BPS - 3rd Ed. Chapter 14 1

Chapter 14

Tests of significance: the basics

BPS - 3rd Ed. Chapter 14 2

What would happen if we repeated the sample or experiment many times?

How likely would it be to see the observed results if the claim was untrue?

Do the data give evidence against a claim?

Reasoning of Tests of Significance

BPS - 3rd Ed. Chapter 14 3

The statement being tested is called the null hypothesis

The null hypothesis is a statement of “no effect” or “no difference”

The test is designed to assess the strength of evidence against the null hypothesis.

Stating HypothesesNull Hypothesis, H0

BPS - 3rd Ed. Chapter 14 4

The statement we are trying to find evidence for is called the alternative hypothesis.

The alternative hypothesis usually indicates “an effect” or “difference”

The alternative hypothesis expresses the hopes or suspicions we bring to the data.

Stating HypothesesAlternative Hypothesis, Ha

BPS - 3rd Ed. Chapter 14 5

Case Study I: “Weight Gain”

Suppose we know that weight gain after the age of 30 varies from individual to individual according to a Normal distribution with standard deviation = 1 lbs

The symbol represents the mean or expected weight gain for all individuals (the parameter)

Ten individuals between the age of 30 and 40 yield an average gains of 1.02 lbs.

Do these data provide sufficient evidence that people in this age range tend to gain weight each year?

x

BPS - 3rd Ed. Chapter 14 6

Case Study I: “Weight Gain”

If the claim that = 0 is true (no average weight gain), the sampling distribution of from 10 individuals is Normal with = 0 and standard deviation

The data yielded = 1.02, which is more than three standard deviations from = 0. This provides strong evidence that people gain weight.

If the data yielded = 0.3, which is less than one standard deviations from = 0, there would be less convincing evidence that individuals gains weight.

x

0.31610

1

n

σ

x

x

BPS - 3rd Ed. Chapter 14 7

Case Study IWeight gain

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Statistical hypotheses

Null: H0: = 0

One sided alternatives

Ha:

Ha:

Two sided alternative

Ha: 0

BPS - 3rd Ed. Chapter 14 9

Weight gain

Case Study I

The null hypothesis is “no average weight gain”

The alternative hypothesis “yes, average weight gain”

H0: = 0 Ha: > 0

We use a one-sided test because we are interested only in determining weight gain (and not weight loss)

BPS - 3rd Ed. Chapter 14 10

Take an SRS of size n from a Normal population with unknown mean and known standard deviation .

The test statistic for null hypothesis H0: = 0 is

Test StatisticTesting the Mean of a Normal Population

nσ

μxz 0

BPS - 3rd Ed. Chapter 14 11

Weight Gain

Case Study I

The test statistic for no average weight gain is:

This shows that the sample mean is more than 3 standard deviations above the hypothesized mean of 0. This provides strong evidence against H0.

3.23

101

01.020

nσ

μxz

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The P-value provides the probability that the test statistic would take a value as extreme or more extreme than the value observed if the null hypothesis were true.

The smaller the P-value, the stronger the evidence the data provide against the null hypothesis

P-value

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P-value for Testing Means Ha: > 0

P-value is the probability of getting a value as large or larger than the observed test statistic (z) value.

Ha: < 0 P-value is the probability of getting a value as small or

smaller than the observed test statistic (z) value.

Ha: 0 P-value is two times the probability of getting a value as

large or larger than the absolute value of the observed test statistic (z) value.

BPS - 3rd Ed. Chapter 14 14

Weight Gain

Case Study I

For test statistic z = 3.23 and alternative hypothesisHa: > 0, the P-value is:

P-value = P(Z > 3.23) = 1 – 0.9994 = 0.0006

Interpretations: If H0 is true, there is only a 0.0006 (0.06%) chance that we would see results at least as extreme as those in the sample we therefore have evidence against H0 and in favor of Ha.

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Weight gain

Case Study I

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If the P-value is as small or smaller than the significance level (i.e., P-value ≤ ), then we say that data are statistically significant at level .

If we choose = 0.05, we are requiring that the data give evidence against H0 so strong that it would occur no more than 5% of the time when H0 is true.

If we choose = 0.01, we are insisting on stronger evidence against H0, evidence so strong that it would occur only 1% of the time when H0 is true.

Statistical Significance

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The four steps in carrying out a significance test:1. State the null and alternative hypotheses.2. Calculate the test statistic.3. Find the P-value.4. State your conclusion.

The procedure for Steps 2 and 3 is on the next page.

Tests for a Population Mean

BPS - 3rd Ed. Chapter 14 18

BPS - 3rd Ed. Chapter 14 19

Weight gain problem

Case Study I

1. Hypotheses: H0: = 0Ha: > 0

2. Test Statistic:

3. P-value: P-value = P(Z > 3.23) = 1 – 0.9994 = 0.0006

4. Conclusion:

Since the P-value is smaller than = 0.01, there is strong evidence that people gain weight in this age range.

3.23

101

01.020

nσ

μxz

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Studying Job Satisfaction

Case Study II

Does the job satisfaction of assembly workers differ when their work is machine-paced rather than self-paced? A matched pairs study was performed on a sample of workers, and each worker’s satisfaction was assessed after working in each setting. The response variable is the difference in satisfaction scores, self-paced minus machine-paced.

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Studying Job Satisfaction

Case Study II

The null hypothesis “no average difference” in the population of assembly workers. The alternative hypothesis (that which we want to show is likely to be true) is “there is an average difference in scores” in the population of assembly workers.

H0: = 0 Ha: ≠ 0

This is considered a two-sided test because we are interested determining a difference in either direction.

BPS - 3rd Ed. Chapter 14 22

Studying Job Satisfaction

Case Study II

Suppose job satisfaction scores follow a Normal distribution with standard deviation = 60. Data from 18 workers gave a sample mean score of 17. If the null hypothesis is µ0 = 0, the test statistic is:

1.20

1860

0170

nσ

μxz

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Studying Job Satisfaction

Case Study II

For test statistic z = 1.20 and alternative hypothesisHa: ≠ 0, the P-value would be:

P-value = P(Z < -1.20 or Z > 1.20)

= 2 P(Z < -1.20) = 2 P(Z > 1.20)

= (2)(0.1151) = 0.2302

If H0 is true, there is a 0.2302 (23.02%) chance that we would see results at least as extreme as those in the sample. Therefore do not have good evidence against H0 and in favor of Ha.

BPS - 3rd Ed. Chapter 14 24

Studying Job Satisfaction

Case Study II

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Studying Job Satisfaction

Case Study II

1. Hypotheses: H0: = 0Ha: ≠ 0

2. Test Statistic:

3. P-value: P-value = 2P(Z > 1.20) = (2)(1 – 0.8849) = 0.2302

4. Conclusion:

Since the P-value is larger than = 0.10, there is not sufficient evidence that mean job satisfaction of assembly workers differs when their work is machine-paced rather than self-paced.

1.20

1860

0170

nσ

μxz

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Confidence Intervals & Two-Sided Tests

A level two-sided significance test

rejects the null hypothesis H0: = 0

exactly when the value 0 falls outside a

level 1 – confidence interval for .

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Case Study II

A 90% confidence interval for is:

Since 0 = 0 is in this confidence interval, it is plausible that the true value of is 0. Thus, there is insufficient evidence(at = 0.10) that the mean job satisfaction of assembly workers differs when their work is machine-paced rather than self-paced.

40.26 to 6.26

23.261718

601.64517

n

σzx

Studying Job Satisfaction