Chapter V Hypothesis.ppt

Hypothesis

Quantitative

Techniques and

Simulation

Chapter V

Senior Lectures by:

RESEARCH

METHOD

Formulating the HYPOTHESIS

Test of the Hypothesis

Statement of the PROBLEM

1. Recognition of the FACTS 2. Discovery of the Problem

3. Problem Formulation

1. Design Test

What is a Hypothesis?

Hypothesis Testing for One Population Value:

–

Population mean

(



a.

 (population standard deviation) is given (known):



Use z/standard normal/bell shaped distribution

b.

 (pop std dev) is not given but s (sample std dev) is given



Use student’s t distribution

–

Population proportion (



)

–

Population Variance (







Use 

(Chi-Square) distribution. Population Standard Deviation =

Example: The proportion of the students of AUCA with cell phones is p = .80



Use z/standard normal/bell shaped distribution

Hypothesis

Hypothesis: A statistical hypothesis is a statement on a

probabilistic model and a hypothesis test is a method to determine

the possibility of that statement based on a sample.

Presumptions thus often provide the occasion for an

investigation. For this reason it is called research hypothesis.

I a

ssume

the m

ean

M

onthly

cell ph

one bill

the stu

dents o

f

AUCA

Purpose of hypothesis testing

•

The purpose of hypothesis testing is to

determine whether there is enough statistical

evidence in favor of a certain belief about a

parameter.

Example

: Is there statistical

evidence in a random

sample of potential

customers, that support the

hypothesis that more than

10% of the potential

• States the assumption (numerical) to be tested. This

hypothesis is assumed to be true, and the collected data

will be analyzed to see

if it is contradictory to the null

hypothesis.

Research Hypothesis

“

The mean monthly cell phone bill the student of AUCA is

less than 10000 Rwf”

Example of the Null Hypothesis:

The mean monthly cell phone bill the student of AUCA is at least ten

thousand Rwf.

H

:  10000

•

Always contains “=” , “≤” or “



” sig

•

May or may not be rejected

The Alternative Hypothesis, H

a

or

H

1

• Is the opposite of the null hypothesis

Example

– The mean monthly cell phone bill the student of

AUCA is less than 10000 Rwf

( H

:



< 10000)

• Never contains the “=” , “≤” or “



” sign

• May or may not be accepted

• Is generally the hypothesis that is believed (or needs

to be supported) by the researcher.

This is

what

you w

ant to

Examples: Give the null hypothesis and the alternative

hypothesis

• Is there statistical evidence in a random sample of potential

customers, that support the hypothesis that more than 10% of the

potential customers will purchase a new products?

• You want to show that people find the new design for a recliner

chair more comfortable than the old design.

• You are trying to show that cigarette smoke has an effect on the

quality of a person’s life.

• The mean age of the students enrolled in evening classes at a

certain college is greater than 36 years.

• The mean weight of packages shipped on Air Express during the

past month was less than 36.7 lb.

The critical concepts are these:

1. There are two hypotheses: the null and the alternative hypotheses.

2. The procedure begins with the assumption that the null hypothesis

is true.

3. The goal is to determine whether there is enough evidence to infer

that the alternative hypothesis is true, or the null is not likely to be

true.

4. There are two possible decisions:

Reject the null: To conclude that there is enough evidence to

infer that the alternative hypothesis is true.

Fail to reject the null: To conclude that there is insufficient

evidence to support the alternative hypothesis.

Claim:

the mean life

expectancy in Africa is

over than 50.1 years

is 60:

x = 60

years

Is X=60 likely if Ho:



≤ 50.1

REJECT

Null Hypothesis

If not likely,

Hypothesis Testing Process

Suppose the

sample mean of the

Life expectancy

H0:



≤ 50.1

Ha:



> 50.1

Sample

Sample

Population

Sampling Distribution of x



≤ 50.1

If

H

is true

If it is unlikely that we

would get a sample

mean of this value ...

... then we reject the null

hypothesis that



≤

50.1

... if in fact this were

the population mean…

x=60

Reject H₀ Do not reject H₀

Level of Significance,

a

•

Defines unlikely values of sample statistic if null hypothesis is true

– Defines rejection region of the sampling distribution.

