Download 7.1 Statistical Hypotheses

Chapter 7 – Hypothesis Testing with One Sample
7.1 Statistical Hypotheses
A statistical hypothesis is a mathematical claim about a population parameter.
Examples:
 The mean height of women is less than 65 inches tall.   65 ,
 The percentage of Floridians favoring a bullet train is 57%. p  .57
 The average distance driven per year by Americans is more than 10,000 miles.
  10, 000
 At least 5% of Americans earn more than $100,000 per year.
p  .05
Hypothesis Testing - Basic Procedure
If we wanted to know whether any of the above hypotheses are true, we would conduct
a hypothesis test. When we test a statistical hypothesis, we follow the following basic
procedure:
1. Draw a random
sample for the random variable in question.
2. Determine if the results from the sample data are consistent or inconsistent
with the hypothesis.
3. If the sample data is significantly
different from the null
hypothesis (Ho), we would reject the null hypothesis (as
being false), and support the alternative hypothesis (Ha).
If
the data is not significantly different, we would not reject the null hypothesis –
there would not be any evidence to support the alternative hypothesis.
We can only “support” a claim if the null is rejected and we support the alternative.
The goal, then, is to reject the Ho and say Ha is correct. The Ho will always contain
the condition of equality.
1
Formal Hypothesis Tests
In a formal hypothesis test, the opposite claims are given the names null hypothesis
hypothesis is denoted by H 0 and the
alternative hypothesis is denoted by H a . The null and alternative
and alternative hypothesis. The null
hypotheses need to be assigned as follows (see also p 367):
The null


hypothesis is the hypothesis being tested.
be the hypothesis we want to reject
contain the condition of equality (, , )
The alternative
H a must:
H0
must:
hypothesis is always the opposite of the null hypothesis.
 be the hypothesis we want to support
 not contain the condition of equality (, ,  )
A formal hypothesis test will always conclude with a decision to reject H 0 based
on sample data, or the decision that there is not strong enough evidence to reject H 0 .
Example: A battery manufacturer claims that the average life of its batteries is at least
300 minutes. To test the claim, a sample of 100 batteries is drawn. The sample is
tested and the mean battery life of the sample is 294 minutes with a sample standard
deviation of 20 minutes.
a) Write the null and alternative hypotheses.
b)
If the null hypothesis were rejected, what would you conclude?
c)
If the null hypothesis were not rejected, what would you conclude?
2
Example: A tire manufacturer claims its tires last on average more than 35,000 miles.
If we think the claim is false, then we would write the claim as H 0 , remembering to
include the condition of equality. Our hypothesis for this test would be:
H o:
H a:
We would then hope that our sample data would allow the rejection of the null
hypothesis, refuting the company’s claim.
Example: On the other hand, if we worked for the tire company and wanted to gather
evidence to support their claim, then we would make the company’s claim H a , and
remember that equality should not be included in the claim. Our hypothesis test would
then use the hypotheses:
H o:
H a:
If the sample data were able to support the rejection of H 0 , this would be strong
evidence to support the claim H a which is what the company believes to be true.
Note: For simplicity, in your text and on all quizzes and tests, it the condition of
equality is inferred by the wording, that will automatically be the null hypothesis.
(Look at the example above and decide which would be the best “as worded”.)
3
Types of Error
Whenever sample data is used to make an estimate of a population parameter, there is
always a probability of error due to drawing an unusual sample. There
are two main types of error that occur in hypothesis tests.
Type I Error – A sample is chosen whose sample data leads to the
rejection of the null hypothesis when, in fact, H 0 is true.
Type II Error – A sample is chosen whose sample data leads to not
rejecting the null hypothesis when, in fact, H 0 is false.
Summarizing:
True
Type I Error
Correct Decision
False
Correct Decision
Type II Error
H0
H0
H0
Rejected
Not Rejected
H0
In layman’s terms, Type I error is to think you are right (because you
have rejected the null hypothesis) when you are wrong, and Type II
error is to think you may be wrong (because you did not reject the
null hypothesis) when you are actually right. (Also see p 368.) We
would rather make a Type II than a Type I error.
Exercise: Go back and determine what a Type I Error and a Type II Error would be for our tire examples.
4
Level of Significance
In hypothesis tests, a conservative
approach is usually taken toward
the rejection of the null hypothesis. That is, we want the probability of
making a Type I Error to be small. (We’d rather actually be right and think we’re
wrong than think we’re right and actually be wrong.)
The maximum acceptable probability (of making a Type I error) is usually chosen at
the beginning of the hypothesis test, and is called the level of significance for the
test. The level of significance is denoted by  , and the most commonly used values
are   .10 ,   .05 , and   .01 . Each of these represents the ________________
____________________________________________.
The probability of making a Type II Error in a hypothesis test is denoted by  . Once
 is determined, the value of  is also fixed, but the calculation of this value is
beyond the scope of this course.
5
Types of Tests
There are three
basic types of hypothesis tests (see p 338 text, old book p 326):
The location of the “tail” is based on what we actually want to prove (Ha). The
HYPOTHESISED parameter (mean or proportion) will be “in the middle.”
Left-tailed Test –
used when the null hypothesis being tested is a claim that the
population parameter at least () a given value. Note that the alternative
hypothesis then claims that the parameter is less than (<) the
value.
Right-tailed Test – used when the null hypothesis being tested is a claim that the
population parameter is at most () a given value. Note that the alternative
hypothesis then claims that the parameter is greater than (>) the
value.
Two-tailed Test –
used when the null hypothesis being tested is a claim that the
population parameter is equal to (=) a given value. Note that the alternative
hypothesis then claims that the parameter is not equal to () the
value.
6
Given the three examples below, decide what type of test would be used: a LEFTTAILED, RIGHT-TAILED or TWO-TAILED test.
Example: The Census Bureau claims that the percentage of Tampa Area residents
with a bachelor’s degree or higher is 24.4%. We would write the null and alternative
hypotheses for this claim as:
H 0 : p  .244
H a : p  .244
In this case, we would have to reject H 0 if our sample percentage was either
significantly more than 24.4%, or significantly less than 24.4%. That is, if our sample
proportion was in ________________________ of the distribution of all sample
proportions.
Example: From our first tire manufacturer example:
H 0 :   35,000
H a :   35,000
We would reject H 0 in the case above if our sample mean were significantly less than
35,000. That is, if our sample mean was in _____________________ of the
distribution of all sample means.
Example: From our second tire manufacturer example:
H 0 :   35,000
H a :   35,000
We would reject H 0 in this case if our sample mean were significantly more than
35,000. That is, if our sample mean was in _______________________ of the
distribution of all sample means.
7
7.2 Testing a Claim about the Mean Using a Large Sample
When a hypothesis test involves a claim about a population mean, then we will draw a
sample and look at the sample mean to test the claim. If the sample drawn is large
enough ( n  30 ), then the Central Limit Theorem applies, and the distribution of sample
means is approximately normal. As usual, we also have that: x   and  x 

