Lecture 7 Hypothesis testing.pptx

shakirRahman10 593 views 45 slides Jul 15, 2023
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About This Presentation

Objectives:
Understand the elements of hypothesis testing for testing a population mean (for large sample):
Identify appropriate null and alternative hypotheses
Select a level of significance
Compute the value of test statistic
Locate a critical or rejection region
Interpret the appropriate conclusi...


Slide Content

1 Hypothesis Testing: Null & Alternative Hypotheses & Significance level Shakir Rahman BScN , MScN , MSc Applied Psychology, PhD Nursing (Candidate ) University of Minnesota USA. Principal & Assistant Professor Ayub International College of Nursing & AHS Peshawar Visiting Faculty Swabi College of Nursing & Health Sciences Swabi Nowshera College of Nursing & Health Sciences Nowshera

3 Objec t ives By the end of this session, the learners would be able to: Understand the elements of hypothesis testing for testing a population mean (for large sample ): Identify appropriate null and alternative hypotheses Select a level of significance Compute the value of test statistic Locate a critical or rejection region Interpret the appropriate conclusion

S ta t i s t i cs Descripti v e I n fe r e n t i al Esti m atio n H y pothesis testing Point Estimation Interval Estimation (CI) 4

5 Statistical Inference Inferential Statistics It consists of methods for measuring and drawing conclusion about a population based on information obtained from a sample Estimation (Point & Interval Estimation) Significance/ Hypothesis Testing

HYPOTHESIS: A CLAIM Claim 6

7 Why Hypothesis testing? To answer…. Will a new medication lower a person’s blood pressure? Is a new teaching strategy better than a traditional one? Will seat belts reduce the severity of injuries caused by accidents? Do college students spend more than 2 hours watching TV each day?

7 Hypothesis Testing : Tentative assumption related to certain phenomenon which a researcher want to verify Allows us to use sample data to test a claim about a population, such as testing whether a population mean equals s a me number . Example1 : Does an average box of cereal contain 368 grams of cereal? 368 gm.

Hypothesis Testing : Allows us to use sample data to test a claim about a population, such as testing whether a population mean equals s a me number. Example 2 : Do BSc. Nursing students spend more than 6 hours in Library studying Biostatistics per week? 9

Hypothesis Hypothesis: An informed guess or a conjecture about a population parameter, which may or may not be true. It tests whether a population parameter is less than , greater than , or equal to a specified value (hypothetical). Population S a mp l e Inference S t a t i s t i c Parameter 10

Frequent users of narcotics have a mean anger expression score higher than for non-users. (40). A cardiologist claimed that men between the age 40-60 years, who have had a myocardial infarction have mean serum cholesterol level different from general population (242mg/dl.). The mean balance score to assess muscle function among rheumatoid arthritis (RA) patients is lower than osteo-arthritis (OA) patients (4). 10 What is hypothesis? A stat e m e nt of b e li e f us e d i n t h e e v a lu a t i o n of a such a s the m e a n of a pop u l a t i o n para m e ter population (  ). Examples:

11 1 2 3 4 5 State null and Alt ernative hypotheses Level of si g n ificance = α Test Statistics Crit ic al region Steps of Hypothesis Testing Conclusion

Types of hypotheses There are two types of hypotheses: Null hypothesis (H ) - A claim that there is no difference between the population parameter and the hypothesized value. For example, the mean of a population (  ) equals the hypothesized value  . Alternative or Researcher hypothesis (H a or H 1 ) - A claim that disagrees with the null hypothesis. For example, the mean of a population (  ) is not equal to the hypothesized value  .

Ex a m p l e s of N u l l & Alte r native hypotheses Null hypothesis: Frequent users of narcotics have a mean anger expression score lower than or equal for nonusers of narcotics (40). Men between the age 40-60 years, who have had a myocardial infarction have mean serum cholesterol level similar to the general population (242mg/dl.). The mean balance score to assess muscle function among rheumatoid arthritis (RA) patients is greater than or equal to the osteo-arthritis (OA) patients (4). Alternative or Researcher hypothesis: Frequent users of narcotics have a mean anger expression score higher than for nonusers of narcotics (40). A cardiologist claimed that men between the age 40-60 years, who have had a myocardial infarction have mean serum cholesterol level different from general population (242mg/dl.). The mean balance score to assess muscle function among rheumatoid arthritis (RA) patients is lower than the osteo-arthritis (OA) patients (4).

