The document provides instruction on hypothesis testing using confidence intervals and the chi-square test. It gives the steps to conduct hypothesis tests using confidence intervals, including determining the confidence level and interval, calculating the test statistic, and comparing it to the interval to make a conclusion. It also provides the steps for chi-square tests, including calculating the chi-square statistic, determining degrees of freedom, finding the critical value from the chi-square table, and comparing the calculated statistic to make a conclusion. Several examples are worked through to demonstrate both methods.

1. LESSON 08: Hypothesis Testing (continued)
Lesson Objectives:
Student will test hypotheses using confidence intervals.
Student will test observed values versus expected values using
the chi-square test.
Testing Hypotheses with a Confidence Interval
Recall that a
confidence interval allows the researcher to find a probability
with a certain level of confidence
. Remember that the
empirical rule states that 68% of the data will fall within 1
standard deviation of the mean in either direction, 95% of the
data will fall within 2 standard deviations of the mean, and that
99.7% of the data will fall within 3 standard deviations of the
mean as it pertains to the area under the normal distribution
curve
. When we are testing a hypothesis we can set the confidence
level and find the probability that an event will or will not
occur according to the confidence interval. If we know the
confidence level then we can calculate the confidence interval
and determine the
rejection region
. If the test statistic falls in either of the rejection regions then
we can reject the null hypothesis. When testing a large sample
we use the z-score and there are set confidence intervals for
each confidence level when using z-scores and they are as
follows:
Confidence Level
Corresponding z-score
Corresponding z-score rejection region

2. 90%
1.645 z < -1.645 or z
> 1.645
95%
1.96 z < -1.96 or z
< 1.96
99%
2.575 z < 2.575 or z
> 2.575
Recall that the confidence interval formula is the sample mean
+/- z(standard deviation/√n)
For t-scores, it is a little different since the t-score depends on n
- 1 degrees of freedom. The first row of the t-table gives alpha
(a) which is 1 - confidence level. The 2nd column has .40
which is equal to the amount of area left in the two tails of the
normal distribution if the confidence level is set at 60%. The
3rd column represents 75% and so on. Below is a list of the
confidence levels and the corresponding alpha for each one so
that you will know which column holds the correct t-score based
on the given confidence level. Remember that the table gives
you the total area left after you have considered the confidence
level and it will need to be divided by two to come up with the
area in each tail of the normal distribution.
Confidence level
Corresponding a-value for t-table
60% .40
75% .25
90% .10
95% .05
97.5% .025
99% .01
99.5% .005
99.95% .0005
Recall that the confidence interval formula is the sample mean

3. +/- t(standard deviation/√n)
Let's practice:
The average yearly income of families in a particular state is
$40,000. However a sample of 100 people show that the
average income is $45,000 with a population standard deviation
of $500. Conduct a hypothesis test at confidence level 95% to
see if the true average income of families reported by the state
is correct.
Step 1
: State the null and alternative hypothesis
H
0
= sample mean = population mean
H
1
= sample mean ≠ population mean
Step 2
: Determine the confidence interval. (we will use a z-score
since this is a large sample)
Since the confidence level is 95%, the confidence interval is
-1.96< z <1.96
Step 3: Determine the z-score
z =
45000-40000
/500=10
Step 3:
Compare this z-score to the confidence interval and make a
conclusion.
This is outside of the confidence interval which means that we
must reject the null hypothesis and conclude that the average
income reported by the state is incorrect.
A particular school gives annual standardized tests at the end of
the year and last year's average score was 70 with a standard
deviation of 5. A sample of 10 students' tests were pulled and

