chapter 15: tests of significance: the basics stat 1450

Chapter 15: Tests of Significance: The Basics

STAT 1450

Connecting Chapter 15 to our Current Knowledge of Statistics

▸ Chapters 10 & 12 used information about a population to answer

questions about a sample. (e.g., 20% of people are smokers, what is

the probability that a random sample of 2 people smoke).

▸ Now with inference, we have statistics problems where we use the

information about a sample to answer questions concerning the

population.

15.0 Tests of Significance

Connecting Chapter 15 to our Current Knowledge of Statistics

▸ If we want to estimate a population parameter, then we should use

statistics to create a confidence interval.

▸ If, on the other hand, we want to assess the evidence provided by

data about some claim concerning a population parameter, we need

to do a hypothesis test.

15.0 Tests of Significance

The Reasoning of Tests of Significance

▸ We are now inquiring about a behavior of an event if a phenomena

was repeated numerous times. We will begin by working with simple

random samples of data from Normal populations with known standard

deviations.

15.1 The Reasoning of Tests of Significance


▸ Situation: People drink coffee for a variety of professional, and now, social reasons. Coffee used to merely be a beverage option on the menu. Now, it is the main attraction for a growing number of restaurants and shoppes. The standard “cup of coffee” is 8 oz. However, even a Tall at Starbuck’s is 12 oz.



Please answer the following questions:

a. How many ounces of coffee do you think people typically drink each day? ____

b. How many ounces of coffee do you drink daily?

____

Note: We will only consider the population of regular coffee drinkers.


Null and Alternative Hypotheses

▸ We have two possible hypotheses about this situation:1. The mean amount of coffee consumed daily is not different than the value listed in a). 2. The mean amount of coffee consumed daily is different from the value listed in a).

▸ These hypotheses have names:

The null hypothesis (denoted H0) is the claim tested about the population

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

null hypothesis. Usually the null hypothesis is a statement of “no effect” or “no

difference.” It commonly assumes the “benefit of the doubt.”

The alternative hypothesis (denoted Ha) is the claim about the population

parameter that we are trying to find evidence for.

15.2 Stating Hypotheses

One- and Two-sided Alternative Hypotheses

▸ An alternative hypothesis is one-sided if it states that a parameter is

larger than or smaller than the null hypothesis value.

▸ It is two-sided if it states that the parameter is different from the null

value. (It could be either smaller or larger.)

▸ Question: Is the alternative hypothesis in our situation one-sided or

two-sided?


One- and Two-sided Alternative Hypotheses

▸ An alternative hypothesis is one-sided if it states that a parameter is

larger than or smaller than the null hypothesis value.

▸ It is two-sided if it states that the parameter is different from the null

value. (It could be either smaller or larger.)

▸ Question: Is the alternative hypothesis in our situation one-sided or

two-sided?

Testing a “difference” implies that we are interested in whether the mean daily coffee

consumption amount is either greater than, or, less than our estimate..


Example: Coffee Consumption

▸ Let’s use one of the values from a) to compose the null and alternative

hypotheses.



▸ Let’s use one of the values from a) to compose the null and alternative

hypotheses.

Assuming a value of _20__. H0: m= _20_ vs. Ha: m _20__



▸ Suppose it is known that the standard deviation for daily coffee consumption is 9.2 oz. The average amount of coffee consumed daily for a random sample of 48 people is 26.31 oz.

▸ True or False: We know with certainty that the average amount of coffee consumed daily is different from our hypothesized value.



▸ Suppose it is known that the standard deviation for daily coffee consumption is 9.2 oz. The average amount of coffee consumed daily for a random sample of 48 people is 26.31 oz.

▸ True or False: We know with certainty that the average amount of coffee consumed daily is different from our hypothesized value.

▸ FALSE. It may be likely given our sample size, but not with certainty. There is variability in sampling. It is possible that the next random sample of 48 could yield a different average.


Stating Hyptotheses

▸ Important note: Base your alternative hypothesis on your question of

interest—do not base it on the data.


Definition: Test Statistic

▸ A test statistic calculated from the sample data measures how far

the data departs from what we would expect if the null hypothesis

were true.

▸ The further this statistic is from 0, the more the data contradicts the null

hypothesis.