H :    H a :    H :    H a :    H :    H a :    Hypothesis Forms for Means

Test of Hypothesis H0: µ = 45 H0: µ≤ 45 H0: µ ≥45 Ha: µ≠45 Ha: µ>45 Ha: µ<45

Two-tailed for H a 16 Directional or Non-directional Hypotheses One tailed hypotheses are directional ; two-tailed hypothesis is otherwise non-directional Alternative hypothesis (H a ): One-tailed, upper tail  H a :  >  One-tailed, lower tail  H a :  <  Two-tailed  H a :    Null hypothesis (H ): One-tailed, upper tail for H a  H :    One-tailed, lower tail for H a  H :     H :  = 

Null & Alternative Hypothesis The mean number of patients per day in a clinic during the month of July is 150 On the average, children attend schools within 3.1 kilometers of their homes in Hyderabad The average stay in the burn-ward of Civil hospital is at least 24 days The average life span in Pakistan is at least 55 yrs Has the average community level of suspended particulate exceeded 35 units per cubic meter for this month? Does mean age of onset of a certain acute disease for school children differ from 10.4? A psychologist claims that the average IQ of a sample of 50 children is significantly above the normal IQ of 100

Underlying assumptions for testing of hypothesis for population mean The sample has been randomly selected from the population or process. The underlying population is normally distributed (or if not normally distributed, then n is large say greater than or equal to 30). Population variance (  2 ) either known or sample variance (s 2 ) assumed to be approximately equal to population variance (  2 ), when n is large.

Basic Elements of Testing Hypothesis Null Hypothesis Alternative Hypothesis (Researcher Hypothesis) Choice of appropriate level of significance (  ) Assumptions Test Statistic (Formula): Application of sample results in the formula to calculate the value of test statistic use for decision purpose. Rejection Region (Critical Region): Based on alternative hypothesis and level of significance (  ). Conclusion: If the calculated value of the test statistic falls in the rejection region, reject H in favor of H a , otherwise fail to reject H .

Steps of Hypothesis Testing Solving Hypothesis Testing Problems (Traditional Method) Step 1 State the hypothesis and identify the claim Step 2 State the Level of Significance Step 3 Compute the test value (Test Statistics) Step 4 Make the decision to reject or fai l to reject the null hypothesis ( Critical region using Z-table) Step 5 Summarize the results (Conclusion)

Steps of Hypothesis testing Step 1 H :    H a :    Write down the appropriate null (H ) and alternative (H a ) hypotheses. (Write H0 and Ha as mathematical statements H :    H a :    H :    H a :   

Steps of Hypothesis Testing (Contd.) Step 2: State the level of significance  Note: Apply the same level of significance (  ) used at the time of sample size determination. Step 3: Find the test Statistic Perform the calculations to compute test statistic (Z-score) Test statistic:  n z  x   

Steps of Hypothesis Testing (Contd.) Step 4: Determine the critical region using Z- table The critical region is also known as Rejection region. When the computed test statistic falls in the rejection region we reject the null hypothesis in favor of the alternative hypothesis Reject Ho if, Zcal < -Ztab Zcal > Ztab Zcal > Ztab OR Zcal < -Ztab H a :    a H :    H a :   

Steps in a Hypothesis Test ( C ontd.) Step 5: Conclusion Every hypothesis test ends with the experimenters (you and I) either Rejecting the Null Hypothesis, or Failing to Reject the Null Hypothesis As strange as it may seem, you never ACCEPT the Null Hypothesis. The best you can ever say about the Null Hypothesis is that you don’t have enough evidence, based on a sample, to reject it! How: Decision is made on the basis of test statistic and critical region. If the calculated value of the test statistic falls in the rejection region, reject H in favor of Ha, otherwise fail to reject H .