n

s
.
n
Note: Since s and n are known from the sample data, we have a good estimate of  x ,
but we do not know  since this is the parameter we are testing a claim about. In order
to have a value for  , we will always
assume that the null hypothesis is
true in any hypothesis test.
Since the null hypothesis must be of one of the following types:
  0 ,
  0 ,
where 0 is a constant, we will always assume
or   0
for the purpose of our test that
  0 .
That is, the hypothesized parameter (mean or proportion) will be in
the “middle” of our normal distribution sketch.
8
The Standardized Test Statistic
There are three methods of determining if you should reject or not reject the null
hypothesis:
Confidence intervals
(is the hypothesized statistic within the confidence interval for the sample)
P- values
(how probable it is that the sample statistic will be at/below/above the
hypothesized parameter - that is, the probability of rejecting H0) This means
that as the P-value gets smaller, the ________ likely we are to make a Type
I error (reject the null hypothesis when it is actually true.)
Critical Values
(how many standard deviations the sample is away from the hypothesized
parameter) The more standard deviations from the hypothesized parameter,
the ____________ likely we are to reject H0.
Note: The same conclusion should be made regardless of the method chosen.
Let’s go back to our previous example to see how this might work . . .
9
Example: A battery manufacturer claims that the average life of its batteries is at least
300 minutes (   300 ). To test the claim a sample of n = 100 batteries is drawn. The
sample of batteries is tested and the mean battery life in the samples found to be
x  294 minutes with a sample standard deviation of s = 20 minutes. Is this data
sufficiently different from the manufacturer’s claim to justify rejecting the claim as
false?
If the manufacturers claim is correct, then  x    300 , and so we will assume that
 x    300 . From our sample, we can also estimate that  x 
The observed value
x  294
s
20

