Chapter 15: Tests of Significance: The
Basics
STAT 1450
15.0 Tests of Significance
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.
15.1 The Reasoning of 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
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.
15.1 The Reasoning of Tests of Significance
The Reasoning of Tests of Significance
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.
____
15.2 Stating Hypotheses
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?
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?
 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..
15.2 Stating Hypotheses
Example: Coffee Consumption
▸ Let’s use one of the values from a) to compose the null and alternative
hypotheses.
15.2 Stating Hypotheses
Example: Coffee Consumption
▸ 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__
15.2 Stating Hypotheses
Example: Coffee Consumption
▸ 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.
15.2 Stating Hypotheses
Example: Coffee Consumption
▸ 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.
15.2 Stating Hypotheses
Stating Hyptotheses
▸ Important note: Base your alternative hypothesis on your question of
interest—do not base it on the data.
15.3 P-values & Statistical Significance
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.
15.3 P-values & Statistical Significance
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.
15.3 P-values & Statistical Significance
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.”
15.3 P-values & Statistical Significance
Tests of Significance & the Justice System
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…)
-OrThe jury rules that the defendant is ‘guilty.’
15.3 P-values & Statistical Significance
Tests of Significance & the Justice System
▸ Note: Our jury system assumes innocent until proven guilty.
▸ The actual truth of whether the person indeed committed the crime
may never be known.
15.3 P-values & Statistical Significance
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.
15.3 P-values & Statistical 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.
15.3 P-values & Statistical Significance
Significance Level
▸ After choosing an appropriate level of significance, we can make a
decision about H0.
P-Value vs. Decisions about H0
α
P-value > α
Ho
P-value ≤ α
Ho
15.3 P-values & Statistical Significance
Significance Level
▸ After choosing an appropriate level of significance, we can make a
decision about H0.
P-Value vs. Decisions about H0
α
P-value > α
Fail to Reject Ho
P-value ≤ α
Reject Ho
15.3 P-values & Statistical Significance
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.
15.4 Tests for a Population Mean
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
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
𝑥−𝜇
𝑧=
𝜎 𝑛
has the z distribution.
▸ To test the hypothesis 𝐻0 : 𝜇 = 𝜇0 , compute the one-sample z statistic
𝑥 − 𝜇0
𝑧=
𝜎 𝑛
15.4 Tests for a Population Mean
Tests for a Population Mean
Key Words
Alternative
Hypothesis
Rejection Region
P-value
(“extremely unlikely values”)
15.4 Tests for a Population Mean
Tests for a Population Mean
Key Words
“more than”
“increased”
Alternative
Hypothesis
P-value
𝐻𝑎 : 𝜇 > 𝜇0
𝑃(𝑍 ≥ 𝑧)
Rejection Region
(“extremely unlikely values”)
15.4 Tests for a Population Mean
Tests for a Population Mean
Alternative
Hypothesis
P-value
“more than”
“increased”
𝐻𝑎 : 𝜇 > 𝜇0
𝑃(𝑍 ≥ 𝑧)
“less than”
“reduced”
𝐻𝑎 : 𝜇 < 𝜇0
𝑃(𝑍 ≤ 𝑧)
Key Words
Rejection Region
(“extremely unlikely values”)
15.4 Tests for a Population Mean
Tests for a Population Mean
Alternative
Hypothesis
P-value
“more than”
“increased”
𝐻𝑎 : 𝜇 > 𝜇0
𝑃(𝑍 ≥ 𝑧)
“less than”
“reduced”
𝐻𝑎 : 𝜇 < 𝜇0
𝑃(𝑍 ≤ 𝑧)
“different”
“is not”
𝐻𝑎 : 𝜇 ≠ 𝜇0
Key Words
Rejection Region
2 ∗ 𝑃(|𝑍| ≥ 𝑧)
(“extremely unlikely values”)
15.4 Tests for a Population Mean
Example: Coffee Consumption
▸ 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.)
15.4 Tests for a Population Mean
Example: Coffee Consumption
Is the average amount of coffee consumed daily not 20 oz.?
15.4 Tests for a Population Mean
Example: Coffee Consumption
Is the average amount of coffee consumed daily not 20 oz.?
(Hypotheses)
𝐻𝑜 : 𝜇 = 20 𝐻𝑎 : 𝜇 ≠ 20
15.4 Tests for a Population Mean
Example: Coffee Consumption
Is the average amount of coffee consumed daily not 20 oz.?
(Hypotheses)
𝐻𝑜 : 𝜇 = 20 𝐻𝑎 : 𝜇 ≠ 20
(Conditions)
Random sample:
We are told that this is a random sample of 48.
15.4 Tests for a Population Mean
Example: Coffee Consumption
Is the average amount of coffee consumed daily not 20 oz.?
(Hypotheses)
𝐻𝑜 : 𝜇 = 20 𝐻𝑎 : 𝜇 ≠ 20
(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.
15.4 Tests for a Population Mean
Example: Coffee Consumption
Is the average amount of coffee consumed daily not 20 oz.?
