test of hypothesis (z)

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Statistical Test of Hypothesis

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Page 1: Test of hypothesis (z)

Statistical Test of Hypothesis

Page 2: Test of hypothesis (z)

STATISTICS

DATA

central tendency variability

distributions

graphicalrepresentation

HYPOTHESIS TESTING

http://www.RapidLearningCenter.com

is the study of

has

can be classifiedinto

are displayed with

is tested with

Page 3: Test of hypothesis (z)

Statistical test of hypothesis

step by step process

used to verify a claim

using data

from a scientific study

Page 4: Test of hypothesis (z)

Statistically significant a result that has been predicted as

unlikely to have occurred by chance alone, according to a significance

level.

Significance levell α (read as alpha)

a pre-determined threshold probability for a

test of hypothesis.

Definitions

Page 5: Test of hypothesis (z)

hypothesis is a tentative explanation for an observation, phenomenon, or scientific problem that can be tested by further investigation.

In other words a scientific guess.

Two classifications

Null Hypothesis (HO) Alternative Hypothesis (HA)

status quo or default positionmaintained hypothesis or

research hypothesis

-no relationship between two measured phenomena

- that a treatment has no effect-no difference when compared

- relationship exist- treatment has effect

-significant difference when compared

already accepted accepted only when the null hypothesis is rejected

denoted by

= equality

denoted by

≠ only concerned that null hypothesis is not true.

>, < direction is important

Page 6: Test of hypothesis (z)

ProcessProblem or Claim is posted

Null Hypothesis Alternative Hypothesis

are formulated

step 1

significance levelstep 2 is agreed upon

critical value* is determined

step 3 Appropriate Test* is identified and applied

step 4 Conclusionsare made by comparing test result and critical value"Null hypothesis is either accepted or rejected"

step 5 INTERPRETATION problem is solved or claim is verified

Page 7: Test of hypothesis (z)

Interpretation of Significance level

Significance level Interpretation

α≤ .01 very strong evidence against the null hypothesis

.01< α ≤ .05 moderate evidence against the null hypothesis

.05< α ≤ .10 suggestive evidence against tthe null hypothesis

α> .10 little or no real evidence against tthe null hypothesis

Page 8: Test of hypothesis (z)
Page 9: Test of hypothesis (z)

Large-Sample Tests of Hypothesis

Population Mean vs Sample Mean

Population

samplesample

xHo :

xHa

xHa

xHa

:

:

:Assumptions:Large sample (n ≥ 30)Sample is randomly selected

Page 10: Test of hypothesis (z)

Testing Population Mean

Example:

Test the hypothesis that weight loss in a new diet program exceeds 20 pounds during the first month.

Sample data : n = 36, mean = 21, s = 5, μ0 = 20, α = 0.05

Assumptions:Large sample (n ≥ 30)Sample is randomly selected n

s

xz 0

test statistics

Solution:

Step1 : H0 : μ = 20 (μ is not larger than 20)

Ha : μ > 20 (μ is larger than 20)Step2 : α = 0.05 zα (critical value) =1.645

Page 11: Test of hypothesis (z)

20.1

)6(5

1

365

2021

0

z

z

z

ns

xz

Testing Population MeanStep3 : Z test since n ≥ 30

Step4: Decision: Accept Ho

Conclusion: At 5% significance level there is insufficient statistical evidence to concludethat weight loss in a new diet program exceeds 20 pounds per first month.

Page 12: Test of hypothesis (z)

population 2 orsample 2

population 2 orsample 2

population 1 orsample 1

Test Concerning Two Means

Population vs Population or Sample vs Sample

population 1 orsample 1

21

21

:

:

xxHo

Ho

2121

2121

2121

:,:

:,:

:,:

xxHaHa

xxHaHa

xxHaHa

Page 13: Test of hypothesis (z)

Comparing Two Population Means

Example: A random sample of 35 baby boys showed a mean birth weight of 7.4 lbs with a standard deviation 1.18 lbs while 40 baby girls showed a mean birth weight of 6.5 lbs with a standard deviation 1.5 lbs. Test if there are gender differences at 1% level of significance.

Assumptions:

1. Large samples ( n1 ≥ 30; n2 ≥ 30)2. Samples are randomly selected3. Samples are independent 2

22

1

21

21

nn

xxz

test statistics

Solution:

Step1 : H0 : μ1 = μ2 (no gender difference)

Ha : μ1 ≠ μ2 (there is gender difference) Step2 : α = 0.01 zα (critical value) =+/-2.58

Page 14: Test of hypothesis (z)

Comparing Two Population MeansStep3 : z test ( n1 ≥ 30; n2 ≥ 30)

2.90420.0960

9.0

40(1.5)

35(1.18)

6.57.422

2

22

1

21

21

nn

xxz

Step4: Decision: Reject Ho

Conclusion:There is sufficient evidence to conclude that there is a significantdifference in the birthweight between boys and girls at 1% level of significance

Page 15: Test of hypothesis (z)

Large-Sample Tests of Hypothesis

• Other tests– Testing a Population Proportion

– Comparing Two Population Proportions

are left as part of research

Page 16: Test of hypothesis (z)

End of

First Part of the Discussions

Page 17: Test of hypothesis (z)

Review Exercises

Page 18: Test of hypothesis (z)

Review Exercises

1. Ambulatory Services Inc. claims that their average response time is within 30 minutes of receipt of call. The response time for a random sample of 64 cases were recorded, with a sample mean of 34 minutes and a standard deviation of 21 minutes.

(i) Is there sufficient evidence to conclude that the actual response time is

larger than what is claimed by Ambulatory Services Inc.? Use α = .05

2. A chemist from a university claimed that he has invented a new spray that will keep the flowers fresh longer. He based his claim on a test when he selected 500 blossoms of a single type of flower and placed into two groups. One group (consisting of 250 blossoms) was sprayed with his formulation and the other with no spray. For the treatment he found that the average wilting time was 7.2 days with a standard deviation of 1.2 days, while for the control group, 3.6 days with a standard deviation of 1.1 days. Do you agree with the claim of the chemist that the spray actually keeps the flowers fresh longer? Use α = .01

3. A pharmaceutical company claims that they have developed a new drug that will provide immediate relief for persons suffering from vertigo. VERTIPLUS is claimed to provide relief within 5 minutes. A clinical trial was undertaken to test this claim and out of 36 tests the mean relief time is recorded at 6.7 minutes with a standard deviation of 1.38 minutes. Is there sufficient evidence to uphold the claim? α = .01

.

Page 19: Test of hypothesis (z)

THANK YOU!