 2.
n
100
has a z-score (standardized test statistic) of
z
x  x
x

294  300
 3.0
2
Looking in the Standard Normal Table, we find that P( x  294)  .0013 .
Now, one of the following must be true:
Our assumption that   300 is incorrect.
OR
We have drawn a sample whose mean is so small that only 13 in 10,000
samples have a mean as low.
The likelihood of the second statement being true is quite small (.0013). Thus, we
have strong evidence to believe that the first statement is true, and hence that the
manufacturer overstated the mean lifetime of its batteries.
Now let’s reflect on this . . .
How could each of the following be represented in the example above?
P- values
Critical Values
10
Example: Suppose we believe that the mean body temperature of healthy adults is less
than the commonly accepted measurement of 98.6 F. A sample of 60 healthy adults is
drawn with an average temperature of x  98.2 F and with a sample standard deviation
of s  1.1 .
Our hypotheses in this case would be:
H0 : 
98.6
Ha : 
98.6
So we have a ________-tailed test with 0  98.6 .
Based on our sample data, our standardized test statistic is:
z
x  0
s/ n

98.2  98.6 0.4

 2.82
.142
1.1/ 60
How unusual is this? Do you think we should or should not reject the null hypothesis?
Exercise: Suppose we decide to make the stronger claim that the mean body
temperature was less than 98.4. Find the new standardized test statistic.
Exercise: A researcher believes that the average commuting time for Tampa
commuters is more than 25 minutes. A random sample of 100 Tampa commuters finds
that the average commuting time is x  25.5 minutes with a standard deviation of s  11.5
minutes. If we wish to support the researcher’s claim, determine the null and
alternative hypotheses, the type of test, and the standardized test statistic.
11
The P-Value Method
The P-value (probability value) of a test is the probability of drawing a random sample
whose standardized test statistic is at least as contrary to the claim of the null
hypothesis as that observed in the sample group.
Example: In the hypothesis test for body temperature given above, we had:
H 0 :   98.6
H a :   98.6
Our sample had a mean temperature of x  98.2 which is contrary to the null hypothesis.
Only a sample group with an average temperature less than 98.2 would be stronger
evidence against H 0 . Thus the P-value of this test is P( x  98.2) .
Since the z-score of x is just our standardized test statistic z which has the Standard
Normal Distribution, P( x  98.2)  P( z  2.82)  .0024 .
Since the probability of drawing a sample as contrary to the null hypothesis as the
observed sample (assuming H 0 is true) is small, we would decide to reject H 0 .
12
Calculating P-Values
In the example above, we calculated the P-value of the test by finding the area to the
left of the standardized test statistic z on the standard normal curve. Notice that the
example above was also a left-tailed test, and that any hypothesis test which is lefttailed will have the P-value calculated exactly as above. (See diagram on p. 326 old 348 new)
Similarly, for a right-tailed test, we would calculate the
P-value by finding the area to the right of the standardized test statistic.
For a two-tailed test, the null hypothesis is always claiming that   0 , and so the
sample data is contrary to this claim if the sample mean is either much higher or much
lower than 0 . The P-value for a two-tailed test then is the area in both tails of the
normal distribution more extreme than the standardized test statistic. Since the normal
distribution is symmetric, this is just twice the area in one tail. (See diagram on p. 354 old
386 new)
Example: The P-value of the battery life example (p 10 notes) would be:
P( z  3.0)  .0013
Example: A university claims that the average SAT score for its incoming freshmen
is 1080. A sample of 56 freshmen at the university is drawn and the average SAT
score is found to be x  1044 with a sample standard deviation of s  94.7 points.
Our hypotheses in this case would be:
H0 : 
1080
H : 
1080
So we have a _________-tailed test with 0  1080 .
Our standardized test statistic is:
z
x  0
s/ n