(Hypotheses)
𝐻𝑜 : 𝜇 = 20 𝐻𝑎 : 𝜇 ≠ 20
(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 sample:
Not told that coffee consumption follows
a Normal distribution.
However, n = 48  40, so we can use the CLT.
15.4 Tests for a Population Mean
Example: Coffee Consumption
(Test Statistic) 𝑧 =
𝑥−𝜇0
𝜎 𝑛
15.4 Tests for a Population Mean
Example: Coffee Consumption
(Test Statistic) z =
𝑥−𝜇0
𝜎 𝑛
𝑧=
(26.31−20)
(9.2 48)
15.4 Tests for a Population Mean
Example: Coffee Consumption
(Test Statistic) 𝑧 =
𝑥−𝜇0
𝜎 𝑛
𝑧=
(26.31−20)
(9.2 48)
= 4.75
15.4 Tests for a Population Mean
Example: Coffee Consumption
(Test Statistic) z =
𝑥−𝜇0
𝜎 𝑛
𝑧=
(26.31−20)
(9.2 48)
= 4.75
15.4 Tests for a Population Mean
Example: Coffee Consumption
(Test Statistic) z =
(P-value)
𝑥−𝜇0
𝜎 𝑛
𝑧=
(26.31−20)
(9.2 48)
= 4.75
2*P( Z > 4.75) = 2*.0001 = .0002
15.4 Tests for a Population Mean
Example: Coffee Consumption
(Test Statistic) 𝑧 =
(P-value)
(Decision)
𝑥−𝜇0
𝜎 𝑛
𝑧=
(26.31−20)
(9.2 48)
= 4.75
2*P( Z > 4.75) = 2*.0001 = .0002
.0002 is an extremely small probability.
There is enough evidence to conclude the
mean amount of coffee consumed is not 20 oz.
15.4 Tests for a Population Mean
Technology Tips – Conducting Tests of
Significance (σ known)
▸ TI-83/84. STAT  TESTS  ZTest  Enter.
 Select Stats. Enter 𝜇0 , 𝜎, 𝑥 and n. Select Calculate.
 (Note: Select Data when 𝑥 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 𝜇0 , 𝜎, and click on OK.
15.4 Tests for a Population Mean
Technology Tips – Conducting Tests of
Significance (σ known)
▸ TI-83/84. STAT  TESTS  ZTest  Enter.
 Select Stats. Enter 𝜇0 , 𝜎, 𝑥 and n. Select Calculate.
 (Note: Select Data when 𝑥 and n are not provided. Then enter the list where the
data are stored
(for this example)
 Inpt >> STATS
 m0: 20 >> s: 9.2 >>
 Calculate (ENTER)
𝒙 : 26.31 >> n : 48
>> m : ≠
15.4 Tests for a Population Mean
Technology Tips – Conducting Tests of
Significance (σ known)
▸ TI-83/84. STAT  TESTS  ZTest  Enter.
 Select Stats. Enter 𝜇0 , 𝜎, 𝑥 and n. Select Calculate.
 (Note: Select Data when 𝑥 and n are not provided. Then enter the list where the
data are stored
(for this example)
 Inpt >> STATS
 m0: 20 >> s: 9.2 >>
 Calculate (ENTER)
𝒙 : 26.31 >> n : 48
>> m : ≠
z=4.7518 p=2.018 E -6
=> .000002 ≈ 0.
Reject H0, given the p-value ≈ 0.
15.4 Tests for a Population Mean
Technology Tips – Conducting Tests of
Significance (σ known)
▸ TI-83/84. STAT  TESTS  ZTest  Enter.
 Select Stats. Enter 𝜇0 , 𝜎, 𝑥 and n. Select Calculate.
 (Note: Select Data when 𝑥 and n are not provided. Then enter the list where the
data are stored
(for this example)
 Inpt >> STATS
 m0: 20 >> s: 9.2 >>
 Calculate (ENTER)
𝒙 : 26.31 >> n : 48
>> m : ≠
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.
15.4 Tests for a Population Mean
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) )
15.4 Tests for a Population Mean
Tests for a Population Mean: 4-Step Process
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 …)
15.4 Tests for a Population Mean
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 =.05 level?
15.4 Tests for a Population Mean
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 =.05 level?
STATE: Do affluent communities have a mean IQ score above 100?
15.4 Tests for a Population Mean
Example: IQ Scores
Plan:
a.
Identify the parameter
m = average IQ score for affluent communities
15.4 Tests for a Population Mean
Example: IQ Scores
Plan:
a.
Identify the parameter
m = average IQ score for affluent communities
b. List all given information from the data collected.
_n=35, _s= 15,
𝑥 = 112
15.4 Tests for a Population Mean
Example: IQ Scores
Plan:
a.
Identify the parameter
m = average IQ score for affluent communities
b. List all given information from the data collected.
_n=35, _s= 15,
𝑥 = 112
c. State the null (H0) and alternative (HA) hypotheses. H0: _m = 100 HA: _m > 100
15.4 Tests for a Population Mean
Example: IQ Scores
Plan:
a.
Identify the parameter
m = average IQ score for affluent communities
b. List all given information from the data collected.
