population with mean μ = 120, then we would expect 95% of all sample means of sample size n = 42 to be within ± Z α/2 = ± 1.96

5.  / 2  5% Confidence Interval 95% Level of Significance  5%       95%

6. Calculate Upper and Lower Bounds on X X Lower = μ – Z α/2 (s /√n) = 120 – 1.96(1.09) =117.9 X Upper = μ + Z α/2 (s /√n) = 120 + 1.96(1.09) =122.1

7.  / 2  5% Confidence Interval 95% Level of Significance  5%       X Lower = 117.9 X Upper = 122.1 μ = 120 95%

8. Hypothesis Testing Comparisons Compare our sample mean X = 122.4 To the Upper and Lower Limits.

9.  / 2  5% Confidence Interval 95% Level of Significance  5%       X Lower = 117.9 X Upper = 122.1 μ = 120 95% X = 122.4

10. Hypothesis Testing Conclusions Note: Our sample mean X = 122.4 falls outside of the 95% Confidence Interval. We can reach one of two logical conclusions: One, that we expect this to occur for 2.5% of the samples from a population with mean μ = 120. Two, our sample came from a population with a mean μ ≠ 120. Since 2.5% = 1/40 is a rather “rare” event; we opt for the conclusion that our original null hypothesis is false and we reject H 0 : μ = 120 and therefore accept vs. H 1 : μ ≠ 120.

11. Confidence Interval 95% Level of Significance  5% X = 122.4 μ ≠ 120 Conclude μ ≠ 120

12. Alternate Method Rather than compare the sample mean to the 95% lower and upper bounds, one can use the Z Transformation for the sample mean and compare the results with ± Z α/2. Z 0 = ( X – μ  ) / (s / √n) = (122.4 – 120) / 1.09 = 2.20

13.  / 2  5% Confidence Interval 95% Level of Significance  5%       95% Z 0 = 2.20

14. Alternate Method Note: Since Z 0 = 2.20 value exceeds Z α/2 =1.96, we reach the same conclusion as before; Reject H 0 : μ = 120 and Accept H 1 : μ ≠ 120.

15. Alternate Method - Extended We can quantify the probability (p-Value) of obtaining a test statistic Z 0 at least as large as our sample Z 0. P( |Z 0 | > Z ) = 2[1- Φ (|Z 0 |)] p-Value = P( |2.20| > Z ) = 2[1- Φ (2.20)] p-Value = 2(1 – 0.9861) = 0.0278 = 2.8% Compare p-Value to Level of Significance If p-Value < α, then reject null hypothesis Since 2.8% < 5%, Reject H 0 : μ = 120 and conclude μ ≠ 120.