CONCLUSION Presumed innocent: Defendant is innocent = Null hypothesis Remained innocent: Do not feel guilty Fail to reject null hypothesis Blame proven: Culprit behind the bar Reject null hypothesis

Two tailed test with 5% (α) 1- α = 95% i.e. 0.95 α = 5% i.e. 0.05 α/2 = 0.025 1- α = 95% α/2 = 0.025 α/2 = 0.025 0.475 0.475 - 1.96 1.96

Two tailed test with 1% (α) 1- α = 99% i.e. 0.99 α = 1% i.e. 0.01 α/2 = 0.005 1- α = 99% α/2 = 0.005 α/2 = 0.005 0.495 0.495 - 2.58 2.58

Right tailed test with 5% (α) 1- α = 95% i.e. 0.95 α = 5% i.e. 0.05 1- α = 95% α = 0.05 0.45 1.64

Right tailed test with 1% (α) 1- α = 99% i.e. 0.99 α = 1% i.e. 0.01 1- α = 99% α = 0.01 0.49 2.33

Left tailed test with 5% (α) 1- α = 95% i.e. 0.95 α = 5% i.e. 0.05 1- α = 95% α = 0.05 0.45 - 1.64

Left tailed test with 1% (α) 1- α = 99% i.e. 0.99 α = 1% i.e. 0.01 1- α = 99% α = 0.01 0.49 - 2.33

Trying to prove the parameter s “less than ” 29 One-tailed, lower tail  Ha:  <   Z> -Z 

Trying to prove t he parameter i s more than 34 One-tailed, upper tail  Ha:  >   Z>Z 

Trying to prove the parameter is “not equal to ” or“ different than ” 31 Two-tailed  Ha:     Z>Z  /2 and Z< -Z  /2

Common Values of   Z  Z  /2 0.10 1.28 1.645 0.05 1.645 1.96 0.01 2.32 2.575

Example: Mean APTT among DVT P atients A researcher assumes that activated partial thromboplastin time (APTT) of population of patients diagnosed with deep vein thrombosis (DVT) is approximately normally distributed with standard deviation of 7 seconds. A random sample of 30 hospitalized patients suffering from DVT had a mean APTT of 50 seconds. Use a 5 percent level of significance. – Does the data provide sufficient evidence to conclude that mean APTT for DVT patients is different from 53 seconds? Let  = Mean APTT of all hospitalized DVT patients. (True/Actual)

36 Example: Mean APTT among DVT patients (Contd.) Hypothesis  Description Step 1: Stating null & alternative hypothesis H o :  =53 seconds  The mean APTT of DVT patients is equal to 53 seconds. H a :  ≠53 seconds  The mean APTT of DVT patients is different from 53 seconds. Step 2: Level of significance (α) = 0.05 Step 3: Test Statistic n  3 ; x  5   2 . 3 5 1 . 2 8 5  5 3 7  3   5 3 ;   7 Z    n 3 z  x   

Table Value:  0.5 – 0.025 = 0.4750 α / 2  Z = ±1.96 (Critical Value) Significance Level: α = 0.05  α/2=0.05/2=0.025 (Area above or below Critical Value) . 5 . 4 . 3 . 2 . 1 -3 -2 -1 0 1 2 3 z Area above Z 3 7

Criti c a l Regi o n

Step 4: Critical Region H a :   53  Z>1.96 and Z<-1.96 Step 5: Conc l u s i on The calculated value of the test statistic (Z cal =-2.35) falls in the rejection region, so, we are rejecting our null hypothesis. i .e. The mean APT T o f DVT pat i ents i s d if fe r e n t f r om 53 seconds. Example: Mean APTT among DVT patients (Contd.) -1.96 1 . 9 6

References Bluman, A. (2004). Elementary statistics: A step by step approach. Boston: Mc Graw Hill.

Acknowledgements Dr Tazeen Saeed Ali RM, RM, BScN, MSc ( Epidemiology & Biostatistics), Phd (Medical Sciences), Post Doctorate (Health Policy & Planning) Associate Dean School of Nursing & Midwifery The Aga Khan University Karachi. Kiran Ramzan Ali Lalani BScN, MSc Epidemiology & Biostatistics (Candidate) Registered Nurse (NICU) Aga Khan University Hospital