Transcript Hypothesis Testing - about NumberBender

Hypothesis Testing
When “p” is small, we reject the Ho.
Warm-up: state your
hypotheses
a) Larry's car averages 26 miles per gallon
on the highway. He switches to a new brand
of motor oil that is advertised to increase gas
mileage. After driving 3000 highway miles
with the new oil, he wants to determine if the
average gas mileage has increased.
 = the mean gas mileage for Larry’s car on the highway.
Ho:  = 26 mpg
Ha:  > 26 mpg
Warm-up: state your
hypotheses
(b) A May 2005 Gallup Poll report on a national survey
of 1028 teenagers revealed that 72% of teens said
they rarely or never argue with their friends. You
wonder whether this national result would be true in
your school. So you conduct your own survey of a
random sample of students at your school.
p= the proportion of teens in your school who rarely or
never fight with their friends.
Ho: p= .72
Ha: p≠ .72
Warm-up: state your
hypotheses
The city manager noted that paramedics arrived within 8
minutes after 78% of all calls involving life-threatening
injury last year. Based on this year's random sample of
400 calls, she wants to determine whether the
paramedics have arrived within 8 minutes more
frequently this year.
p = proportion of calls involving life–threatening injuries for
which the paramedics arrived within 8 minutes.
H0: p = 0.78
Ha: p > 0.78
Warm-up: state your
hypotheses
A national study reports that households spend an
average of 30% of their food expenditures in restaurants.
A restaurant association in your area wonders if the
national results apply locally. They interview a sample of
households and ask about their total food budget and the
amount spent in restaurants.
μ = mean percent of local household food expenditures
used for restaurant meals.
H0: μ = 30
Ha: μ ≠ 30
Why is “p” always gets to
decide which hypothesis
needs to go
Small P-values are evidence against H0
because they say that the observed
result is unlikely to occur when H0 is true.
Large P-values fail to give evidence
against H0.
 = the mean gas mileage for Larry’s car on the highway.
Ho:  = 26 mpg
Ha:  > 26 mpg
Two-tailed test
One study chose 18 subjects at random from a group of people
who assembled electronic devices. Half of the subjects were
assigned at random to each of two groups. Both groups did
similar assembly work, but one work setup allowed workers to
pace themselves, and the other featured an assembly line that
moved at fixed time intervals so that the workers were paced by
machine. After two weeks, all subjects took the Job Diagnosis
Survey (JDS), a test of job satisfaction. Then they switched
work setups and took the JDS again after two more weeks. This
is a matched pairs design. The response variable is the
difference in JDS scores, self-paced minus machinepaced.Suppose we know that differences in job satisfaction scores
in follow a Normal distribution with standard deviation σ = 60,
perform a hypothesis testing for this case.
Part1: Hypotheses
Ho: µ = 0, there is NO difference in job satisfaction
between the two work environments
Ho: µ ≠ 0, there is a difference in job satisfaction
between the two work environments
Part 2: Conditions
Because the workers were chosen without
replacement, randomly sampled from the
assembly workers, and then randomly
assigned to each group, the requirements of
SRS and independence are met. The
question states that the differences in job
satisfaction follow a Normal distribution.
Part 3: Test Statistic
x = 17
µ=0
∂ = 60
n = 18
z= x-ℳ
σ/√n
z= 17-0
60/√18
z= 1.20
The two-sided p-value for
z=1.20 is
p = 2(.1151)
p = .2302
Part 4: Conclusion
p = .2302
Ho: µ = 0, there is NO difference in job satisfaction
between the two work environments
Ho: µ ≠ 0, there is a difference in job satisfaction
between the two work environments
With a p-value of .2302, we have a fairly strong
evidence against the alternative hypothesis
therefore we can conclude that there is no
significant job difference satisfaction between
the two work environments set by the company.
Your turn!
Job satisfaction with a larger
sample: Suppose that the job satisfaction study
had produced exactly the same outcome x = 17
as in the previous example but from a sample of
75 workers rather than just 18 workers. perform
a hypothesis testing for this new case.
Z = 2.45
P = .0142
A p=value of .0142 gives a strong evidence against the null
hypothesis, therefore we can conclude that there is a
significant job difference satisfaction between the two work
environments set by the company.
The company medical director institutes a health
promotion campaign to encourage employees to
exercise more and eat a healthier diet. One measure
of the effectiveness of such a program is a drop in
blood pressure. The director chooses a random
sample of 50 employees and compares their blood
pressures from physical exams given before the
campaign and again a year later. The mean change in
systolic blood pressure for these n = 50 employees is
x = −6. We take the population standard deviation to
be σ = 20. The director decides to use an α = 0.05
significance level.
Step 1: Hypotheses We want to know if the health
campaign reduced blood pressure on average in the
population of all employees at this large company.
Taking μ to be the mean change in blood pressure
for all employees, we test
H0: μ = 0
Ha: μ < 0
Step 2: Conditions Since σ is known, we will use a onesample z test for a population mean. Now we check
conditions.
•SRS The medical director took a “random sample” of 50
company employees. Our calculation method assumes that
an SRS was taken.
•Normality The large sample size (n = 50) guarantees
approximate Normality of the sampling distribution of x,
even if the population distribution of change in blood
pressure isn't Normal.
•Independence There must be at least (10)(50) = 500
employees in this large company since the medical director
is sampling without replacement.
Step 3: Test Statistic
P = P(Z ≤ −2.12) = 0.0170
μ is the mean change in blood pressure for all employees
H0: μ = 0
Ha: μ < 0
P = 0.0170
α = 0.05
Step 4: Interpretation Since our P-value, 0.0170, is
less than α = 0.05, this result is statistically
significant. We reject H0 and conclude that the mean
difference in blood pressure readings from before
and after the campaign among this company's
employees is negative. In other words, the data
suggest that employees' blood pressure readings have
decreased on average.