1. Hypothesis Testing 1
Copyright by Winston S. Sirug, Ph.D.

2. Hypothesis Testing
 It is a statistical method that is used in making statistical decisions using experimental data. 
 It is basically an assumption that we make about the population parameter.

3. Hypothesis Testing
Hypothesis testing was introduced by
 Sir Ronald Fisher,
 Jerzy Newman
 Karl Pearson
 Egon Pearson (Karl Pearson’s son)

4. Hypothesis Testing
3 Methods to Test the Hypothesis
 traditional method
 p-value
 confidence interval

5. Two Types of Statistical Hypothesis
Null Hypotheses, symbolized by H0, is a statistical hypothesis testing that assumes that the observation is due to a chance factor.
Alternative hypothesis, symbolized by H1 it states that there is a difference between two population means (or parameters)

6. Level of Significance
The level of significance refers to the degree of significance in which we accept or reject the null hypothesis.
Level of significance is the maximum probability of committing a Type I error.
That is, P(Type I error) = α.

7. Level of Significance
The critical or rejection value is the range of the values of the test value that indicates that there is significant difference and that the null hypothesis (H0) should be rejected.
noncritical or nonrejection region is the range of the values of the test value that indicates that the difference was probably due to chance and that the null hypothesis (H0) should not be rejected.

8. One Tailed versus Two Tailed
A one-tailed test shows that the H0 be rejected when test value is in the critical region on one side of the mean.
A two-tailed test, the H0 should be rejected when the test value is in either of the two critical regions.

9. Hypothesis and the Tails of the Test
Two-tailed test
Left-tailed test
Right-tailed test
Signs in the H0
H0: μ = k
H0: μ = k or H0: μ  k
H0: μ = k or H0: μ  k
Signs in the H1
H1: μ ≠ k
H1: μ  k
H1: μ  k
Rejection Region
In both tails
In the left tail
In the right tail

10. Critical Value Approach to Hypothesis Testing
Rejection Region
α = 0.05
0.9500
Critical Values for
α = H1: μ  k
(Right-tailed test)
0.5000
0.4500
z  Critical value
Nonrejection Region
0.9500
Critical Values for
α = H1: μ  k
(Left-tailed test)
Rejection Region
α = 0.05
0.4500
0.5000
Critical value  z
Nonrejection Region

11. Critical Value Approach to Hypothesis Testing
Testing the Hypothesis about the Mean (σ Known) at 0.05 Significance Level H1: μ ≠ k
2.5%
95%
2.5%
– z
Rejection Region Nonrejection Region Rejection Region
Critical value
Critical value

12. Possible Outcome of a Hypothesis Test
Statistical Decision
H0 True
H0 False
Do not
reject H0
Correct decision
Confidence = 1 – α
Type II error
P(Type II error) = β
Reject H0
Type I error
P(Type I error) = α
Power = 1 – β

13. Steps in Conducting Hypothesis Testing
 State the null hypothesis (H0) and the alternative hypothesis (H1).
 Choose the level of significance, α, and the sample size.
 Determine the test statistic & sampling distribution.
 Determine the critical values that divide the rejection and nonrejection regions.
 Collect the data and compute the value of the test statistic.
 Make a statistical decision.
 State the conclusion.

14. One Sample z Test
The one sample z test is a statistical test for the mean of a population. It is used when n  30, or when the population is normally distributed and population standard deviation is known.
when  is known
When  is unknown

15. Assumptions in One Sample z test
 Subjects are randomly selected.
 Population distribution is normal.
 The population should be known.
 Cases of the samples should be independent.
 Sample size should be greater than or equal to 30.

16. Procedure for One Sample z test
H0: μ = specified value
H1: μ , ,  specified value
 Calculate the sample mean for one sample z test .
 Calculate the value of the one sample z test
 Statistical decision for hypothesis testing
If zcomputed  zcritical, do not reject H0.
If zcomputed  zcritical, reject H0.

17. Example 1: One Sample z Test
A researcher reports that the average salary of College Deans is more than ₧63,000. A sample of 35 College Deans has a mean salary of ₧65,700. At α = 0.01, test the claim that the College Deans earn more than ₧63,000 a month. The standard deviation of the population is ₧5,250.