4. the average score was 85. Considering a 99% confidence level,
did the school report the correct overall average for the
students' test scores?
State the null and alternative hypotheses
.
H
0
= sample mean = population mean
H
1
= sample mean
≠ population mean
Determine the confidence level(we will use a t-score since the
sample size is small)
10 -1 = 9 degrees of freedom
99% confidence level = t
.005 with 9 degrees of freedom
= 3.24 or 3.24/2 = 1.62
So the confidence interval is -1.62 > t > 1.62 and the rejection
region is t < -1.62 or t > 1.62
Determine the t-score
t = 85 - 70 / 5 = 3
Compare the t-score to the confidence interval and make a
conclusion.
Since the t-score is beyond the confidence interval and within
the rejection region we must reject the null hypothesis.
Using the Chi-Squared Distribution
The
chi-squared statistic (x
2
) is used to compare observed values to the expected values in
an experiment
.
The formula for the chi-squared statistic is the sum of all
(observed values - expected value)

5. 2
/ expected value
. The chi-squared table is much like the t-table meaning that
you have to calculate the degrees of freedom but this time we
use the number of categories -1 rather than n -1. The chi-
squared table has p-values listed in the first row which
represent the area left under the chi-squared distribution as it
relates to the confidence interval.
P=.05 represents a 95% confidence interval, P= .01 represents a
90% confidence interval, and P= .001 represents a 99% interval
. Once you have found the degrees of freedom then you look
over to the corresponding column for the p-value. The number
where that row and that column meet is the chi-squared
statistic. The data must fill two conditions in order to use the
chi-squared distribution: 1) the total observed values must
exceed 20 and 2) the expected value must exceed 4 for each
category. Watch this
video
to see what the chi-squared distribution looks like and how it
works.
Let's Practice:
A high school principal gave a questionnaire to 25 boys and 25
girls to see if gender played a role in the students' responses to
the following statement:
"All gym classes should be scheduled at the end of the day."
1- disagree 2-don't know 3-agree
The principal is only concerned with the number of disagree
responses obtained. The principal's hypotheses and the results
of the questionnaire are listed below:
H
0
= Gender does not play a role in the students' responses.
H
1
= Gender does play a role in the students' responses.

6. Categories (Gender)
Frequency of "disagree" responses
Boys
20
Girls
10
Step 1
: Organize all values to find x
2
Steps to finding x
2
Boys
Girls
Total
Observed Values (o)
20
10
30
Expected Values (e)
15
15
30
o - e
5
-5
( o - e)
2
25
25
(o - e)
2
/ e

7. 1.6
1.6
the sum of all (o -e)
2
/ e
1.6
1.6
3.2
What does this mean?
The observed values are the actual number of males and females
who answered disagree to the question. The expected values is
the average of the observed values(total expected value/# of
categories). Now that we have calculated the chi-squared or
critical value we must compare it to the actual chi-squared
statistic. We find the degrees of freedom by subtracting 1 from
the number of categories. Since we have two categories (male
and female) the degrees of freedom = 2 - 1 = 1. Remember that
the confidence level is 95% so that means that there is a = .05
so we are looking for the number where row 1(degrees of
freedom) and column p = .05 (alpha level) intersect. Find this
number on the
chi-squared table
. The value is 3.84. To conclude we must determine whether x
2
is greater than or equal to that value or less than that value. If
the chi squared that we calculated from the set of data is ≥ the
p-value
(value from chi-squared table)
then we must reject the null hypothesis and if it is less than the
p-value then we can not reject the null hypothesis. Since x
2
= 3.2 and 3.2 < 3.84 we can not reject the null hypothesis. So

8. the principal can not rule out the fact that the genders did not
play a role in the students' responses to the statement.
Additional Resources
chi-squared worked examples
NEXT TEACHER OFFICE HOURS ARE:
Grading Rubric:
Grading for this lesson:
To get a 10
: All answers are correct the first time, or within first revision.
To get a 9
: You can have 1 incorrect answer after your original
submission.
To get an 8
: You can have 2 incorrect answers after your original
submission.
To get a 7
: You can have 3 incorrect answers after your original
submission.
To get a 6
: You can have 4 incorrect answers after your original
submission.
To get a 5
: Cheating- Plagiarism - purposeful or mistaken, which will
lower your finalgrade for the course (so be very careful when
posting your work!); lack of effort, disrespect, or attitude (we
are here to communicate with you if you don't understand
something);
lesson requirements have not been met.
Note: For this class it is necessary to post the questions over
each answer. Failure to do so will result in asking for a
revision.
No grade will be given for incomplete work
.