Note: A test statistic tells us how many standard deviations our value is away from

the hypothesized mean. A positive test statistic is above the mean. A negative one is

below the mean. We then use this information to figure out how likely it is to see

results like ours if the null hypothesis was true.

15.3 P-values & Statistical Significance

Definition: P-value

▸ The probability, computed assuming that the null hypothesis is true,

that the test statistic would take a value as extreme or more extreme

than that actually observed is called the P-value (probability value) of

the test.


P-values

▸ If the P-value is small enough, the data we observed would be

unusual (very unlikely to have happened) if the null hypothesis were

true.

▸ If the P-value is not small enough, the data we observed are

not strange at all (could plausibly have happened due to sampling

variability) if the null hypothesis were true.


Tests of Significance & the Justice System


Tests of Significance Justice System

Null Hypothesis The defendant gets the “benefit of the doubt.” (i.e., they are not guilty).

Alternative Hypothesis They are “guilty.”

Test Statistic Totality of Evidence collected.

P-value The probability of observing data as extreme as what was collected under the assumption that the defendant is, indeed, “not guilty.”

When the evidence collected seems ‘likely’ (based upon the null hypothesis)

Decision Jury rules that the defendant is ‘not guilty.”



Tests of Significance Justice System

When the evidence collected seems ‘likely’ (based upon the null hypothesis)

Decision Jury rules that the defendant is ‘not guilty.”

When, the evidence collected seems ‘extremely unlikely’ (based upon the null hypothesis)

Decision Either we have “bad” data (mistrial, tampering, etc…)

-Or-The jury rules that the defendant is ‘guilty.’



▸ Note: Our jury system assumes innocent until proven guilty.

▸ The actual truth of whether the person indeed committed the crime

may never be known.

Significance Level

▸ Question: What is the cut-off between “likely,” “unlikely,” and

“extremely unlikely?”

▸ If the P-value is as small or smaller than , we say that the data are

▸ statistically significant at level . The quantity is called the

significance level or the level of significance.


Significance Level

▸ Question: What is the cut-off between “likely,” “unlikely,” and

“extremely unlikely?”

Most common is P-value < 0.05.

“Likely” prob. > .10 “Unlikely” 5% to 10% “Extremely

Unlikely” < 5%

▸ If the P-value is as small or smaller than , we say that the data are

▸ statistically significant at level . The quantity is called the

significance level or the level of significance.


Significance Level

▸ After choosing an appropriate level of significance, we can make a

decision about H0.


P-Value vs. α

Decisions about H0

P-value > α Ho

P-value ≤ α Ho

Significance Level

▸ After choosing an appropriate level of significance, we can make a

decision about H0.


P-Value vs. α

Decisions about H0

P-value > α Fail to Reject Ho

P-value ≤ α Reject Ho

Significance Level

▸ Question: Why should a significance level be set before the test has

been done?

Suppose that your P-value = 0.025. If = 0.05, you would reject the null hypothesis;

if = 0.01, you would fail to reject the null hypothesis. If you did not set a

significance level before the test, you might change your mind based on the results

to fit the decision you (might) desire.

▸ The test statistic for hypothesis testing has is based upon our work

from sampling distributions and confidence intervals.


Tests for a Population Mean

▸ Tests of significance, allow researchers to determine the validity of certain hypotheses based upon P-values. There are various parameters that we can test (proportions, standard deviations, etc…).

▸ We will begin with the most common parameter to be tested, the mean; much like how we began our confidence interval discussion by estimating the true mean, µ.

15.4 Tests for a Population Mean


▸ Draw an SRS of size n from a large population that has the Normal

distribution with mean μ and standard deviation σ. The one-sample z

statistic

has the z distribution.

▸ To test the hypothesis , compute the one-sample z statistic




Key WordsAlternative Hypothesis P-value

Rejection Region (“extremely unlikely values”)





“more than”“increased”






“less than”“reduced”






“less than”“reduced”

“different”“is not”


▸ The standard deviation of daily coffee consumption is 9.2 oz. A random

sample of 48 people consumed an average of 26.31 oz. of coffee daily.

Is this evidence that the average amount of coffee consumed daily is

different from our original estimate?

Poll: Using your intuition, do you believe we have enough evidence

against our original claim?

(a) Yes (b) No

(Note: m could theoretically = 20, but based upon our data it is unlikely.)



Is the average amount of coffee consumed daily not 20 oz.?