1044  1080 36

 2.85
94.7 / 56 12.65
Since the test is two-tailed, our P-value is given by:
P( z  2.85 or z  2.85)  2  P( z  2.85)  2  .0022  .0044
13
Exercise: Find the P-value for the test on Tampa commuting times: A researcher
believes that the average commuting time for Tampa commuters is at least 25 minutes.
A random sample of 100 Tampa commuters finds that the average commuting time is
x  25.5 minutes with a standard deviation of s  11.5 minutes. Find the P-value and
think about what you might conclude. (See notes page 11.)
Deciding to Reject the Null Hypothesis
In the examples above, we saw that a very small P-value would lead us to reject the
null hypothesis, and a high P-value would not.
Since the P-value of a test is the probability of randomly drawing a sample at least as
contrary to H 0 as the observed sample, we can also think of the P-value as the
probability that we will be wrong if we choose to reject H based
on our sample data. The P-value then is the probability of making a
Type I Error.
0
Recall that the maximum acceptable probability of making a Type I Error is the level
of significance  , and is usually determined at the outset of the hypothesis test.
The rule we will use to decide whether to reject H 0 is:
Reject H 0 if P  
Do not reject H 0 if P  
Now go back to previous example and decide if we should reject the null hypothesis
based on an alpha level of .01.
14
PROCESS for HYPOTHESIS TESTING:
1.
2.
3.
4.
5.
Write an Ha alternative hypothesis (you want to support)
and a Ho null hypothesis (include equality) (you want to
reject).
Select a level of significance α = 0 .1, 0.05, 0.01 (standard)
Calculate the test statistic
If the P value is < α, reject Ho, conclude Ha
If the P value is > α, do not reject Ho
Write a sentence regarding your conclusion.
Exercise: Is there sufficient evidence to support the researchers claim that the mean
commuting time of Tampa commuters is more than 25 minutes at a   .05 level of
significance? Recall that the sample mean was 25.5 minutes, and the sample standard
deviation was 11.5 min for a sample size of 100. Is there sufficient evidence to support
a claim that the mean commuting time is more than 22 minutes at an   .05 level of
significance?
15
Rejection Regions and Critical Values
A second method to determine whether to reject the null hypothesis is to use rejection
regions and critical values.
A rejection region for a hypothesis test is the range of values for the standardized test
statistic that would cause us to decide to reject the null hypothesis. Critical values
for a hypothesis test are the z-scores that separate the rejection
region(s) from the non-rejection region. The critical values will be
denoted by z0 .
The rejection region for a test is determined
by the type of test
(left/right/two tailed) and the level of significance  for the test.
For a left-tailed test, the rejection region is a region in the left tail of the normal
distribution, for a right tailed test, it is in the right tail, and for a two tailed test, there
are two equal rejection regions in either tail.
(See Diagram p. 341old 354 new)
Since 1- the level of confidence is the maximum acceptable probability of a Type I
Error (see t-distribution), we want the area under the normal curve in the rejection
region to have an area of  . We can use this area to find our critical values.
(See Diagram p. 341 old, p 354 new)
Once we establish the critical values and rejection region, if the
standardized test statistic for a sample data set falls in the rejection
region, we will reject the null hypothesis and conclude that we can
support the alternative hypothesis at the given alpha level (level of
significance.)
16
Example: In our body temperature example, we were using a left-tailed test. If the
level of significance was   .05 , then the rejection region would be the values in the
lowest 5% of the standard normal distribution. Looking up .05 in the standard normal
table (or bottom of t-distribution chart), we see that this corresponds to a z-score of at
most -1.645, so this would be our critical value z0 , and so our rejection region is
z  1.645 . Since our standardized test statistic z  2.82 falls into this rejection region,
we would choose to reject H 0 .
Example: In our SAT example, we had a two-tailed test, so the rejection region will
be the two tails at either end of the normal distribution. If we again want   .05 , then
the area in both rejection regions together should be .05, and so we will look up
 / 2  .025 in the standard normal table to get critical values of z0  1.96 . The rejection
region thus consists of: z  1.96 and z  1.96 . Since our standardized test statistic
z  2.85 falls in the rejection region, we would reject the university’s claim that   1080 .
Exercises: Find the critical values (z0)and rejection region(s):
 A right-tailed test with   .05
 A left-tailed test with
  .01
 A two-tailed test with   .10
 A right-tailed test with   .02
17
Exercise: Mercury levels in fish are considered dangerous to people if they exceed 0.5
mg mercury per kilogram of meat. A sample of 50 tuna is collected, and the mean
level of mercury in these 50 fish is 0.6 mg/kg with a standard deviation of 0.2 mg/kg.
A health warning will be issued if it the claim that the mean exceeds 0.5 mg/kg can be
supported at the   .10 level of significance.
Determine the null and alternative hypotheses in this case, the type of test, and the
critical value(s) and rejection region.
Find the standardized test statistic for this data. Should the health warning be issued?
What is the P-value of this test?
Do p. 392 #
18
7.3 Testing a Claim about the Mean Using a Small Sample
Recall that if a sample of values for a normal random variable is drawn and if the
sample size is less than 30, and the population standard deviation is unknown, then the
random variable
t
x 
s/ n
has the Student t-distribution with n 1degrees of freedom.
When testing a claim about the mean using sample data from a small sample, we
should therefore use the appropriate t-distribution instead of the standard normal
distribution to determine our standardized test statistic, critical values, rejection
regions, and P-values.
The standardized test statistic for a t-distribution test will be given by:
t
x  0
s/ n
Note: Probability values for the t-distributions are not given in the text, so we will
limit our discussion to using the critical value method in this case. However, P-values
for t-distribution tests can be calculated using the TI-83.
19
Finding Critical Values for the t-distributions
The process for locating the rejection regions for a t-distribution hypothesis test are the
same as for the normal distribution tests. The critical values however will be different.
To find the critical value(s) t0 for a test, first determine if the test is one-tailed or twotailed, and the level of significance  . The critical values can be found in the tdistribution table in the front of the book by looking up the entry in the column giving
the level of significance and the row giving the degree of freedom.
Note that:
 For a right-tailed test, t0 is the value in the table
 For a left-tailed test, t0 is the negative of the value in the table
 For a two-tailed test, there are two critical values t0 both the value and its
opposite
Example: SAT Math Scores are normally distributed. A sample of scores for 16
students has mean score of x  522.8 with a sample standard deviation of s  154.5 .
Suppose we wished to support the claim that the average SAT Math score exceeds 500
using a   .05 level of significance.
The null and alternative hypotheses in this case would be:
H0:
Ha:
So the test is ___________-tailed with 0  500 .
Using the t-distribution table with One Tail   .05 and ______ degrees of freedom, we
get a critical value of t0=
. The rejection region is thus: t
.
The standardized test statistic is given by:
t
x  0
s/ n

