1. Carrying Out Significance Tests
Lesson
Carrying Out Significance Tests

2. Knowledge Objectives
Identify and explain the four steps involved in formal hypothesis testing.

3. Construction Objectives
Using the Inference Toolbox, conduct a z test for a population mean.
Explain the relationship between a level α two-sided significance test for µ and a level 1 – α confidence interval for µ .
Conduct a two-sided significance test for µ using a confidence interval.

4. Vocabulary
Hypothesis – a statement or claim regarding a characteristic of one or more populations
Hypothesis Testing – procedure, base on sample evidence and probability, used to test hypotheses
Null Hypothesis – H0, is a statement to be tested; assumed to be true until evidence indicates otherwise
Alternative Hypothesis – H1, is a claim to be tested.(what we will test to see if evidence supports the possibility)
Level of Significance – probability of making a Type I error, α

5. Inference Toolbox
To test a claim about an unknown population parameter
Step 1: Hypotheses
Identify population of interest and parameter
State null and alternative hypotheses
Step 2: Conditions
Check 3 conditions (SRS, Normality, Independence); if met, then continue (or proceed under caution if not)
Step 3: Calculations
State test or test statistic
Use calculator to calculate test statistic and p-value
Step 4: Interpretation
Conclusion, connection and context about P-value and the hypotheses

6. Z Test for a Population Mean
Note: not usually seen in AP world (t-test more common)

7. Hypothesis Testing Approaches
Classical
Logic: If the sample mean is too many standard deviations from the mean stated in the null hypothesis, then we reject the null hypothesis (accept the alternative)
P-Value
Logic: Assuming H0 is true, if the probability of getting a sample mean as extreme or more extreme than the one obtained is small, then we reject the null hypothesis (accept the alternative).
Confidence Intervals
Logic: If the sample mean lies in the confidence interval about the status quo, then we fail to reject the null hypothesis

8. Reject null hypothesis, if
Classical Approach
-zα
-zα/2
zα/2 zα
Critical Regions
x – μ0
Test Statistic: z0 =
σ/√n
Reject null hypothesis, if
Left-Tailed
Two-Tailed
Right-Tailed
z0 < - zα
z0 < - zα/2 or
z0 > z α/2
z0 > zα

9. Reject null hypothesis, if
P-Value Approach z0
-|z0|
|z0| z0
P-Value is the area highlighted
x – μ0
Test Statistic: z0 =
σ/√n
Reject null hypothesis, if
P-Value < α

10. Confidence Interval Approach
x – zα/2 · σ/√n x + zα/2 · σ/√n
Lower Bound
Upper Bound μ0
Reject null hypothesis, if
μ0 is not in the confidence interval

11. Example 1
Assume that cell phone bills are normally distributed. A simple random sample of 12 cell phone bills finds x-bar = $ The mean in 2004 was $ Assume σ = $ Test if the average bill is different today at the α = 0.05 level. Use each approach.
Step 1: Hypothesis H0:  = $ (mean cell phone bill is unchanged) Ha:  ≠ $ (mean cell phone bill has changed)
Step 2: Conditions SRS: stated in problem Normality: population is normally distributed Independence: far more than 120 cell phones

12. Example 1: Classical Approach
A simple random sample of 12 cell phone bills finds x-bar = $ The mean in 2004 was $ Assume σ = $ Test if the average bill is different today at the α = 0.05 level. Use the classical approach.
not equal  two-tailed
X-bar – μ –
Z0 = = = = 2.69
σ / √n /√
Using alpha, α = 0.05 the shaded region are the rejection regions. The sample mean would be too many standard deviations away from the population mean. Since z0 lies in the rejection region, we would reject H0.
Zc = 1.96
Zc (α/2 = 0.025) = 1.96

13. Example 1: P-Value
A simple random sample of 12 cell phone bills finds x-bar = $ The mean in 2004 was $ Assume σ = $ Test if the average bill is different today at the α = 0.05 level. Use the P-value approach.
not equal  two-tailed
X-bar – μ –
Z0 = = = = 2.69
σ / √n /√
The shaded region is the probability of obtaining a sample mean that is greater than $65.014; which is equal to 2(0.0036) = Using alpha, α = 0.05, we would reject H0 because the p-value is less than α.
-Z0 = -2.69
P( z < Z0 = -2.69) = (double this to get p-value because its two-sided!)

14. Using Your Calculator: Z-Test
For classical or p-value approaches
Press STAT
Tab over to TESTS
Select Z-Test and ENTER
Highlight Stats
Entry μ0, σ, x-bar, and n from summary stats
Highlight test type (two-sided, left, or right)
Highlight Calculate and ENTER
Read z-critical and/or p-value off screen
From previous problem: z0 = and p-value =

15. Example 1: Confidence Interval
A simple random sample of 12 cell phone bills finds x-bar = $ The mean in 2004 was $ Assume σ = $ Test if the average bill is different today at the α = 0.05 level. Use confidence intervals.
Confidence Interval = Point Estimate ± Margin of Error
= x-bar ± Zα/2 σ / √n
= ± 1.96 (18.49) / √12
= ±
Zc (α/2) = 1.96
50.64
x-bar
54.55
75.48
The shaded region is the region outside the 1- α, or 95% confidence interval. Since the old population mean lies outside the confidence interval, then we would reject H0.

