PLG500 L5 One Sample T-test

8/10/2019 PLG500 L5 One Sample T-test

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Hypothesis testing

Dr Norizan


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Dr Norizan

Meaning: to infer something about a population parameter

based on a sample statistic

Two types of statistical inference: confidence interval: estimates the value of a

population parameter with an interval ofplausible values.

hypothesis test: assesses the evidence provided

by the data against a particular hypothesis aboutthe population parameter(s)


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Dr Norizan

Test a given theory or belief about a

population parameter

Find out if a claim about a population

parameter is trueMake a decision about a population

parameter based on the value of a sample

statistic


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Dr Norizan

Example:

A company claims that the weight of a fruit bar it

produces is 200g

How can we test that this claim is true? We cannot check all the fruit bars the company

produces

So we take 100 fruit bars at random and find the

mean weight

Then we compare the two values


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We need to compare the population

parameter (mean weight = 200g) against the

sample statistic (mean weight of sample)

Dr Norizan


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Dr Norizan

General procedure:

Choose a specific hypothesis to be tested

:This is called the null hypothesis

example: H0: = 200


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In the event that we reject the nullhypothesis, We have an alternative hypothesis to establish

Example: Ha: 200

alternatively Ha: 200 or Ha: 200

The alternative hypothesis states what wesuspect to be true about the populationparameter


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We use a test statistic to do this

The test statistic is t

In this case, we carry out a one sample t test

We must also be sure (95%)that our decisionis correct

Dr Norizan


9/41Dr Norizan

General procedure:

Choose a test statistic to evaluate the null

hypothesis (e.g. t statistic)

Choose a random sample, and make

measurements (e.g. mean)

Use the measurements to calculate the t statistic

and determine the likelihood of the hypothesis


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General procedure:

Determine the probability of obtaining a test

value as extreme as the observed value

The null hypothesis is rejected if the observed

significance level small enough


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It is the probability, if the null hypothesis

were true, that the sample outcome would

be as or more extreme as the one actually

observed It can be calculated from a sampling

distribution

The smaller the p value, the stronger the

evidence against the null hypothesis


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A significance level is a cut-off for how

small the p value must be in order for the

sample data to be considered decisive

Common values are .05 and .01


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1. State the null hypothesis

2. State the alternative hypothesis

3. Determine the test statistic

4. Determine the significance level5. Identify sample distribution

6. Identify critical region

7. Make decision


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If the p value is less than ,

then the sample result is said to be

statistically significant at the level


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One should not regard pre-specified

significance levels like =.05 as magical cut-

off values distinguishing significance from

insignificance

Rather, p values represent a continuum of

varying degrees of the evidences strength

against the null hypothesis


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How much time did you spend to study

PLG500 in a week?

Write the null hypothesis

Record the number of hours per week youtake to study for PLG500

Enter your data into an SPSS file

Conduct a one sample t test (what is the

actual mean?)


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Dr Norizan

Student Hours studying PLG500


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Dr Norizan

5

n = 36

mean = 3.77

p =.000

.025

.025.025

p = .000 is smaller than p = .05

Therefore we reject H0


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Dr Norizan

3

n = 36mean = 3.77

p =.009

.025.025

p = .009 is smaller than p = .05

Therefore we reject H0


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Dr Norizan

3.5

n = 36

mean = 3.77

p =.404

.025.025

p = .404 is larger than p = .05

Therefore we fail to reject H0


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Dr Norizan

Conclusions made do not have 100% certainty

Conclusions made are associated with

particular levels of significance

This tells us how confident we are that theconclusions made are very close to the real

situation


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Dr Norizan

One must consider the practical significanceof the result

Example: a new teaching method improvesperformance of a group of students by 5

marks p value for the t test = 0.03

Statistically, this is significant

However, does an increase of 5 marks meananything?


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Dr Norizan

Sample size plays an important role in tests

of significance.

A large sample can detect even a very small

difference or effect

A small sample may fail to detect even a large

difference or effect


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Dr Norizan

Assumptions is important to define the

sampling distribution of a test statistic

Correct significance levels can only be

calculated when the distribution is defined Tests of assumptions should be incorporated

as part of the hypothesis testing procedure


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Dr Norizan


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Dr Norizan


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To determine whether the mean IQ of

adopted children differs from the mean for

the general population of children (known to

be 100)


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Null Hypothesis, Ho: =100

Set =.05 (the commonly chosen value)

Data collected from a random sample of

n=25 adopted children, mean = 108,=15

If the probability (p) is less than .05 () Howill be rejected at the .05 level of

significance.

If p>.05, Ho is not rejected


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Dr Norizan

The area under the normal curve = 100%

100% = 100% 100 =1

For = .05, we want to have 95%

confidence that our decision is correct

this represents 95% of the area under the

normal curve

95% = 95% 100 = .95

1= .95


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Dr Norizan

= .05

The area shaded

red is .05


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Dr Norizan

= .05 p = .02

p = .02

The area

shaded red

is .02


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Dr Norizan

If the p value is less than ,

The sample is statistically significant

at significance level


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Dr Norizan

Example: p = .02, = .05

The sample mean is statistically significant

at significance level of .05

= .05

p = .02


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1. Analyze

2. Compare means

3. One sample t test

4. Move selected variable to the testvariable box

5. Select test value (= population mean)

6. Options 95% confidence interval -continue

7. OK

Dr Norizan


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Dr Norizan

One-Sample Statistics

103 56.57 25.941 2.556Exam Performance (%)

N Mean Std. Deviation Std. ErrorMean


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Dr Norizan


103 56.57 25.941 2.556Exam Perf ormance (%)

N Mean Std. DeviationStd. Error

Mean

One-Sample Test

-1.341 102 .183 -3.427 -8.50 1.64Exam Perf ormance (%)

t df Sig. (2-tailed)

Mean

Dif f erence Lower Upper

95% Confidence

Interv al of the

Diff erence

Test Value = 60


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Dr Norizan

t = - 1.34, d.f. = 102, p = .183

p > .05,

Therefore we fail to reject the null

hypothesis

Thus the sample mean is not significantlydifferent from the population mean

Make decision


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Dr Norizan

One-Sample Test

-8.383 102 .000 -21.427 -26.50 -16.36Exam Perf ormance (%)

t df Sig. (2-tailed)

Mean

Diff erence Lower Upper

95% Confidence

Interv al of the

Diff erence

Test Value = 78


103 56.57 25.941 2.556Exam Perf ormance (%)

N Mean Std. DeviationStd. Error

Mean


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t = - 8.38, d.f. = 102, p = .00

p < .05,

Therefore we rejectthe null hypothesis

Thus the sample mean is significantlydifferent from the population mean

Make decision

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PLG500 L5 One Sample T-test