522.8  500
22.8

 .59
38.625
154.5 / 16
Because the standardized test statistic is _______ in the rejection region, we would
_______ reject H 0 , and so the sample data is _________ sufficient to support the claim
that the mean exceeds 500 at the .05 level of significance.
20
Example: A biologist measures the weights of anesthetized female grizzly bears
during winter. A sample of 14 bears is found to have a mean weight of x  376.6 lbs.
with a sample standard deviation of s  32.5 lbs. Is there sufficient evidence to support
the claim that the mean weight of all female bears in the area is less than 400 lbs. using
an   .01 level of significance? (Assume that the weights of female grizzly bears are
normally distributed)
The null and alternative hypotheses in this case would be:
So the test is ________-tailed with
0  .
Using the t-distribution table with One Tail   .01 and _____ degrees of freedom, we
get a critical value of t0  . The rejection region is thus:
t
The standardized test statistic is given by:
t
x  0
s/ n

The standardized test statistic is/is not in the rejection region, so we would/would not
reject H 0 , which supports/is insufficient to support the claim that the mean bear weight
is less than 400 lbs.
Exercise: A statistics professor gives his class their second exam. In order to estimate
the mean score, the professor chooses 8 exams at random and grades them. The mean
grade for the eight exams chosen is x  84 and they have a standard deviation of s  7 .
Assume that the students’ test scores are normally distributed.
The professor believes the average score will be higher than the mean of 80 on the first
exam. Conduct a hypothesis test using the   .05 level of significance, to support the
professor’s claim.
21
7.4 Testing a Claim about a Population Proportion
Recall that if a random sample is drawn and if the sample proportion p̂ is measured
and npˆ  5 , nqˆ  5 , then the distribution of p̂ is approximately normal with  p̂  p and
 pˆ  pq / n .
When testing a claim about a population proportion, the null hypothesis has one of the
following forms:
p  p0 , p  p0 , or p  p0
So as with the mean, we will assume p  p0 and
we will use the following standardized test statistic for a proportion test:
pˆ   pˆ
z
 pˆ

pˆ  p0
p0 q0 / n
Note: p0 and q0 are based on the hypothesis.
This random variable should have the standard normal distribution, and so we will
calculate all of our rejection regions, critical values, and P-values using the standard
normal distribution as we did when testing a mean using a large sample.
Example: A computer chip manufacturer tests microprocessors coming off the
production line. In one sample of 577 processors, 37 were found to have defects. The
company wants to claim that the proportion of processors that are defective is only 4%.
Can the company’s claim be rejected at the   .01 level of significance?
The null and alternative hypotheses in this case would be:
H 0 : p  .04
H a :   .04
So the test is two-tailed with p0  .04 .
Since
pˆ  37 / 577  .064 ,
the standardized test statistic is given by:
z
Looking up
pˆ  p0
p0 q0 / n