•

Is designated by a , (level of significance)

– Typical values are .01, .05, or .10

•

Is selected by the researcher at the beginning.

•

Provides the critical value(s) of the test .

a

Normal

distribution

If Alpha ( )equals

0.1

0.05

0.01

One tail

Critical region

2.33

1.64

1.28

Two-tailed

H

: μ ≥ 50.1

H

: μ < 50.1

₀

H

:

μ

≤ 50.1

H

:

μ

> 50.1

H

:

μ

= 50.1

H

:

μ

≠ 50.1

a

a

/2

Represents

critical value

Level of significance =

a

0

0

a

/2

a

Two-tailed test

Rejection

region is

shaded

Interpreting the p-value…

The smaller the p-value, the more statistical evidence exists

to support the alternative hypothesis.

•

If the p-value is

less than 1%,

there is overwhelming

evidence that supports the alternative hypothesis.

•

If the p-value is

between 1% and 5%,

there is a strong

evidence that supports the alternative hypothesis.

•

If the p-value is

between 5% and 10%

there is a weak

evidence that supports the alternative hypothesis.

Interpreting the p-value…

Overwhelming

Evidence

(Highly

Significant)

Strong Evidence

(Significant)

Weak Evidence

(Not Significant)

No Evidence

(Not Significant)

Actual

situation

Our

decision

Null (Ho)

hypothesis is

false

Null (Ho)

hypothesis is

true

Reject the null

(Ho) hypothesis

Correct

decision

Type I

error (α)

Called Level of Significance

Conclusions of a Test of Hypothesis…

If we reject the null hypothesis, we conclude

that there is enough evidence to infer that the

alternative hypothesis is true.

If we fail to reject the null hypothesis, we

conclude that there is not enough statistical

evidence to infer that the alternative

hypothesis is true.

This does not mean that we

n

s

μ

x

t

n



1





The test statistic is:

Using Small Samples

(The

population

must be

approximately

normal)

Steps in Hypothesis Testing

• Specify the population value of interest.

•

Assumptions

: Randomization, quantitative variable,

normal population distribution (robustness?)

• Formulate the appropriate null and alternative

hypotheses.

• Specify the desired level of significance.

• Determine the rejection region or p_value

Example 1: Lower Tail t Test for Mean

Test the claim that the true mean monthly cell phone bill the student of

AUCA is less than ten thousand Rwf. For testing this hypothesis we

took a sample of 20 students and we asked them, how much they

spend a month in charging their cell phone? The answers are shown in

the table below

1. Specify the population value of interest

Mean monthly cell phone bill the student of AUCA

2. Formulate the appropriate null and alternative

hypotheses



Ho: μ



10000



Ha: μ < 10000 (This is a lower tail test)

3. Specify the desired level of significance



Suppose that

a

= .05 is chosen for this test

Monthly cellphone

10000

800

5000

5800

4500

4873

7000

5801

5500

10000

3500

9500

2000

7800

1500

2570

500

6531

Step in SPSS

First fill the data in SPSS, them to do this, click on

Analyze

, and then

Compare means

followed by

Output from SPSS

Making decision: Reject null hypothesis because Sig. = .000 < a = .05

Interpretation: The difference found is highly significant therefore we can conclude

at 5% of significance level, there is

evidence

that supports the alternative

hypothesis, i.e. students invest in charge their phone less than ten thousand Rwf.

One-Sample Statistics

  N Mean Std. Deviation

Std. Error  Mean

Monthly Cell Phone _{20 5194.4500 2842.31 635.56}

One-Sample Test

Test Value = 10000

t df Sig. (2-tailed)

Mean  Difference

95% Confidence Interval of  the Difference

Lower Upper

Monthly Cell Phone _{-7.561 19 .000 -4805.55 -6135.79 -3475.31}

•

Decision with Sig. or p_value with

a

=0.05

Example 2:

A pharmaceutical company conducts research on the efficacy of a vaccine against measles. The variable considered is the antibody titers produced by the vaccine. The vaccine produced by another laboratory reports an average titer of antibodies 1.9. To test whether the new vaccine is more effective than the older vaccine, the shot was given to 16 volunteers and obtained the following results:

Average titer of antibody 3 2.5 2.4 1.9 1.8 1.5 2.6 2.7 3.1 1.7 2.3 2.2 2.4 2.2

Steps using SPSS

To do this, click on

Analyze

, and then

Compare means

followed by

One-Sample T test

One-Sample Statistics

N Mean Std. Deviation Std. Error Mean Average titer

of antibody 16 2.225 0.5183 0.1296

One-Sample Test

Test Value = 1.9

t df

Sig.

