1. Click the mouse button or press the Space Bar to display the answers.
5-Minute Check on Chapter 9-3a
How do we use a confidence interval on a test of population mean?
In a matched pair design, what is the data that we use?
What is the “power” of an inference test?
What type of error is associated with the power of a test?
Two-side: If μ0 is in the interval, then FTR
If μ0 is outside the interval, then REJ
The data is the difference data (after – before)
Power is defined to be: 1 – P(β)
Type 2 error: FTR H0 when Ha is true
Click the mouse button or press the Space Bar to display the answers.

2. Review of Testing a Claim
Lesson 9 - R
Review of Testing a Claim

3. Objectives Explain the logic of significance testing.
List and explain the differences between a null hypothesis and an alternative hypothesis.
Discuss the meaning of statistical significance.
Use the Inference Toolbox to conduct a large sample test for a population mean.
Compare two-sided significance tests and confidence intervals when doing inference.
Differentiate between statistical and practical “significance.”
Explain, and distinguish between, two types of errors in hypothesis testing.
Define and discuss the power of a test.

4. Vocabulary
none new

5. What you Learned
State the null and alternative hypotheses in a testing situation when the parameter in question is a population mean µ.
Explain in nontechnical language the meaning of the P-value when you are given the numerical value of P for a test.
Calculate the one-sample z-statistic and the P-value for both one-sided and two-sided tests about the mean µ of a Normal population.
Assess statistical significance at standard levels α by comparing P to α.
Recognize that significance testing does not measure the size or importance of an effect.
Recognize when you can use the z test and when the data collection design or a small sample from a skewed population makes it inappropriate.
Explain Type I error, Type II error, and power in a significance-testing problem.

6. 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

7. Determining Ho and Ha
Ho – is the status quo; what the situation is currently the claim made by the manufacturer
Ha – is the alternative that you are testing; the new idea the thing that proves the claim false
H0 and Ha always refer to population parameters

8. Inference Toolbox Step 1: Hypothesis Step 2: Conditions
Identify population of interest and parameter
State H0 and Ha
Step 2: Conditions
Check appropriate conditions
Step 3: Calculations
State test or test statistic
Use calculator to calculate test statistic and p-value
Step 4: Interpretation
Interpret the p-value (fail-to-reject or reject)
Don’t forget 3 C’s: conclusion, connection and context

9. Conditions for Significance Tests
SRS
simple random sample from population of interest
Independence
Population, N, such that N > 10n
Normality
For means:
population normal or large enough sample size for CLT to apply or use t-procedures
t-procedures: boxplot or normality plot to check for shape and any outliers (outliers are a killer)
For proportions: np ≥ 10 and n(1-p) ≥ 10

10. 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
X-bar – μ
Z0 =
σ / √n

11. Hypothesis Testing: Four Outcomes
Reality
H0 is True
H1 is True
Conclusion
Do Not Reject H0
Correct Conclusion
Type II Error
Reject H0
Type I Error
H0: the defendant is innocent
H1: the defendant is guilty
Type I Error (α): convict an innocent person
Type II Error (β): let a guilty person go free
Note: a defendant is never declared innocent; just not guilty
decrease α  increase β
increase α  decrease β

12. Increasing the Power of a Test
Four Main Methods:
Increase significance level, 
Consider a particular alternative that is farther away from 
Increase the sample size, n, in the experiment
Decrease the population (or sample) standard deviation, σ
Only increasing the sample size and the significance level are under the control of the researcher

13. Summary and Homework Summary Homework
Inference testing has an H0 and an Ha to evaluate
H0 is the status quo and cannot be proven
Ha is the claim that’s being tested
Small p-values (tails), or p < , are in favor of Ha
P(Type I error – reject H0 when H0 is true) = 
P(Type II error – FTR H0 when Ha is true) = 
Power (of a test) = 1 - 
Increase power by increasing n or 
Homework
pg 738 – 9; – 74

14. Problem 1
When performing inference procedures for means we sometimes use the normal distribution and sometimes use Student’s t distribution. How do you decide which to use?
We use normal distributions in inference procedures for population proportions always and for population means if we know the population standard deviation (very rare).
We use the Student’s t distribution for population means when we don’t know the population standard deviation and have to use the sample standard deviation as its estimate.