9. Assignment:
For questions 1- 5 use confidence intervals to test the
hypothesis.
1) A light bulb producing company states that its lights will
last an average of 1200 hours with a standard deviation of 200
hours. A sample of 100 light bulbs from the company were
tested and the researcher found that the average life of each
light bulb was 1050 hours. At a 95% confidence level,
determine whether these light bulbs are in compliance with the
company's claim.
2) A company's human resource department claims that all
employees are present on the average 4 days out of the work
week with a standard deviation of 1. They hired an outside
company to do an audit of their employees' absences. The
company took a sample a 10 people and found that on the
average the employees were present 3 days per week. With a
95% confidence level, determine whether the company's claim
is true based on the data from the sample.
3) A teacher claims that all of her students pass the state
mandated test with an average of 90 with a standard deviation
of 10. The principal gave the test to 20 of her students to see if
the teacher's claim was true. He found that the average score
was 75. With a 95% confidence level, determine whether the
teacher is making the correct claim about all of her students.
4) The lifeguard's at a local pool have to be able to respond to
a distressed swimmer at an average of 10 seconds with a
standard deviation of 4 in order to be considered for
employment. If a sample of 100 lifeguards showed that their
average response time is 15 seconds, with a confidence level of
95% determine whether this group may be considered for
employment.
5) It is believed that an average of 20 mg of iodine is in each
antibiotic cream produced by a certain company with a standard
deviation of 5 mg. The company pulled 150 of its antibiotic
creams and found that on the average each cream contained 29
mg of iodine. Determine with a 95% confidence level whether

10. or not these creams are in compliance with the company's
belief?
For questions 6 - 10 use the chi-squared distribution to test the
hypothesis.
6) A restaurant owner wants to see if the business is good
enough for him to purchase a restaurant. He asks the present
owner for a breakdown of how many customers that come in for
lunch each day and the results are as follows: Monday - 20,
Tuesday - 30, Wednesday - 25, Thursday - 40 and Friday - 55.
The prospective owner observes the restaurant and finds the
following number of customers coming for lunch each day:
Monday- 30, Tuesday - 15, Wednesday- 7, Thursday 40, and
Friday - 33. At a 95% confidence level determine whether the
present owner reported the correct number of customers for
lunch each day.
7) An employer polled its employers to see if they agree with
the proposed new store hours and whether or not their present
shift made a difference in their answers. The customers
answered 1 for agree, 2 for don't know, and 3 for disagree.
Nine first shift employees answered "agree", 15 second shift
employees answered "agree", and 20 third shift employees
answered agree. With a 95% confidence level determine
whether or not the employees' present shift played a role in
their responses to the poll.
8) A politician surveyed 100 citizens to determine if their job
title had anything to do with the way they responded to the
following statement: "A city-wide curfew will be put into
place. Select the time that you think it should be put into
place. 8pm, 9pm, or 10pm". He is mostly concerned with the
10 pm responses. 25 teachers chose 10pm, 40 doctors chose
10pm, and 35 police responded 10pm. With a 95% confidence
level, determine whether job title plays a role in how the
citizens responded to the statement.
9) A meter reader did an experiment to see if there is a
relationship between the number of tickets she writes and the
number of blocks she is away from the park that is considered

11. the heart of the city. At 0 blocks from the park she writes 35
tickets, at 1 block away from the park she writes 25 tickets, at 2
blocks from the park she writes 20 tickets and at 3 blocks from
the park she writes 25 tickets. Use a 95% confidence level.
10) A high school principal asks his students to respond to the
following statement: "School should start at 9:00am rather than
7:00am. Answer 1 for agree, 2 for don't know, and 3 for
disagree." There were 90 seniors who answered agree, 35
juniors, 30 sophomores, and 25 freshmen. Help the principal
decide with a 95% confidence level that the students' status
played a role in how they responded to the question.
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