Statistics Presentation week 7

STATISTICS & PROBABILITY

Hypothesis Testing

What is a Hypothesis?

I assume the mean GPA of this class

is 3.5!

• an assumption about

the population

parameter

• an educated guess

about the population

parameter

Hypotheses Testing: This is the process of making an inference or generalization on

population parameters based on the results of the study on samples.

Statistical Hypotheses: It is a guess or prediction made by the researcher

regarding the possible outcome of the study.

Reject?Accept?

Hypotheses Testing

is deciding between what is REALITY and what

is COINCIDENCE!

Null Hypothesis (Ho): is always hoped to be rejectedAlways contains “=“ sign

Alternative Hypothesis (Ha): •Challenges Ho•Never contains “=“ sign•Uses “< or > or “•It generally represents the idea which the researcher wants to prove.

Types of Statistical Hypotheses

The Null Hypothesis: Ho

Ha: The average GPA of this class is

a) higher than 3.5 (Ha: 3.5)

b) lower than 3.5 (Ha: 3.5)

c) not equal to 3.5 (Ha: 3.5)

The Alternative Hypothesis: Ha

Ex. Ho: The average GPA of this class is 3.5

H0: = 3.5

Types of Hypotheses Tests

1. One-tailed left directional test– this is used if Ha uses symbol

Rejection regionArea = 0.05

Acceptanceregion

= 0.05

Critical value is obtained

from the table

Types of Hypotheses Tests2. One-tailed right directional test

– this is used if Ha uses symbol

Rejection regionArea= 0.05

Acceptanceregion

Critical value is obtained

from the table = 0.05

Types of Hypotheses Tests

3. Two-tailed test: Non-directional – this is used if Ha uses symbol

Rejection region Rejection regionArea=.025 Area=.025

Acceptanceregion

= 0.05/2Critical value is

obtainedfrom the table

Level of Significance, and the Rejection Region

means the probability of being right is 95% , and the probability of being wrong is 5%. So what is = 0.01?

.

05.0

Acceptance Region

Rejection regionArea is 0.05


means the researcher is taking a 1% risk of being

wrong and a 99% risk of being right.So, what is = 0.05?

0.01

Acceptance Region

Rejection regionArea is 0.01


means the probability of committing Type I error is 5%.

Acceptanceregion

= 0.05/2= 0.025

Rejection region Rejection regionArea=.025 Area=.025

= 0.05, since it is 2-T, then


To summarize:

means the researcher is taking a 1% risk of being

wrong and a 99% risk of being right.So, what is = 0.05?

means the probability of being right is 95% and the probability of being wrong is 5%. So what is = 0.01?

means the probability of committing Type I error is 5%.

So what is = 0.01?

© 1984-1994 T/Maker Co.

Errors in Hypothesis Testing

Errors

Errors in Decisions

Errors in Conclusions

Type I ( error )Rejecting a true

Ho!

Ho: ERAP is not guilty

If the court convictsERAP, when in facthe is not guilty, the

court commitsType I error!

Type I is the same as the or the level of significance.

Errors in Hypothesis Testing

Errors

Errors in Decisions

Errors in Conclusions

Ho: ERAP is not guilty

If the court acquitsERAP, when in fact

he is guilty, thecourt commitsType II error!

Type II ( error )Accepting a false

Ho!

© 1984-1994 T/Maker Co.

Decisions made regarding Ho(Reject Ho/Do not reject Ho)

If we reject Ho, it means it is wrong!

If we do not reject Ho, it doesn’t mean it is correct,

we just don’t have enough evidence

to reject it!

Testing the Significance of Difference Between Means

Z-testn 30 is known

t-testn < 30

is unknown

F-test(ANOVA)

3 or more s

Testing the Significance of Difference Between Means“n is large or when n 30 & is known.”

• Hypothesized/population mean VS Sample mean and population standard deviation is known.

nx

Z

x - is the sample mean - is the population meann - is the sample size - is the population std. dev.