Is the average amount of coffee consumed daily not 20 oz.?(Hypotheses)



Is the average amount of coffee consumed daily not 20 oz.?(Hypotheses) (Conditions) Random sample: We are told that this is a random sample of 48.




Large enough population:sample ratio: Population of coffee drinkers > 20*48 =960.




Large enough population:sample ratio: Population of coffee drinkers > 20*48 =960.

Large enough sample: Not told that coffee consumption follows a Normal distribution. However, n = 48 40, so we can use the CLT.



(Test Statistic)



(Test Statistic) = 4.75




(P-value) 2*P( Z > 4.75) = 2*.0001 = .0002




(P-value) 2*P( Z > 4.75) = 2*.0001 = .0002

(Decision) .0002 is an extremely small probability. There is enough evidence to conclude the mean amount of coffee consumed is not 20 oz.


Technology Tips – Conducting Tests of Significance (σ known)

▸ TI-83/84. STAT TESTS ZTest Enter.

Select Stats. Enter and n. Select Calculate.

(Note: Select Data when and n are not provided. Then enter the list where the data

are stored.)

▸ JMP. Enter the data. Analyze Distribution.

“Click-and-Drag” (the appropriate variable) into the ‘Y, Columns’ box. Click on OK.

Click on the red upside-down triangle next to the title of the variable from

the‘Y,Columns’ box. Proceed to ‘Test Mean.’ Enter and click on OK.






are stored

(for this example)

Inpt >> STATS

m0: 20 >> s: 9.2 >> : 26.31 >> n : 48 >> m : ≠

Calculate (ENTER)






are stored

(for this example)

Inpt >> STATS

m0: 20 >> s: 9.2 >> : 26.31 >> n : 48 >> m : ≠

Calculate (ENTER) z=4.7518 p=2.018 E -6 => .000002 ≈ 0.Reject H0, given the p-

value ≈ 0.






are stored

(for this example)

Inpt >> STATS

m0: 20 >> s: 9.2 >> : 26.31 >> n : 48 >> m : ≠

Calculate (ENTER) z=4.7518 p=2.018 E -6 => .000002 ≈ 0.Reject H0, given the p-

value ≈ 0. There is enough

evidence to conclude the mean amount of coffee consumed is not 20 oz.


Tests for a Population Mean: 4-Step Process

1. State: What is the practical question that requires a statistical test?

2. Plan:

a. Identify the parameter

b. List all given information from the data collected.

c. State the null (H0) and alternative (HA) hypotheses.

d. Specify the level of significance.

e. Determine the type of test. (Left-tailed, Right-tailed, Two-Tailed)

f. Sketch the region(s) of “extremely unlikely” test statistics. (Rejection Region(s) )



3. Solve:

a. Check the conditions for the test you plan to use.

(Random sample? Large population:sample ratio? Large enough for sample?)

b. Calculate the test statistic.

c. Determine (or estimate) the P-Value.

4. Conclude:

a. Make a decision about the about the null hypothesis.

(Reject H0 or Fail to reject H0).

b. Interpret the decision in the context of the original claim.

(i.e.,“There is enough (or not enough) evidence at the α level of significance that …)


Example: IQ Scores

▸ Recall that IQ scores from Chapter 14 followed a Normal Distribution with s= 15. You suspect that persons from affluent communities have IQ scores above 100. A random sample of 35 residents of an affluent community had an average IQ score of 112. Is there significant evidence to support your claim at the a=.05 level?


Example: IQ Scores

▸ Recall that IQ scores from Chapter 14 followed a Normal Distribution with s= 15. You suspect that persons from affluent communities have IQ scores above 100. A random sample of 35 residents of an affluent community had an average IQ score of 112. Is there significant evidence to support your claim at the a=.05 level?

STATE: Do affluent communities have a mean IQ score above 100?


Example: IQ Scores

Plan:

a. Identify the parameter m = average IQ score for affluent

communities


Example: IQ Scores

Plan:


communities

b. List all given information from the data collected. _n=35, _s= 15,


𝑥=112

Example: IQ Scores

Plan:


communities


c. State the null (H0) and alternative (HA) hypotheses. H0: _m = 100 HA: _m > 100


𝑥=112

Example: IQ Scores

Plan:


communities


c. State the null (H0) and alternative (HA) hypotheses. H0: _m = 100 HA: _m > 100__

d. Specify the level of significance. a =.05


𝑥=112

Example: IQ Scores

Plan:


communities






𝑥=112

Example: IQ Scores

Plan:


communities





f. Sketch the region(s) of “extremely unlikely” test statistics. (Rejection Region(s) )


𝑥=112


3. Solve:


Random sample? Large population : sample ratio? Large enough for

sample?

