Transcript 8.2 Testing the Difference Between Means (Small, Independent

Statistics
8.2 Testing the Difference
Between Means (Small,
Independent Samples)
Statistics
Mrs. Spitz
Spring 2009
Statistics
Objectives/Assignment
• How to perform a t-test for the
difference between two population
means, 1 and 2 using small
independent samples.
Assignment: pp. 384-388 #1-24 all
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Two Sample t-Test for the
Difference Between Means
• So, it is impractical and costly to collect
samples of 30 or more from each of two
populations. What about if both
populations have a normal distribution?
If so, you can still test the difference
between their means. In this section,
you will learn how to use a t-test to test
the difference between two population
means, 1 and 2 using a sample from
each population.
Two Sample t-Test for the
Difference Between Means
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•
To use a t-Test for small independent
samples, the following conditions are
necessary:
1. The samples must be independent, so 1st
sample cannot be related to the sample
selected from the second population.
2. Each population must have a normal
distribution.
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So, the problem meets that
criteria, what next?
• The sampling distribution for x1  x2 ,
the difference between he sample
means, is a t-distribution with mean 1  2
The standard error and the degrees of
freedom of the sampling distribution
depend on whether or not the
population variances  2 and  2 are
1
2
equal.
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Pooled Estimate of the Standard
Deviation
• If the population variances are equal,
information from both samples is combined
to calculate a pooled estimate of the
standard deviation.
Pooled estimate of 
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Pooled Estimate of the Standard
Deviation—VARIANCES EQUAL
• The standard error for the sampling
distribution of x1  x2 is:
And d.f. of n1 + n2 - 2
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Pooled Estimate of the Standard
Deviation-VARIANCES NOT EQUAL
• The standard error for the sampling
distribution of is:
And d.f. smaller of n1 – 1 or n2 - 1
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Requirements for z-test
• The requirements for the z-test described in 8.1 and
the t-test described in this section are compared
below:
If the sampling distribution for x1  x2 is a tdistribution, you can use a two-sample t-test to test the
difference between two populations 1 and 2.
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Ex. 1: A Two-Sample t-Test for the
Difference Between Means
• Consumer Reports tested several types of snow tires
to determine how well each performed under winter
conditions. When traveling on ice at 15 mph, 10
Firestone Winterfire tires had a mean stopping
distance of 51 feet with a standard deviation of 8
feet. The mean stopping distance for 12 Michelin
XM+S Alpine tires was 55 feet with a standard
deviation of 3 feet. Can you conclude that there is a
difference between the stopping distances of the two
types of tires? Use  = 0.01. Assume the
populations are normally distributed and the
population variances are NOT equal.
Gather your information . . .
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Winterfire
Alpin
x1
51
x2 55
s1
8
S2 3
n1
10
n2
12
• Sample Statistics for Stopping Distances
• So now that you have all your relevant data,
now go back and figure it out.
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Solution Ex. 1
• You want to test whether the mean stopping
distances are different. So, the null and
alternative hypotheses are:
Ho: 1 = 2
and Ha: 1  2 (Claim)
Because the variances are NOT equal, and the
smaller sample size is 10, use the d.f. = 10 –
1 = 9. Because the test is a two-tailed test
with d.f. = 9, and  = 0.01, the critical values
are?
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Solution Ex. 1
• Okay, you got me. . . the critical values
are -3.250 and 3.250. The rejection
region is t < -3.250 and t > 3.250. The
standard error is:
Everybody good so far? Questions?
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Solution Ex. 1
• Using the t-test, the standardized test
statistic is:
The graph following shows the location of
the critical regions and the standardized
test statistic, t.
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Solution Ex. 1
• Because t is not in a
rejection region, you
should fail to reject
the null hypothesis.
At the 1% level,
there is not enough
evidence to
conclude that the
mean stopping
distances of the
tires are different.
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Ex. 2: A Two-Sample t-Test for the
Difference Between the Means
• A manufacturer claims that the calling range
(in miles) of its 900-MHz cordless phone is
greater than that of its leading competitor.
You perform a study using 14 phones from
the manufacturer and 16 similar phones from
its competitor. The results are shown on the
next slide. At  = 0.05, is there enough
evidence to support the manufacturer’s
claim? Assume the populations are normally
distributed and the population variances are
equal.
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First your sample Statistics for
Calling Range
• The claim is “the mean range of our cordless
phone is greater than the mean range of
yours.” So, the null and alternative
hypotheses are:
Ho: 1  2
and Ha: 1 > 2 (Claim)
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Solution Ex. 2
• Because the variances are equal,
d.f. = n1 + n2 – 2
= 14 + 16 – 2
= 28
Because the test is a right-tailed test, d.f. =
28 and  = 0.05, the critical value is 1.701.
The rejecetion is t > 1.701.
Solution Ex. 2
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• The standard error is:
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Solution Ex. 2
• Using the t-Test, the standardized test
statistic is:
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Solution Ex. 2
• The graph at the left
shows the location of
the rejection region
and the standardized
test statistic, t.
Because t is in the
rejection region, you
should decide to reject
the null hypothesis. At
the the 5% level, there
is enough evidence to
support the
manufacturer’s claim
that its phone has a
greater calling range
than its competitor’s.