Jeux Dargent Et De Hasard En Ligne Machine De Mise Sous Vide

Objectives
Example
Braking distances for two types of tires were
collected from two independent random samples.
Use a 0.05 significance level to test the claim that
the braking distances of the two types of tires are
different.
Determine whether two samples are
dependent or independent.
Test the difference between two sample
means, using the z test.
n

type A
50
type B
40
54 ft
51 ft
5.3 ft
4.9 ft
8.1
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Example
Use a 0.05 significance level to test the claim that the distances come
from populations with different means.
Are all requirements satisfied?
The samples are independent random samples.
Both samples are large.
Both population standard deviations are known.
State the level of significance, and the null and
alternate hypotheses.
Identify the statistic that is relevant to this test.
The test statistic is z.
Example – Traditional Method
Use a 0.05 significance level to test the claim that the distances come
from populations with different means.
Find the test statistic, the critical values and the
critical region.
This is a two-tailed test, so the critical values are
z = ±1.96.
Example – Traditional Method
Example – P-value Method
Use a 0.05 significance level to test the claim that the distances come
from populations with different means.
Use a 0.05 significance level to test the claim that the distances come
from populations with different means.
Find the test statistic and find the P-value.
Draw a graph and include the test statistic, critical value and
critical region.
Since the test
statistic is in the
critical region the
null hypothesis is
rejected.
For z = 2.78, the standard normal distribution table
shows a two tail P-value of 0.0054 .
z = 2.783
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Example – P-value Method
Example
Use a 0.05 significance level to test the claim that the distances come
from populations with different means.
Use a 0.05 significance level to test the claim that the distances come
from populations with different means.
Draw a graph and show the test statistic and P-value.
Restate the decision in simple, nontechnical terms
and address the original claim.
Since the P-value is
less than or equal to
the significance level 
the null hypothesis is
rejected.
P-value = 0.0054
Conclude that the sample data support the claim
that there is a difference between the braking
distance of the type A tires and type B tires.
Example
Objectives
We wish to test the claim that the mean length of hospital stay
for routine childbirth is different for insured and uninsured
women. A random sample of 14 insured women had a
mean of 2.3 days and a standard deviation of 0.7 day while
12 uninsured women had a mean of 1.9 days and a
standard deviation of 0.4 day. Use  = 0.01.
State the level of significance, and the null and alternate
hypotheses.
Test the difference between two sample
means, using the t test.
Identify the statistic that is relevant to this test and determine
its sampling distribution.
The test statistic is t. df = n2 - 1 = 11
8.2
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Example – Traditional Method
Example – Traditional Method
insured: n = 14, = 2.3, s = 0.7
uninsured: n = 12, = 1.9, s = 0.4
Use a 0.01 significance level to test the claim that insured
and uninsured women have different hospital stays.
insured: n = 14, = 2.3, s = 0.7
uninsured: n = 12, = 1.9, s = 0.4
Use a 0.01 significance level to test the claim that insured
and uninsured women have different hospital stays.
Find the test statistic, the critical values and the critical
region. Draw a graph and include the test statistic, critical
value(s) and critical region(s).
Draw a graph and include the test statistic, critical value and
critical region.
This is a two-tailed test, so the critical values are t = ±3.106 .
Since the test statistic
is not in the critical
region the null
hypothesis is not
rejected.
t = 1.819
Example – P-value Method
Example – P-value Method
insured: n = 14, = 2.3, s = 0.7
uninsured: n = 12, = 1.9, s = 0.4
Use a 0.01 significance level to test the claim that insured
and uninsured women have different hospital stays.
insured: n = 14, = 2.3, s = 0.7
uninsured: n = 12, = 1.9, s = 0.4
Use a 0.01 significance level to test the claim that insured
and uninsured women have different hospital stays.
Find the test statistic and the P-value.
Draw a graph and show the test statistic and P-value.
Since the P-value is
greater than the
significance level the
null hypothesis is not
rejected.
For t = 1.819, the Student’s t distribution table shows a
P-value between 0.05 and 0.10. Using Excel, the P-value
is 0.0962.
P-value = 0.0962
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Example
insured: n = 14, = 2.3, s = 0.7
uninsured: n = 12, = 1.9, s = 0.4
Use a 0.01 significance level to test the claim that insured
and uninsured women have different hospital stays.
Restate the decision in simple, nontechnical terms
and address the original claim.
Conclude that the sample data do not support the
claim that there is a difference between the mean
hospital stay for insured and uninsured mothers.
Example
A sample of 16 insured women had mean hospital stays of 2.3
days with a standard deviation of 0.5 days. 12 uninsured
women had mean hospital stays of 1.9 days with a standard
deviation of 0.6 days. Assume the standard deviations are
equal. At  = 0.01, test the claim that the means are
different.
State the level of significance, and the null and alternate
hypotheses.
Identify the statistic that is relevant to this test.
The test statistic is t. df = n1+ n2 - 2 = 26
Example
insured: n = 16, = 2.3, s = 0.5
uninsured: n = 12, = 1.9, s = 0.6
Find the test statistic and the critical values.
This is a two-tailed test, so the critical values are t = ±2.779.
Assess the results and state a conclusion.
Since -2.779 < 1.92 < 2.779, the null hypothesis cannot be
rejected. The sample data do not support the claim that there
is a difference between the mean hospital stay for insured
and uninsured mothers.
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