Yes Certainly more than 35 We were

informed that affluent

households the data came from a

Normal Distribution.



3. Solve:



sample?







𝑧=𝑥−𝜇0

𝜎 /√𝑛=112−100

15 /√35=4.73


3. Solve:



sample?






c. Determine (or estimate) the P-Value P(Z > 4.73) = .0001


𝑧=𝑥−𝜇0

𝜎 /√𝑛=112−100

15 /√35=4.73


4. Conclude:

a. Make a decision about the null hypothesis. P-value = .0001 ≤ .05 = a (default)




4. Conclude:







4. Conclude:





There is enough evidence at the a=.05 level to conclude that persons from affluent

communities have a mean IQ score above 100.






are stored

(for this example)

Inpt >> STATS

m0: 100 >> s: 15 >> : 112 >> n : 35 >> m : >

Calculate (ENTER) z=4.733 p=1.11 E -6 => .0000011 ≈ 0.

There is enough evidence to conclude affluent

communities have a mean IQ score above 100.


Final Note

▸ Some homework exercises will provide you with raw data.

▸ You are to use the data to compute the

sample mean and/or standard deviation.

▸ Then proceed with computing the confidence interval or performing a

test of significance.


Final Note


Steps for Success-Conducting Tests of Significance

1. Set up your Hypotheses.

Final Note




2. Check your Conditions.

Final Note





3. Compute the Test Statistic.

Final Note






4. Compute the P-Value.

Final Note






4. Compute the P-Value.

5. Make a Decision.

Significance from a Table

▸ The graphing calculator and JMP provide the most accurate P-value

calculations. Tables can also be used to estimate P-values.

▸ There are two methods of determining the P-Value for a z-statistic.

15.5 Significance from a Table

Significance from a Table (Method 1—Table C)

This method uses Table C (page 679 in the textbook).

1. Compare z with the critical values z* and the bottom of Table C.

2. If Z falls between two values of z*, then the P-value falls between

the two corresponding values of P in the “One-sided P” or the “Two-

sided P” row of Table C.


Significance from a Table (Method 2—Table A)

This method uses Table A (page 676 in the textbook).

1. Compute the P-value, which is:

a. P (Z > z) for a Right-tailed test.

b. P (Z < z) for a Left-tailed test.

c. 2*P (Z > |z|) for a two-tailed test.

2. Compare the P-value with α.


Significance from a Table

▸ Using technology to compute P-values is most preferred.

▸ For our purposes, using Table C is a suitable alternative to technology.

This may not produce the same accuracy as the other options, but it

will strengthen estimation skills.


Example: P(Z< -1.45) via Table C

▸ We need to estimate P (Z < -1.45) via Table C.

1. Search the z* row of Table C.

2. 1.45 is between 1.282 and 1.645.


Example: P(Z< -1.45) via Table C

▸ We need to estimate P (Z < -1.45) via Table C.

1. Search the z* row of Table C.

2. 1.45 is between 1.282 and 1.645.

3. Therefore its P-value is between .10 and .05 (similar to a 1-sided test).

4. .05 < P-value < .10


Example: P(Z< -1.45) via Table A

▸ We need to estimate P (Z < -1.45) via Table A.

1. Compute P( Z < -1.45) from Table A.


Example: P(Z< -1.45) via Table A

▸ We need to estimate P (Z < -1.45) via Table A.

1. Compute P( Z < -1.45) from Table A.

2. P(Z < -1.45) = .0735.

▸ Notice that this coincides with the answer from Table C.

▸ Table C will allow us to work with distributions where the population

standard deviation is unknown (this will be covered in Chapter 18).


Five-Minute Summary

▸ List at least 3 concepts that had the most impact on your knowledge of

tests of significance.

___________ _____________

_______________

chapter 15: tests of significance: the basics stat 1450

Documents

tests of significance

reasoning of tests

null hypothesis value

alternative hypotheses

population parameter

significance situation

null value

hypothesis test