1. Parameter Estimation

2. Parameter Estimation Estimating Population Mean μ Based on Sample Data

3. Estimating Population Mean m
Sample Data Sample Mean X = Sample Standard Deviation s = = Sample Size = 42
Assume sample was drawn from mormally istributed population with mena u and standard deviation.
From the 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 ± Zα/2 = ± 1.96

4. PArmater
Suppose we believe the average systolic blood pressure of healthy adults is normally distributed with mean μ = 120 and variance σ2 = 50.
To test this assumption, we sample the blood pressure of 42 randomly selected adults. Sample statistics are
Mean X = 122.4
Variance s2 = 50.3
Standard Deviation s = √50.3 = 7.09
Standard Error = s / √n = 7.09 / √42 = 1.09

5. Central Limit Theorem
The distribution of all sample means of sample size n from a Normal Distribution (μ, σ2) is a normally distributed with Mean = μ Variance = σ2 / n
For our case:
Mean μ = 120 Variance σ2 / 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.

6. Hypothesis Testing Test the assumption
H0: μ = 120 vs. H1: 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 ± Zα/2 = ± 1.96

7. Level of Significance a = 5%
Confidence Interval 95%
Level of Significance a = 5%
95%
a / 2 = 2.5%
a / 2 = 2.5%
-Za/2 = -1.96
+Za/2 = +1.96

8. Calculate Upper and Lower Bounds on X
XLower = μ – Zα/2 (s /√n) = 120 – 1.96(1.09) =117.9
XUpper = μ + Zα/2 (s /√n) = (1.09) =122.1

9. Level of Significance a = 5%
Confidence Interval 95%
Level of Significance a = 5%
95%
a / 2 = 2.5%
a / 2 = 2.5%
μ = 120
-Za/2 = -1.96
+Za/2 = +1.96
X Lower = 117.9
X Upper = 122.1

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

11. Level of Significance a = 5%
Confidence Level 95%
Level of Significance a = 5%
95%
a / 2 = 2.5%
a / 2 = 2.5%
X = 122.4
-Za/2 = -1.96
+Za/2 = +1.96
X Lower = 120.2
X Upper = 124.5

12. Hypothesis Testing Conclusions
Note: Our sample mean X = 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 H0: μ = 120 and therefore accept vs. H1: μ ≠ 120 .

13. Level of Significance a = 5%
Confidence Interval 95%
Level of Significance a = 5%
μ ≠ 120
X = 122.4
Conclude μ ≠ 120

14. 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 ± Zα/2.
Z0 = ( X – μ ) / (s / √n) = (122.4 – 120) / 1.09 = 2.20

15. Level of Significance a = 5%
Confidence Interval 95%
Level of Significance a = 5%
95%
a / 2 = 2.5%
a / 2 = 2.5%
Z0 = 2.20
-Za/2 = -1.96
+Za/2 = +1.96

16. Alternate Method Note: Since Z0= 2.20 value exceeds Zα/2 =1.96, we
reach the same conclusion as before;
Reject H0: μ = 120 and Accept H1: μ ≠ 120.

17. Alternate Method - Extended
We can quantify the probability (p-Value) of
obtaining a test statistic Z0 at least as large as our sample Z0.
P( |Z0| > Z ) = 2[1- Φ (|Z0|)]
p-Value = P( |2.20| > Z ) = 2[1- Φ (2.20)]
p-Value = 2(1 – ) = = 2.8%
Compare p-Value to Level of Significance
If p-Value < α, then reject null hypothesis
Since 2.8% < 5%, Reject H0: μ = 120 and conclude μ ≠ 120.