(2-tailed) Mean Difference

95% Confidence Interval of the Difference

Lower Upper

Average titer of

antibody 2.508 15 0.024 0.325 0.049 0.601

Note : The value of p, or Sig gives us the SPSS default is bilateral, unilateral if we value: Sig /2 (.024/2 = 0.012)

Interpretation: This result indicates that the data are consistent with an average

value greater than 1.9, because the difference found is highly significant

(Sig = 0.012), therefore we can conclude that the new vaccine produced

antibody titers significantly higher than those produced by the old vaccine.

Bivariate Tests of Differences

Bivariate Tests of Differences

Involve only two variables: a variable that acts like a dependent

variable and a variable that acts as a classification variable.



Differences in mean scores between groups or in comparing

how two groups’ scores are distributed across possible response

categories.

Type of

Type of

Measurement

Measurement Differences between two Differences between two

independent groups

independent groups Differences among three or Differences among three or

more independent groups

more independent groups

Ordinal

Mann-Whitney U-test

Wilcoxon test

Kruskal-Wallis test

Nominal

Z

-test (two proportions)

Chi-square test

Chi-square test

Interval and

ratio

Independent groups:

t

-test or

Z

-test

Related or paried

groups

The

t

-Test for Comparing Two Means

Example

: Researchers are

interested in exam anxiety.

They administer an anxiety

inventory to students just

before the final exam in a

Sociology class. They also

administer it before the final

exam in a Business class. To

compare the two sets of

scores,

they

use

t-test for independent sampl

es

Example

: Researchers are

interested in exam anxiety.

They administer an anxiety

inventory on the second day of

class. Then they give it again on

the day of the midterm. To

compare the two sets of

scores,

they use

Determining when an independent samples

t-test is appropriate:

Is the dependent variable interval or ratio?

Can the dependent variable scores be grouped based upon some categorical

variable?

Does the grouping result in scores drawn from independent samples?

Are two groups involved in the research question?

Assumption when we use t test (parametric test)

Remember that for proper use of the distribution "t" or normal distribution "Z", the

data must satisfy the following assumptions:



Assume that the random samples are independent



Level of measurement of dependent variable is interval-ratio



Randomness: samples were selected using a probabilistic method. Otherwise

inference is not applied.



Normality: The variables of analysis, in both populations are normally

distributed. (Boxplot, histogram with normal curve, Normal Q-Q plot,

Shapiro-Wilk, KS, etc.). If not satisfy these conditions do using a nonparametric test or

you transform your variable .



Homogeneity of variances: The population variances are not different. That is:

(Levene test, F, etc.). If not corrected the number of degrees of freedom and

used the t test cuff applies a nonparametric test. When samples are very unequal

Example for hypothesis testing between two independent groups

A cigarette maker analyzes two different brands for determining the nicotine

content. A sample was taken of each brand and got the following results (in

milligrams).

Brand A:

24

26

25

22

23

Brand B:

27

28

25

29

26

Do the above results indicate that there is a difference in the average content of

nicotine in both brands?

Solution

Formulate the appropriate hypotheses:

Step using SPSS (Create data file)

Enter the data in SPSS, with the variable “Nicotine” takes up one column, and the

Brand variable for identifying whether the nicotine data was from brand A or brand

B subject takes up another column.

The “Nicotine” is considered as the dependent, response or outcome variable,

and the “Brand” variable is the independent or factor variable. The two variables

should be created in the way as seen in the data editor on the right. The Brand

variable takes on two possible values, 1 or 2. The value “1” for brand A, and the

value “2” for brand B.