15. Problem 2a
The blood sugar levels of 7 randomly selected rabbits of a certain species are measured 90 minutes after injection of 0.8 units of insulin. The data (in mg/100 ml) are: 45, 34, 32, 48, 39, 45, 51. The researcher would like to use this sample to make inferences about the mean blood sugar level for the population of rabbits.
(a) List the conditions (assumptions) that must be satisfied in order to compute confidence intervals or perform hypothesis tests. Also, provide information to convince me that the condition is satisfied.
Conditions:
SRS – problem states random selection, so assume SRS is ok Independence – assumption is reasonable from problem (more than 70 rabbits)
Normality – n = 7 is too small for CLT to apply so box-plot of data is required to check if normality is not a good assumption and to see if there are any outliers! [no outliers, symmetry not great, normality plot ok]

16. Problem 2b
The blood sugar levels of 7 randomly selected rabbits of a certain species are measured 90 minutes after injection of 0.8 units of insulin. The data (in mg/100 ml) are: 45, 34, 32, 48, 39, 45, 51. The researcher would like to use this sample to make inferences about the mean blood sugar level for the population of rabbits.
(b) Calculate a 90% confidence interval for the mean blood sugar level in the population of all rabbits of this species 90 minutes after insulin injection.
Calculations: n = 7, x-bar = 42 mg/100 ml, s = mg/100 ml, and α/2 = .05
CI = x-bar  t.95,6(s/√n) = 42  1.943(7.165/√7) = 42  5.262
From calculator: [36.738, ]

17. Problem 3
The mean length of the incision used in arthroscopic knee surgery has been 2.0 cm. An article in a recent medical journal reports that a new procedure can reduce the average length of the incision used in such surgeries. The authors of the article reported that in a sample of 36 such surgeries in which the new technique was used, the average incision length was 1.96 cm, with a standard deviation of 0.13 cm. Perform a test at the α = .05 level to determine whether the mean length of incision used in the new procedure is significantly less than 2.0 cm. Assume that any conditions needed for inference are satisfied.

18. Problem 3 Hypothesis: μ = average incision length of old procedure
H0: μ = 2.0 cm no change
Ha: μ < 2.0 cm new procedure better (One sided test)
Conditions: “Assume that any conditions satisfied”
Calculations: μ = 2.0 cm, n = 36, x-bar = 1.96 cm, s = 0.13 cm, and α = .05
TS = (1.96 – 2.0)/(0.13/√36) =
From calculator: TS = p-value =
Interpretation: Since p < , then we have evidence to reject H0 in favor of Ha. The new procedure has smaller average length incision (scars) than the old procedure. Statistically significant result, but not much real practical significance (0.04 cm improvement or about 1/50 of an inch).

19. Problem 4
A company institutes an exercise break for its workers to see if this will improve job satisfaction. A questionnaire designed to assess worker satisfaction was given to 10 randomly selected workers before and after implementation of the exercise program. These scores are provided in the table below: Perform a significance test to determine whether the exercise program was effective in improving job satisfaction. You should state the conditions necessary for the test you perform to be valid, but you do not need to check these conditions. Organize your work clearly and be sure to show all steps of the test below.
Worker # 1 2 3 4 5 6 7 8 9 10
Score before 34 28 29 45 26 27 24 15
Score after 33 36 50 41 37 39 21 20

20. Problem 4 cont
Worker # 1 2 3 4 5 6 7 8 9 10
Score before 34 28 29 45 26 27 24 15
Score after 33 36 50 41 37 39 21 20
Difference -1 -4 11 14
Hypothesis: μ = average difference between after and before program
H0: μdiff = program ineffective
Ha: μdiff > program improves job sat (One sided test)
Conditions: SRS – assume from random; Independent – Employees > 100 (?); Normal – no CLT, but box-plot and normality plots OK
Calculations: μ = 0, n = 10, x-bar = 8.94, s = 6.749, and α = .05
TS = (8.94 – 0)/(6.749/√10) =
From calculator: t = and p-value =
Interpretation: Since p-value is < , we reject H0 in favor of Ha and conclude that the exercise program improves job satisfaction.