.064  .04
.024

 3.0
(.04)(.96) / 577 .008
z  3.00 in
the standard normal table we get a value of .9987, so
P( z  3.00)  1  .9987  .0013 and since we have a two-tailed test, the P-value is just twice
this amount or .0026. Since this is less than   .01 , we can reject the company’s claim.
22
Example: An opinion poll of 1018 randomly chosen American adults finds that only
34% approve of the president’s job performance. (CBS News Poll Feb ’06.) The
president’s political advisors want to know if this is sufficient data to show that less
than 4/10 of Americans approve of the president’s job performance using the   .05
level of significance.
The null and alternative hypotheses in this case would be:
H0 : 
Ha : 
So the test is left-tailed with
Since
p=
p0 
0.34, the standardized test statistic is given by:
z=
For a left-tailed test with   .05 , we will have z0  1.645 , and so since z =
z0
the null hypothesis should not/be rejected, and so the data does/not support the claim p
< 0.4 at the   .05 level of significance.
Exercise: Calculate the P-value of the above test.
Exercise: According to the IRS, 23.8% of all 2003 income tax returns were filed
electronically. To estimate whether the number has increased, a random sample of 500
tax returns for 2004 is selected, and the IRS finds that 132 of the returns were filed
electronically. Conduct a hypothesis test to support the claim that more than 23.8% of
2004 tax returns are filed electronically. What is the test statistic and P-value for the
test at   .01 and at  =.05 ? Write a conclusion and state why you do or do not reject H0.
23
Using the TI-83 to Conduct a Hypothesis Test: Mean – Large Sample
The program Z-Test on the TI-83 can be used to calculate the P-value and standardized test statistic for a hypothesis
test of the mean ( n  30 ). To run the program:



Press [STAT]
Use the arrow keys to highlight the TESTS Menu
Highlight 1: Z-Test from the list and press [ENTER]
As before, if you know n,  , and x , then make sure Inpt: is set to Stats. If you want the TI-83 to calculate from a
data set, you would choose Data for Inpt:
Enter the values of 0 ,  , x , and n.
Next choose the type of test by selecting the form of the alternative hypothesis (, ,  )
When you are finished highlight the word Calculate and press [ENTER]. The test statistic z, P-value, x , and n will
then appear on the screen.
Example: Using the data from our temperature example:
0  98.6   1.1 x  98.2 , and n  60 , the Z-Test program gives: z  2.816715161 with a P-value of
P  .0024259384
Using the TI-83 to Conduct a Hypothesis Test: Mean – Small Sample
The program T-Test on the TI-83 can be used to calculate the P-value and standardized test statistic for a tdistribution test. To run the program:



Press [STAT]
Use the arrow keys to highlight the TESTS Menu
Highlight 1: T-Test from the list and press [ENTER]
As before, if you know n, s, and x , then make sure Inpt: is set to Stats. If you want the TI-83 to calculate from a
data set, you would choose Data for Inpt:
Enter the values of 0 , s , x , and n.
Choose the type of test as with the Z-Test program. And then Calculate.
The test statistic t, P-value, x , s, and n will then appear on the screen.
Example: Using the data from our SAT example:
0  500 s  154.5 x  522.8 , and n  16 , the T-Test program gives: t  0.5902912621 with a P-value of
P  .2818916884
24
Using the TI-83 to Conduct a Hypothesis Test: Proportion
The program 1-PropZTest on the TI-83 can be used to calculate the P-value and standardized test statistic for a
proportion test. To run the program:



Press [STAT]
Use the arrow keys to highlight the TESTS Menu
Highlight 5: 1-PropZTest from the list and press [ENTER]
Enter the values of p0 , x , and n. As usual, if only p̂ is known, then x can be calculated as x  npˆ
Next choose the type of test as with the other tests, and Calculate.
The test statistic z, P-value, p̂ , and n will then appear on the screen.
Example: Using the data from our processor example:
p0  .04 , x  37 , and n  577 , the 1-PropZTest program gives: z  2.957234176 with a P-value of
P  .00310426
25