statistics: unlocking the power of data lock 5 hypothesis testing: conclusions stat 250 dr. kari...

Statistics: Unlocking the Power of Data Lock5

Hypothesis Testing: Conclusions

STAT 250

Dr. Kari Lock Morgan

SECTION 4.3• Significance level (4.3)• Statistical conclusions (4.3)• Statistical significance (4.3)


A formal hypothesis test has only two possible conclusions:

1. The p-value is small: reject the null hypothesis in favor of the alternative

2. The p-value is not small: do not reject the null hypothesis

Formal Decisions

How small?


Significance Level

The significance level, , is the threshold αbelow which the p-value is deemed small enough to reject the null hypothesis

p-value < α => Reject H0

p-value > α => Do not Reject H0

Often = 0.05, unless otherwise specifiedα


p-value < α

Results would be rare, if the null were true

Reject H0

We have evidence that the alternative is true!

p-value ≥ α

Results would not be rare, if the null were true

Do not reject H0

Our test is inconclusive


Statistical Conclusions

In a hypothesis test of

H0: = 10 vs Hμ a: < 10 μ

the p-value is 0.002. With α = 0.05, we conclude:

a) Reject H0

b) Do not reject H0

c) Reject Ha

d) Do not reject Ha




H0: = 10 vs Hμ a: < 10 μ


a) There is evidence that = 10 μb) There is evidence that < 10 μ

c) We have insufficient evidence to conclude anything




H0: = 10 vs Hμ a: < 10 μ


a) Reject H0

b) Do not reject H0

c) Reject Ha

d) Do not reject Ha




H0: = 10 vs Hμ a: < 10 μ


a) There is evidence that = 10 μb) There is evidence that < 10 μ

c) We have insufficient evidence to conclude anything


H0 : X is an elephantHa : X is not an elephant

Would you conclude, if you get the following data?

• X walks on two legs

• X has four legs

Elephant Example


“For the logical fallacy of believing that a hypothesis has been proved to be true, merely because it is not contradicted by the available facts, has no more right to insinuate itself in statistical than in other kinds of scientific reasoning…” -Sir R. A. Fisher

Never Accept H0

•“Do not reject H0” is not the same as “accept H0”!

• Lack of evidence against H0 is NOT the same as evidence for H0!


Statistical Significance

When the p-value is less than , the αresults are statistically significant.

If our sample is statistically significant, we have convincing evidence against H0, in favor of Ha


p-value < α

Results would be rare, if the null were true

Reject H0

We have evidence that the alternative is true!

Results are statistically significant

p-value ≥ α

Results would not be rare, if the null were true

Do not reject H0

Our test is inconclusive

Results are not statistically significant


Question #1 of the Day

Does red wine (resveratrol) promote weight loss?

If so, why?


• Resveratrol, an ingredient in red wine and grapes, has been shown to promote weight loss in rodents. We’ll look at a study on Grey Mouse Lemurs (a primate).

• A sample of lemurs had various measurements taken before and after receiving resveratrol supplementation for 4 weeks:• body mass• resting metabolic rate• locomotor activity• food intake

Red Wine and Weight Loss

BioMed Central (2010, June 22). “Lemurs lose weight with ‘life-extending’ supplement resveratrol. Science Daily.


Study Design

This is a

a) Experimentb) Observational study


Study Design

Can we use this study to make conclusions about causality?

a) Yesb) No



In the test to see if the mean resting metabolic rate is higher after treatment, the p-value is 0.013.

Using = 0.05, is this difference statistically αsignificant? (should we reject H0: no difference?)

a) Yesb) No



In the test to see if the mean resting metabolic rate is higher after treatment, the p-value is 0.013.

Using = 0.05, what can we conclude?α

a) Mean resting metabolic rate is higher after resveratrol supplementation in lemursb) Mean resting metabolic rate is not higher after resveratrol supplementation in lemursc) Nothing



In the test to see if the mean body mass is lower after treatment, the p-value is 0.007.


a) Yesb) No



In the test to see if the mean body mass is lower after treatment, the p-value is 0.007.


a) Mean body mass is higher after resveratrol supplementation in lemursb) Mean body mass is not higher after resveratrol supplementation in lemursc) Nothing



In the test to see if locomotor activity changes after treatment, the p-value is 0.980.


a) Yesb) No



In the test to see if locomotor activity changes after treatment, the p-value is 0.980.


a) Locomotor activity changes after resveratrol supplementation in lemursb) Locomotor activity does not change after resveratrol supplementation in lemursc) Nothing



In the test to see if mean food intake changes after treatment, the p-value is 0.035.


a) Yesb) No


Informal strength of evidence against H0:

Formal decision of hypothesis test, based on = 0.05 :

statistically significant

not statistically significant



Question #2 of the Day

What aspect of sunlight might help protect against multiple

sclerosis?


Multiple Sclerosis and Sunlight• It is believed that sunlight offers some protection

against multiple sclerosis, but the reason is unknown

• Researchers randomly assigned mice to one of:• Control (nothing)• Vitamin D Supplements• UV Light

• All mice were injected with proteins known to induce a mouse form of MS, and they observed which mice got MS

Seppa, Nathan. “Sunlight may cut MS risk by itself”, Science News, April 24, 2010 pg 9, reporting on a study appearing March 22, 2010 in the Proceedings of the National Academy of Science.


Study Design

Can this study be used to make conclusions about causality, at least in mice?

a) Yesb) No


Multiple Sclerosis and SunlightFor each situation below, write down

Null and alternative hypotheses Informal description of the strength of evidence against H0

Formal decision about H0, using α = 0.05 Conclusion in the context of the question

In testing whether UV light provides protection against MS (UV light vs control group), the p-value is 0.002.

In testing whether Vitamin D provides protection against MS (Vitamin D vs control group), the p-value is 0.47.


Multiple Sclerosis and SunlightIn testing whether UV light provides

protection against MS (UV light vs control group), the p-value is 0.002.


Multiple Sclerosis and SunlightIn testing whether Vitamin D provides

protection against MS (Vitamin D vs control group), the p-value is 0.47.


Conclusions

p-value < α

Generic conclusion:Reject H0

Conclusion in context:We have (strong?) evidence

that [fill in alternative hypothesis]

p-value ≥ α

Generic conclusion:Do not reject H0

Conclusion in context:We do not have enough

evidence to conclude that [fill in alternative hypothesis]

H0

Ha


To DoRead Section 4.3

Do HW 4.2, 4.3 (due Friday, 10/23)

statistics: unlocking the power of data lock 5 hypothesis testing: conclusions stat 250 dr. kari...

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