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Session 8

Tests of Hypotheses

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By the end of this session, you will be able to

set up, conduct and interpret results from a test of hypothesis concerning a population mean

explain how means from two populations may be compared, and state assumptions associated with the independent samples t-test

interpret computer output from one or two-sample t-tests, present and write up conclusions resulting from such tests

explain the difference between statistical significance and an important result

Learning Objectives

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Farmers growing maize in a certain area were getting average yields of 2900 kg/ha.

A “new” Integrated Pest Management (IPM) approach was attempted with 16 farmers.

Objective: To determine if the new approach results in an increase in maize yields.

Yields from these 16 farmers (after using IPM) gave mean = 3454 kg/ha, with standard deviation = 672 kg/ha hence s.e. = 168.

Can we determine whether IPM has really increased maize yields?

An illustrative example

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In above example, clearly the sample mean of 3454 kg/ha is greater than 2990 kg/ha

But the question of interest is “does this result indicate a significant increase in the yield

or might it just be a result of the usual random variation of yield”

Hypothesis testing seeks to answer such questions by looking at the observed change relative to the “noise”, i.e.

the standard error in the sample estimate

Is the yield increase real?

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Null hypothesis H0: = 2900where is the true mean yield of farmers in the area using the new approach

The promoters of the new approach are confident that yields with the new approach cannot possibly decrease.

Hence the above null hypothesis needs to be tested against the alternative hypothesis

H1: > 2900

Null H0 & Alternative H1

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Testing the hypothesis

Compute the t test statistic

t = ( - )/(s/n) = (3454 – 2900)/(168) = 3.30

which follows a t-distribution with n-1=15 degrees of freedom. Use values of the t-distribution to find the probability of getting a result, which is as extreme, or more extreme than the one (3.30) observed, given H0 is true.

The smaller this probability value, the greater is the evidence against the null hypothesis.This probability is called the p-value or significance level of the test

x

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Analysis in StataType db ttesti or look for the

One-sample mean comparison calculator on the menu

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Results

Result from the one-sided test done here

t-probabilities from formulae or table

t-value

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Interpretation and conclusions

It is clear from t-tables that the p-value is smaller than 0.01.

Using statistical software, we get the exact p-value as 0.0024.

This p-value is so small, there is sufficient evidence to reject H0.

Conclusion:

Use of the new IPM technology has led to an increase in maize yields (p-value=0.0024)

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Agric Non-agric

156 223

282 131

222 137

172 146

183 130

206 122

210 141

198 192

199 188

211 212

As part of a health survey, cholesterol levels of men in a small rural area were measured, including those working in agriculture and those employed in non-agricultural work.

Aim: To see if mean cholesterol levels were different between the two groups.

An example: Comparing 2 means

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Begin with summarising each column of data.

Agric Non-agricMean= 203.9 162.2

Std. dev. = 33.9 37.6Variance = 1147 1412

There appears to be a substantial difference between the two means.

Our question of interest is:

Is this difference showing a real effect, or could it merely be a chance occurrence?

Summary statistics

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To answer the question, we set up:

Null hypothesis H0:

no difference between the two groups (in terms of

mean response), i.e. 1 = 2

Alternative hypothesis H1:

there is a difference, i.e. 1 2

The resulting test will be two-sided since the alternative is “not equal to”.

Setting up the hypotheses

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• Use a two-sample (unpaired) t-test- appropriate with 2 independent samples

• Assumptions - normal distributions for each sample- constant variance (so test uses a pooled

estimate of variance)- observations are independent

• Procedure - assess how large the difference in means is, relative to the noise in this difference, i.e. the std. error of the difference.

Test for comparing means

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tx x

sn

sn

1 2

2

1

2

2

s

n s n s

n nwith n n f2 1 1

22 2

2

1 21 2

1 1

22

d. .

where s2, the pooled estimate of variance, is given by

The test statistic is:

Test Statistic

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The pooled estimate of variance, is :

= 1279.5

Hence the t-statistic is:

= 41.7/(2x1279.5/10)

= 2.61 , based on 18 d.f.

Comparing with tables of t18, this result is significant at the 2% level, so reject H0.

Note: The exact p-value = 0.018

Numerical Results

2 21 1 2 22

1 2

n 1 s n 1 ss

n n 2

tx x

sn

sn

1 2

2

1

2

2

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Difference of means: 41.7Standard error of difference: 15.99

95% confidence interval for difference inmeans: (8.09, 75.3).

Conclusions: There is some evidence (p=0.018) that the mean cholesterol levels differ between those working in agriculture and others. The difference in means is 42 mg/dL with 95% confidence interval (8.1, 75.3).

Results and conclusions

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Analysis in StataInput the data and do a t-test

Or complete the dialogue as shown below

Or type ttesti 10 203.9 33.9 10 162.2 37.6

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Results

This was a 2-sided test

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Take care to report results according to size of p-value.

For example, evidence of an effect is :• almost conclusive if p-value < 0.001 and could be said to be strong if p-value < 0.010• If 0.01< p-value < 0.05, results indicate some evidence of an effect.• If p-value > 0.05, but close to 0.05, it may indicate something is going on, but further confirmatory study is needed.

General reporting the results

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e.g. Farmers report that using a fungicideincreased crop yields by 2.7 kg ha-1, s.e.m.=0.41

This gave a t-statistic of 6.6 (p-value<0.001)

Recall that the p-value is the probability of rejecting the null hypothesis when it is true.

i.e. it is the chance of error in your conclusion that there is an effect due to fungicide!

Significance: further comments

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In relation to the example on the previous slide,we may find one of the following situations fordifferent crops.

Mean yields: with and without fungicide. 589.9 587.2 Not an important finding! 9.9 7.2 Very important finding!

It is likely that in the first of these results, either too much replication or the incorrect level of replication had been used (e.g. plant level variation, rather than plot level variation used to compare means).

How important are sig. tests?

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e.g. There was insufficient evidence in the data todemonstrate that using a fungicide had any effecton plant yields (p=0.128).

Mean yields: with and without fungicide. 157.2 89.9

This difference may be an important finding, but thestatistical analysis was unable to pick up this differenceas being statistically significant.

HOW CAN THIS HAPPEN? Too small a sample size? High variability in the experimental material? One or two outliers? All sources of variability not identified?

What does non-significance tell us?

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• Statistical significance alone is not enough. Consider whether the result is also scientifically meaningful and important.

• When a significant result if found, report the finding in terms of the corresponding estimates, their standard errors and C.I.s

• (as is done by Stata)

Significance – Key Points