16. Using Your Calculator: Z-Interval
Press STAT
Tab over to TESTS
Select Z-Interval and ENTER
Highlight Stats
Entry σ, x-bar, and n from summary stats
Entry your confidence level (1- α)
Highlight Calculate and ENTER
Read confidence interval off of screen
If μ0 is in the interval, then FTR
If μ0 is outside the interval, then REJ
From previous problem: u0 = and interval (54.552, )
Therefore Reject

17. Example 2
National Center for Health has that the mean systolic blood pressure for males 35 to 44 years of age is The medical director for a company examines the medical records of 72 male executives in the age group and finds that their mean blood pressure is Is there evidence to support that their blood pressure is different?
Step 1: Hypothesis H0:  = 128 (younger male executives’ mean blood pressure is 128) Ha:  ≠ 128 (their blood pressure is different than 128)
Step 2: Conditions SRS: possible issue, but selected from free annual exams Normality: sample size large enough for CLT to apply Independence: have to assume more than 720 young male executives in the company (large company!!)

18. Example 2: P-Value Step 3: Calculations: Step 4: Interpretation:
X-bar – μ –
Z0 = = = = 1.092
σ / √n /√
From calculator: z = p-value =
More than 27% of the time with a sample size of 72 from the general population of males in the age group, we would get blood pressure values this extreme or more  Fail to reject H0; not enough evidence to say that this companies executives differ from the general population.

19. Example 3
Medical director for a large company institutes a health promotion campaign to encourage employees to exercise more and eat a healthier diet. One measure of the effectiveness of such a program is a drop in blood pressure. The director chooses a random sample of 50 employees and compares their blood pressures from physical exams given before the campaign and again a year later. The mean change in systolic blood pressure for these n=50 employees is -6. We take the population standard deviation to be σ=20. The director decides to use an α=0.05 significance level.

20. Example 3 cont Hypothesis: H0: Ha: Conditions: 1: 2: 3:
μ = 0 blood pressure is same
μ < 0 Regime lowers blood pressure
SRS -- stated in the problem statement
Normality -- unknown underlying distribution, but large sample size of 50 says x-bar will be Normally distributed (CLT)
Independence -- since sampling is w/o replacement; assume company has over 500 employees

21. Example 3 cont Calculations: Interpretation: x – μ0 -6 – 0
Z0 = =
σ/√n /√50
= -2.12
P-value = P(z < Z0) = P(z < -2.12)
= (unusual !)
P-value =  so only 1.7% of the time could we get a more extreme value. Since this is less than α = 0.05, we reject H0 and conclude that the mean difference in blood pressure is negative (so the regime may have worked!)

22. Example 4
The Deely lab analyzes specimens of a drug to determine the concentration of the active ingredient. The results are not precise and repeated measurements follow a Normal distribution quite closely. The analysis procedure has no bias, so the mean of the population of all measurements is the true concentration of the specimen. The standard deviation of this distribution was found to be σ= grams per liter.
A client sends a specimen for which the concentration of active ingredients is supposed to be 0.86%. Deely’s three analyses give concentrations of , , and Is their significant evidence at the 1% level that the concentration is not 0.86%? Use a confidence interval approach as well as z-test.

23. Example 4 cont Hypothesis: H0: Ha: Conditions: 1: 2: 3:
μ = 0.86 grams per liter
μ  0.86 grams per liter
SRS -- assume that each analyses represents an observation in a simple random sample
Normality -- stated in the problem that distribution is Normal
Independence -- assume each test is independent for the others

24. Example 4 cont Calculations: Interpretation: x – μ0 0.8404 – 0.86
Z0 = =
σ/√n /√3
= -4.99
P-value = 2P(z < Z0) = 2P(z < -4.99)
= (unusual !)
x-bar  z* σ / √n
 (0.0068) / √3 
(0.8303, )
P-value =  so only .04% of the time could we get a more extreme value. Since this is less than α = 0.01, we reject H0 and conclude that the mean concentration of active ingredients is not 0.86

25. Summary and Homework Summary Homework
A hypothesis test of means compares whether the true mean is either
Equal to, or not equal to, μ0
Equal to, or less than, μ0
Equal to, or more than, μ0
There are three equivalent methods of performing the hypothesis test
The classical approach
The P-value approach
The confidence interval approach
Homework
pg 713 ;
X-bar – μ
Z0 =
σ / √n