B A

Ho

:







B A

Step in SPSS for hypothesis testing between two independent groups

To do this, click on

Analyze

, and then

Compare means

followed by

independent samples T test

and then continue the followed steps as

shown in the figure below < continue < OK

Interpretation: the report shows the descriptive statistics, the average content of

nicotine of Brand A is less than the average of Brand B, and standard deviation

for both are similar; but do not know whether this difference observed is

significant.

So we ask the t test for independent samples, which we gives t = -3.00, looking at the next Sig. (2-tailed) the value is .017, lower than .05.

Decision rule: Sig <0.05, therefore the level of significance of 5% we can say the results

indicate that there is a difference significant in the average content of nicotine in both brands,

Assumption

Basic assumptions:

To check if the variable is normally distributed, the following steps in the SPSS

Analyze <descriptive statistics <explore <follow the steps as shown in the figure below <accept

Box plot (to check if there are no outlier values and if

the boxes behave

Assumption

Assumption

Tests of Normality

Cigarette Brand

Kolmogorov-Smirnova _Shapiro-Wilk

Statistic df Sig. Statistic df Sig. Nicotine Brand A _{.136 5 .200}* _{.987 5 .967}

Brand B _{.136 5 .200}* _{.987 5 .967}

a. Lilliefors Significance Correction

Making a Decision and Interpreting the Result of the Test

We observed Shapiro-Wilk statistic given that the samples are small

The p_values or (Sig.)

Brand A: Sig, 967

Brand B: Sig, 967

Decision: From Shapiro-Wilk test of normality are both Sig. greater than 0.05,

therefore we don’t reject Null hypothesis, which imply that it is acceptable

to assume that the average content of nicotine distributions for

Brand A and Brand B populations are both normal (or bell-shaped).

Hypothesis testing to determine the normality

Assumption

Assumption of Homogeneity

Through the Levene test can see if this assumption very important to compare

groups met.

The report of SPSS gives without asking

Ho: the variances are equal

Ha: not assume equal variances

Example data from SPSS (This example uses the file

creditpromo.sav from SPSS

Analyze > Compare Means >

Independent>

► Select $

spent during promotional

period as the test variable.►

Select Type of mail insert

received as the grouping

variable.► Click Define

Groups.

T test > Independent-Sample t

test > Testing Two

independent Sample Means>

Running the Analysis

An analyst at a department store wants to evaluate a recent credit card

promotion. To this end, 500 cardholders were randomly selected. Half received

an ad promoting a reduced interest rate on purchases made over the next three

months, and half received a standard seasonal ad.

SPSS Report

Group Statistics

Type of mail insert

received N Mean DeviationStd. Std. Error Mean

$ spent during promotional period

Standard 250 1566.3890 346.67305 21.92553 New Promotion 250 1637.5000 356.70317 22.55989

The Descriptive table displays the sample size, mean, standard deviation, and standard error for both groups. On average, customers who received the interest-rate promotion charged about $71 more than the comparison group, and they vary a little more around their average.

The procedure produces two tests of the difference between the two groups. One test assumes that the variances of the two groups are equal. The Levene statistic tests this assumption.

Levene's Test for Equality of Variances

F Sig. $ spent during

promotional period

Equal variances

assumed 1.19 0.276 Equal variances not

Example Cont’d

Independent Samples Test

Levene's Test for Equality of

Variances t-test for Equality of Means

F Sig. t df

Sig.

(2-tailed) DifferenceMean

Std. Error Difference

$ spent during promotional period

Equal variances

assumed _{1.19 0.276 -2.26 498 0.024 71.11095 31.45914} -Equal variances

not assumed -2.26 497.595 0.024 -71.11095 31.45914

The “t” column displays the observed t statistic for each sample, calculated as the ratio of the difference between sample means divided by the standard error of the difference (t = -2.26).

The df column displays degrees of freedom (498). For the independent samples t test, this equals the total number of cases in both samples minus 2.

The column labeled Sig. (2-tailed) displays a probability from the t distribution with 498 degrees of freedom (Sig = .024). The value listed is the probability of obtaining an absolute value greater than or equal to the observed t statistic, if the difference between the sample means is purely random.

