Transcript 10-2 z test for Hypothesis Testing

Chapter 10
Section 2
Z Test for Mean
1
z Test for a Mean
The z test is a statistical test for the mean of a population.
It can be used when n > 30, or when the population is
normally distributed and
is known.
The formula for the z test is

X 
z
 n
where
X = sample mean
μ = hypothesized population mean
= population standard deviation
n = sample size

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Days on Dealers Lots
A researcher wishes to see if the mean
number of days that a basic, low-price,
small automobile sits on a dealer’s lot is
29. A sample of 30 automobile dealers
has a man of 30.1 days for basic, lowprice, small automobiles. At a = 0.05m
teat the claim that the mean time is greater
than 29 days. The standard deviation of
the population is 3.8 days.
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Cost of Rehabilitation
The Medical Rehabilitation Education Foundation reports
that the average cost of rehabilitation for stroke victims is
$24,672. To see if the average cost of rehabilitation is
different at a particular hospital, a researcher selects a
random sample of 35 stroke victims at the hospital and
finds that the average cost of their rehabilitation is
$25,226. The standard deviation of the population is
$3251. At α = 0.01, can it be concluded that the average
cost of stroke rehabilitation at a particular hospital is
different from $24,672?
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Cost of Rehabilitation
“reports that the average cost of rehabilitation for stroke
victims is $24,672”
“a random sample of 35 stroke victims”
“the average cost of their rehabilitation is $25,226”
“the standard deviation of the population is $3251”
“α = 0.01”
Step 3: Find the test value.
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Hypothesis Testing
The P-value (or probability value) is the probability of
getting a sample statistic (such as the mean) or a more
extreme sample statistic in the direction of the
alternative hypothesis when the null hypothesis is true.
P-Value
Test Value
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Cost of Men’s Shoes
A researcher claims that the average cost of men’s
athletic shoes is less than $80. He selects a random
sample of 36 pairs of shoes from a catalog and finds the
following costs (in dollars). (The costs have been rounded
to the nearest dollar.) Is there enough evidence to support
the researcher’s claim at α = 0.10? Assume
= 19.2.
60 70 75 55 80 55 50 40 80 70 50 95
120 90 75 85 80 60 110 65 80 85 85 45
75 60 90 90 60 95 110 85 45 90 70 70

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Hypothesis Testing

In this section, the traditional method for
solving hypothesis-testing problems compares
z-values:
 critical value
 test value

The P-value method for solving hypothesistesting problems compares areas:
 alpha
 P-value
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Procedure Table
Solving Hypothesis-Testing Problems
(P-Value Method)
Step 1 State the hypotheses and identify the claim.
Step 2 Compute the test value.
Step 3 Find the P-value.
Step 4 Make the decision.
Step 5 Summarize the results.
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Cost of College Tuition
A researcher wishes to test the claim that the average
cost of tuition and fees at a four-year public college is
greater than $5700. She selects a random sample of 36
four-year public colleges and finds the mean to be $5950.
The population standard deviation is $659. Is there
evidence to support the claim at a 0.05? Use the P-value
method.
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Wind Speed
A researcher claims that the average wind speed in a
certain city is 8 miles per hour. A sample of 32 days has
an average wind speed of 8.2 miles per hour. The
standard deviation of the population is 0.6 mile per hour.
At α = 0.05, is there enough evidence to reject the claim?
Use the P-value method.
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Guidelines for P-Values With No α

If P-value  0.01, reject the null hypothesis. The
difference is highly significant.

If P-value > 0.01 but P-value  0.05, reject the
null hypothesis. The difference is significant.

If P-value > 0.05 but P-value  0.10, consider
the consequences of type I error before
rejecting the null hypothesis.

If P-value > 0.10, do not reject the null
hypothesis. The difference is not significant.
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Significance



The researcher should distinguish between
statistical significance and practical
significance.
When the null hypothesis is rejected at a
specific significance level, it can be concluded
that the difference is probably not due to chance
and thus is statistically significant. However, the
results may not have any practical significance.
It is up to the researcher to use common sense
when interpreting the results of a statistical test.
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Example 5
A report in USA TODAY stated that the average age of
commercial jets in the United States is 14 years. An
executive of a large airline company selects
a sample of 36 planes and finds the average
age of the planes is 11.8 years. The
standard deviation of the sample is
2.7 years. At  = 0.01, can it be
concluded that the average age
of the planes in his Company is
less than the national average?
Example 6
25 32
35 25 30 26.5 26 25.5 29.5 32
30 28.5 30 32 28 31.5 29 29.5
30
29 32
29 29.5
27 28 33
28
27
32
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The average one-year-old (both sexes) is 29 inches
tall.
A random sample of 30 one-year-olds in a large day
care franchise resulted in the following
heights. At  = 0.05, can it be concluded
that the average height differs
from 29 inches?
Example 7
To see if young men ages 8 through 17 years
spend more or less than the national average
of $24.44 per shopping trip to a local mall, the
manager
surveyed 33 young men and found the
average amount spent per visit was $22.97.
The standard deviation of the sample was
$3.70. At  = 0.02, can it be concluded
that the average amount spent at a local
mall is not equal to the national
average of $24.44.
Example 8
A study found that the average stopping distance of a
school
bus traveling 50 miles per hour was 264 feet (Snapshot,
USA TODAY, March12, 1992). A group of automotive
engineers decided to conduct a study of its
school buses and found that for 20 buses,
the average stopping distance of buses
traveling 50 miles per hour was 262.3 feet.
The standard deviation of the population
was 3 feet. Test the claim that the average
stopping distance of the company’s
buses is actually less than 264 feet.
Find the P-value. On the basis of the
P-value, should the null hypothesis be
rejected at  = 0.01? Assume that the
variable isnormally distributed.