Download 1 Chapter 8.2: Tests About a Population Mean Instructor: Dr. Arnab

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Chapter 8.2: Tests About a Population Mean
Instructor: Dr. Arnab Maity
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In Chapter 8.1, we learned about the following.
• Statistical hypothesis
– Null hypothesis: H0
– Alternative hypothesis: H1
• Errors in hypothesis testing
– Type I (Reject H0 when H0 is true)
– Type II (Fail to reject H0 when H0 is in fact not true)
• Test statistic and rejection regions
• For fixed test statistic and sample size n, there is no rejection region that will simultaneously make both type I error level α and type II error level β small.
– In general, we first specify the largest value of type I error level α that can be
tolerated, and then find an rejection region that has the specified value of α. This
makes β as small as possible subject to the bound on α.
Proposition: Suppose an experiment and a sample size are fixed and a test statistic
is chosen. Then decreasing the size of the rejection region to obtain a smaller value
of α results in a larger value of β for any particular parameter value consistent
with Ha .
Significance level: the largest α that can be tolerated, e.g., 0.01, 0.05, 0.10.
The more serious the Type I error is, the smaller level should be chosen.
Level 0.05 test: A test for which Type I error probability is controlled at
level 0.05. Similarly we have level 0.1 test, level 0.01 test, etc.
In this chapter, we will learn about how to construct such rejection regions for testing
different types of hypothesis about a population mean, µ. We will consider three scenarios,
as follows.
• Population distribution is normal with known variance σ 2 .
• Population distribution is normal with unknown variance σ 2 .
• Population distribution may not be normal with unknown variance σ 2 .
– requires large sample size.
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We will need the following results.
• Result 1: If X1 , . . . , Xn is a random sample from a normal distribution N (µ, σ 2 ), then
the sample mean X̄ has a normal distribution N (µ, σ 2 /n).
– The standardized variable Z =
X̄−µ
√
σ/ n
has a N (0, 1) distribution.
– The standardized variable (with sample standard deviation) T =
distribution (t distribution with n − 1 degrees of freedom).
X̄−µ
√
S/ n
has a tn−1
• Result 2: If X1 , . . . , Xn is a random sample from any distribution with mean µ and
variance σ 2 , then for large sample size n, the sample mean X̄ approximately has a
normal distribution N (µ, σ 2 /n).
– The standardized variable Z =
X̄−µ
√
σ/ n
has approximately N (0, 1) distribution.
– The standardized variable (with sample standard deviation) T =
proximately N (0, 1) distribution.
– Requires large sample size.
Steps for testing hypothesis about a parameter:
1. Identify the parameter of interest.
2. Determine the null hypothesis H0 .
3. State the appropriate alternative hypothesis Ha .
4. Calculate the value of the test statistic.
5. State the rejection region for the selected significance level α.
6. Decide whether you reject H0 or not, and state the conclusion.
X̄−µ
√
S/ n
has ap-
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Case 1: A normal population with known σ.
• Null hypothesis: H0 : µ = µ0 , where µ0 is known.
• Test statistic (standardized X̄ using the known σ):
Z=
X̄ − µ0
√
σ/ n
– The test statistic computes the difference between the sample mean and the hypothesized value of the population mean, and scales the difference by the standard
deviation of X̄.
• If H0 is indeed true (that is, if µ = µ0 ) that we know that Z ∼ N (0, 1).
Alternative hypotheses:
Ha : µ 6= µ0 (two-tailed)
We reject H0 when test statistic Z is either sufficiently small of sufficiently large.
Rejection region: Z ≤ −zα/2 or Z ≥ zα/2
Ha : µ < µ0 (lower-tailed)
Ha : µ > µ0 (upper-tailed)
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1. A manufacturer of sprinkler systems used for fire protection in office buildings claims
that the true average system-activation temperature is 130o F. A sample of n = 9
systems, when tested, yields a sample average activation temperature of 131.08o F.
Suppose the distribution of activation temperature is normal with standard deviation 1.5o F. Do the data contradict the manufacturer’s claim at the significance level
α = 0.01? Perform a two-sided test.
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Case 2: A Normal Population Distribution with unknown σ.
• Null hypothesis: H0 : µ = µ0 , where µ0 is known.
• Test statistic:
T =
X̄ − µ0
√ .
S/ n
where S is the sample standard deviation.
• We know that when H0 is true, the test statistic T has a tn−1 distribution.
• Hypothesis testing:
Alternative Hypothesis
Ha : µ 6= µ0
Ha : µ < µ0
Ha : µ > µ0
Rejection Region for a Level α Test
t ≤ −tα/2,n−1 OR t ≥ tα/2,n−1 (two-tailed)
t ≤ −tα,n−1 (lower-tailed)
t ≥ tα,n−1 (upper-tailed)
Case 3: Large-Sample test (n ≥ 40).
• Null hypothesis: H0 : µ = µ0 , where µ0 is known.
• Test statistic:
Z=
X̄ − µ0
√ ,
S/ n
• if H0 is indeed true (that is, if µ = µ0 ) that we know that Z approximately has
a N (0, 1).
The rejection regions are the same as for Case 1:
Alternative Hypothesis
Ha : µ 6= µ0
Ha : µ < µ0
Ha : µ > µ0
Rejection Region for a Level α Test
z ≤ −zα/2 OR z ≥ zα/2 (two-tailed)
z ≤ −zα (lower-tailed)
z ≥ zα (upper-tailed)
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2. An environmentalist collects a liter of water from 45 different locations along the banks
of a stream. He measures the amount of dissolved oxygen in each specimen and observes
that mean oxygen level is 4.62 mg, with a standard deviation of 0.92 mg. A water
purifying company claims that the mean level of oxygen in the water is 5 mg. Conduct
a hypothesis test at a significance level of 0.01 to determine whether the mean oxygen
level is less than 5 mg.
3. A car manufacturer claims that, when driven at a speed of 50 miles per hour, the
mileage of a certain model follows a normal distribution with mean 30 miles per
gallon. A random sample of 10 cars yields x̄ = 29.4 miles per gallon, with sample
standard deviation s = 4 miles per gallon. Is there a reason to believe that the
manufacturer is overestimating average mileage (with significance level of 0.01)?
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4. A sample of 42 radon detectors of a certain type was selected, and each was exposed
to 100pCi/L of radon. The resulting readings gave x̄ = 98.37, and s = 6.11. Does
this data suggest that the population mean reading under these conditions differs from
100? State and test the appropriate hypotheses using α = 0.05.