_n=35, _s= 15,
𝑥 = 112
c. State the null (H0) and alternative (HA) hypotheses. H0: _m = 100 HA: _m > 100__
d. Specify the level of significance.
 =.05
15.4 Tests for a Population Mean
Example: IQ Scores
Plan:
a.
m = average IQ score for affluent communities
Identify the parameter
b. List all given information from the data collected.
_n=35, _s= 15,
𝑥 = 112
c. State the null (H0) and alternative (HA) hypotheses. H0: _m = 100 HA: _m > 100__
d. Specify the level of significance.
e.
Determine the type of test.
 =.05
(Left-tailed, Right-tailed, Two-Tailed)
15.4 Tests for a Population Mean
Example: IQ Scores
Plan:
a.
m = average IQ score for affluent communities
Identify the parameter
b. List all given information from the data collected.
_n=35, _s= 15,
𝑥 = 112
c. State the null (H0) and alternative (HA) hypotheses. H0: _m = 100 HA: _m > 100__
d. Specify the level of significance.
 =.05
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) )
15.4 Tests for a Population Mean
Tests for a Population Mean: 4-Step Process
3. Solve:
a.
Check the conditions for the test you plan to use.
Random sample?
Yes
Large population : sample ratio?
Certainly more than 35
affluent households
Large enough for sample?
We were informed that
the data came from a
Normal Distribution.
15.4 Tests for a Population Mean
Tests for a Population Mean: 4-Step Process
3. Solve:
a.
Check the conditions for the test you plan to use.
Random sample?
Yes
Large population : sample ratio?
Certainly more than 35
affluent households
Large enough for sample?
We were informed that
the data came from a
Normal Distribution.
b. Calculate the test statistic.
𝑥 − 𝜇0 112 − 100
𝑧=
=
= 4.73
𝜎 𝑛
15 35
15.4 Tests for a Population Mean
Tests for a Population Mean: 4-Step Process
3. Solve:
a.
Check the conditions for the test you plan to use.
Random sample?
Yes
Large population : sample ratio?
Large enough for sample?
Certainly more than 35
affluent households
We were informed that
the data came from a
Normal Distribution.
b. Calculate the test statistic.
𝑥 − 𝜇0 112 − 100
𝑧=
=
= 4.73
𝜎 𝑛
15 35
c. Determine (or estimate) the P-Value
P(Z > 4.73) = .0001
15.4 Tests for a Population Mean
Tests for a Population Mean: 4-Step Process
4. Conclude:
a. Make a decision about the null hypothesis. P-value = .0001 ≤ .05 =  (default)
(Reject H0 or Fail to reject H0).
15.4 Tests for a Population Mean
Tests for a Population Mean: 4-Step Process
4. Conclude:
a. Make a decision about the null hypothesis. P-value = .0001 ≤ .05 =  (default)
(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 …)
15.4 Tests for a Population Mean
Tests for a Population Mean: 4-Step Process
4. Conclude:
a. Make a decision about the null hypothesis. P-value = .0001 ≤ .05 =  (default)
(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 …)
 There is enough evidence at the =.05 level to conclude that persons from affluent
communities have a mean IQ score above 100.
15.4 Tests for a Population Mean
Technology Tips – Conducting Tests of
Significance (σ known)
▸ TI-83/84. STAT  TESTS  ZTest  Enter.
 Select Stats. Enter 𝜇0 , 𝜎, 𝑥 and n. Select Calculate.
 (Note: Select Data when 𝑥 and n are not provided. Then enter the list where the
data are stored
(for this example)
 Inpt >> STATS
 m0: 100
>> s: 15 >>
 Calculate (ENTER)
𝒙 : 112
>> n : 35
z=4.733 p=1.11 E -6
>> m : >
=> .0000011 ≈ 0.
There is enough evidence to conclude affluent
communities have a mean IQ score above 100.
15.4 Tests for a Population Mean
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.
15.4 Tests for a Population Mean
Final Note
Steps for SuccessConducting Tests of Significance
1. Set up your Hypotheses.
15.4 Tests for a Population Mean
Final Note
Steps for SuccessConducting Tests of Significance
1. Set up your Hypotheses.
2. Check your Conditions.
15.4 Tests for a Population Mean
Final Note
Steps for SuccessConducting Tests of Significance
1. Set up your Hypotheses.
2. Check your Conditions.
3. Compute the Test Statistic.
15.4 Tests for a Population Mean
Final Note
Steps for SuccessConducting Tests of Significance
1. Set up your Hypotheses.
2. Check your Conditions.
3. Compute the Test Statistic.
4. Compute the P-Value.
15.4 Tests for a Population Mean
Final Note
Steps for SuccessConducting Tests of Significance
1. Set up your Hypotheses.
2. Check your Conditions.
3. Compute the Test Statistic.
4. Compute the P-Value.
5. Make a Decision.
15.5 Significance from a Table
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.
15.5 Significance from a Table
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.
15.5 Significance from a Table
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.
15.5 Significance from a Table
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.
15.5 Significance from a Table
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
15.5 Significance from a Table
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.
15.5 Significance from a Table
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.
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