Z-testn 30

is known

Using PHStat: Go to… “One-Sample Tests; Z-Test for the Mean:Sigma Known”

• Sample mean 1 VS Sample mean 2 and population standard deviation is known.


21

21

11nn

xxZ

1x - is the mean of sample 1


21 & nn - are the sample sizes - is the population std. dev.

Z-testn 30 is known

• Sample mean 1 VS Sample mean 2 and 2 sample standard deviations are known.


2

22

1

21

21

ns

ns

xxZ



21 & nn - are the sample sizes21 & ss - are the sample std. devs.

Z-testn 30 is unknown

Using Microsoft Excel: Go to…“Z-Test: Two-Sample For Means”

The Critical Value Approach in Testing the Significance of Difference

Between MeansThe 5-step solution

Step 1. Formulate Ho and Ha

Step 2. Set the level of significance , usually it is given in the problem.

Step 3. Formulate the decision rule (when to reject Ho); Find the critical value/P-value.

Step 4. Make your decision.

Step 5. Formulate your conclusion.

Approaches inHypothesis Testing

Critical valueapproach

p- valueapproach

Computed vs. Critical5-step solution1.Ho: ___________ Ha: ___________2. = ___; Cri-value= ______ 3. Decision rule: Reject Ho if 4. Decision:5. Conclusion:

valueCrivalueComp

p-value vs. 5-step solution1.Ho : ___________ Ha : ___________2. = ___; p- value=________ 3. Decision rule: Reject Ho if p- value 4. Decision:5. Conclusion:

Z=1.65

What Is Z Given = 0.05?

Finding Critical Values: One-Tailed

.4505.44951.6

.4394.43821.5

54Z = .05

.45

.05

Critical value

Critical Values: Z - Table

1.96 2.58Two-T

1.652.33One-T

0.05.01 Type

You will refer to this table to get the critical value of Zor the .tabularZ

CRITERION:

1. One-tailed test (right directional)

“Reject H0 if Zc ≥ Zt “

2. One-tailed test (left directional)

“Reject H0 if Zc ≤ Zt

3. Two-tailed test (Zc = +)

“Reject H0 if Zc ≥ Zt “

4. Two-tailed test (Zc = -)

‘Reject H0 if Zc ≤ Zt “

EXERCISES:

1. Past records showed that the average final examination grade of students in Statistics was 70 with standard deviation of 8.0. A random sample of 100 students was taken and found to have a mean final examination grade of 71.8. Is this an indication that the sample grade is better than the rest of the students? Test at 0.05 level of significance.

2. A certain type of battery is known to have a mean life of 60 hours. In random sample of 40 batteries, the mean life was found to be 58 hours with a standard deviation of 4.5 hours. Does it indicate that the mean lifetime of such battery has been reduced? Test at 0.01 level of significance.

3. The manager of a rent-a-car business wants to know whether the true average numbers of cars rented a day is 25 with a standard deviation of 6.9 rentals. A random sample of 30 days was taken and found to have an average of 22.8 rentals. Is there a significance between the mean and the sample mean? Test at 0.05 level of significance.

4. Advertisements claim that the average nicotine content of a certain kind of cigarette is 0.30 milligram. Suspecting that this figure is too low, a consumer protection service takes a random sample of 50 of these cigarette from different production slots and find that their nicotine content has a mean of 0.33 milligram with a standard deviation of 0.18 milligram. Use the 0.05 level of significance to test the null hypothesis µ = 0.30 against the alternative hypothesis µ < 0.30.

5. An experiment was planned to compare the mean time (in days) required to recover from common cold for person given a daily doze of 4 mgs. of vitamin C versus those who were not given a vitamin supplement. Suppose that 35 adults were randomly selected for each treatment category and that the mean recovery times and standard deviations for the 2 groups were as follows:

n X δ W/ vit. C 35 5.8 1.2 W/o vit. C 35 6.9 5.8 Suppose your research objective is to show that the use

of vit. C increases the mean time required to recover from common cold. Test using α = 0.05.

Technology

Statistics Presentation week 7