18. Solution 1: Given: = 65,700, μ = 63,000, σ = 5,250, n = 35
Step 1: State the hypotheses and identify the claim.
H0: μ  63,000 (claim)
H1: μ  63,000
Step 2: Level of significance is α = 0.01.
Step 3: The t critical value is (one-tailed test)
Step 4: Compute the one sample z test.

19. Solution 1: Step 5: Decision rule. Step 6: Conclusion.
Reject H0
Step 6: Conclusion.
We can conclude that there is enough evidence to support the claim that the monthly salary of College Deans is more than ₧63,000.

20. One Sample t Test
One Sample t test is a statistical procedure that is used to know the mean difference between the sample and the known value of the population mean.

21. Assumptions in One Sample t test
 The population must be approximately normally distributed.
 Samples drawn from the population should be random.
 Cases of the samples should be independent.
 Sample size should be less than 30.
 The population mean should be known.

22. Procedure for One Sample t test
H0: μ = specified value
H1: μ , ,  specified value
 Calculate the sample mean and standard deviation.
 Calculate the value of the one sample t test
 Calculate the degrees of freedom. df = n – 1
 Statistical decision for hypothesis testing
If tcomputed  tcritical, do not reject H0.
If tcomputed  tcritical, reject H0.

23. Example 1: One Sample t Test
One of the undersecretary of the Department of Labor and Employment (DOLE) claims that the average salary of civil engineer is ₧18,000. A sample of 19 civil engineer’s salary has a mean of ₧17,350 and a standard deviation of ₧1,230. Is there enough evidence to reject the undersecretary’s claim at α = 0.01?

24. Solution 1: Given: = 17,350, μ = 18,000, s = 1,230, n = 19
Step 1: State the hypotheses and identify the claim.
H0: μ = 18,000 (claim)
H1: μ  18,000
Step 2: Level of significance is α = 0.01 & df = 19-1 = 18.
Step 3: The t critical value is 2.878 (two-tailed test)
Step 4: Compute the one sample t test.

25. Solution 1: Step 5: Decision rule. Step 6: Conclusion.
Do not Reject H0
Step 6: Conclusion.
We can conclude that the starting salary of civil engineers is ₧18,000 .

26. Using z Test and t Test Is n  30? Is σ known? Use: Use: Use: Yes No

27. z for Proportion
This test can be considered as a binomial experiment when there are only two outcomes and the probability of success does not change from trial to trial (the outcomes of each trial are independent).

28. Assumptions in z Test for Proportion
 Subjects are randomly selected.
 Population distribution is normal.
 Observations are dichotomous.

29. Procedure for One Sample z test
H0: p = specified value
H1: p , ,  specified value
 Calculate the sample proportion.
 Calculate the value of the z test for proportion.
 Statistical decision for hypothesis testing
If zcomputed  zcritical, do not reject H0.
If zcomputed  zcritical, reject H0.

30. Example 1: z Test for Proportion
A recent survey done by Philippine Housing Authority found that 35% of the population owns their homes. In a random sample of 240 heads of households, 78 responded that they owned their homes. At the 0.01 level of significance, does that indicate a difference from the national proportion?

31. Solution 1: Given: X = 78, n = 240, p = 35% = 0.35
Step 1: State the hypotheses and identify the claim.
H0: p = 0.35 (claim)
H1: p  0.35
Step 2: Level of significance is α = 0.01.
Step 3: The t critical value is 2.576 (two-tailed test)
Step 4: Compute the
p = 0.35
q = 1 – p = 1 – 0.35 = 0.65

32. Solution 1: Step 5: Decision rule. Step 6: Conclusion.
Do not Reject H0
Step 6: Conclusion.
We can conclude that there is not enough evidence to reject the claim that 35% of the Filipinos owned their homes.

33. Do not put your faith in what statistics say until you have carefully considered what they do not say.
– William W. Watt
Copyright by Winston S. Sirug, Ph.D.

38. exercises
1. The environmental Protection Agency releases figures on urban air soot in selected cities in the United States. For the city of St. Louis, the EPA claims that the average number of micrograms of suspended particles per cubic meter of air is 82. Suppose St. Louis officials have been working with businesses, commuters, and industries to reduce this figure. These city officials hire an environmental company to take random measures of air soot over a period of several weeks. The resulting data follow. Use these data to determine whether the urban air soot in St. Louis is significantly lower than it was when the EPA conducted its measurements. Let =.01.

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