The Mean Difference (-71.11095) is obtained by subtracting the sample mean for group 2 (the New Promotion group) from the sample mean for group 1.

Hypothesis test for Dependent sampling – matched pairs

t-test (paried or related samples)

One of the most common experimental designs is the "pre-post" design. A study of this type often consists of two measurements taken on the same subject, one before and one after the introduction of a treatment or a stimulus. The basic idea is simple. If the treatment had no effect, the average difference between the measurements is equal to 0 and the null hypothesis holds. On the other hand, if the treatment did have an effect (intended or unintended!), the average difference is not 0 and the null hypothesis is rejected.

Example for paired comparisons or related

Athlete

Weight

before

Weight

after

1

127

135

2

195

200

3

162

160

4

170

175

5

143

147

6

205

200

7

168

172

8

175

186

9

197

194

10

136

130

10 athletes were subjected to a program of intensive physical training

by the coach. Their weights were recorded (in pounds) before and after

the training with the following results:

Does it affect the program the average

Output from SPSS

Paired Samples Statistics

Mean

N

Std. Deviation Std. Error Mean

Pair 1

After

172.70

10

23.386

7.395

Before

167.80

10

26.578

8.405

Interpretation: the report shows the descriptive statistics, the average being (before) is less than the average after implementing the training, but do not know whether this difference observed is significant.

Paired Samples Test

Paired Differences t df Sig. (2-tailed) Mean Std. Deviation Std. Error Mean 95% Confidence Interval of the

Difference Lower Upper Pair 1 Weight_Before

- Weight_After -4.900 6.740 2.132 -9.722 -.078 -2.299 9 .047

1. A firm is to buy a fleet of cars for use by its salesmen and wishes to chose between two alternative models, A and B. it places an advertisement in a local paper offering 20 liters of petrol free to anyone who has bought a new car of either model in the last year. The offer is conditional on being willing to answer a questionnaire and to note how far the car goes, under typical driving conditions, on the free petrol supplied. The following data were obtained.

Km driven on 20 liters of petrol

Model A

187

218

173

235

Model B

157

198

154

184

202

174

146

173

Conduct the appropriate statistical test of your hypothesis, using a .05 statistical significance level.

Review problems of chapter

2. A group of ten patients who were newly detected diabetes was observed to determine whether an educational program was effective in increasing their knowledge of diabetes. A test was applied before and after the educational program on self-related aspects of the disease. The test results were as follows:

Patient 1 2 3 4 5 6 7 8 9 10

Before 75 62 67 70 55 59 60 64 72 59

After 77 65 68 72 62 61 60 67 75 68

3. In 2014, consumer reports gave the following prices for a sample of 19 cellular cell phones:

Review problems of chapter

4. A company wants to study the effect of the break from work on the productivity of their workers. Select 5 workers and their productivity measured in an ordinary day, and then measure the productivity of workers in the same 5 day granting the break from work. The measured productivity figures are as follows:

Do they indicate these results that the break from work increase productivity?

600 300 289 499 615 279

255 612 353 530 322 375

475 425 445 580 250 399

Assuming a normally distributed population, test the hypothesis at the .05 level that the population mean price for cellular phones at the time of this survey was more than $350 •Conduct the appropriate statistical test of your hypothesis, using a .05 statistical significance level.

Worker Without pause With pause

1 23 28

2 35 38

3 29 39

4 33 37

Bivariate Tests of Differences

5. They have total cholesterol levels of a sample of eight patients before and after participating in a

diet-exercise program. Can be concluded that the program had positive impact?

Patient Before After

1 201 200

2 231 236

3 221 216

4 260 233

5 228 224

6 237 216

7 326 296

8 235 195

Basketball players 90 70 55 60

Football players 95 60 45 49

6. Do athletes in different sports vary in terms of intelligence? Bellow is reported

College Board scores of random samples of colleges’ basketball and football players. Is there a significant difference?

Do your best to present yourself to God as one approved, a worker who

does not need to be ashamed and who correctly handles the word of truth