Economics 173 Business Statistics

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Economics 173Business Statistics

• Lecture 5
• Fall, 2001
• Professor J. Petry
• http//www.cba.uiuc.edu/jpetry/Econ_173_fa01/

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10.3 Testing the Population Mean When the
Population Standard Deviation is Known

• Example 10.1
• A new billing system for a department store will
  be cost- effective only if the mean monthly
  account is more than 170.
• A sample of 400 monthly accounts has a mean of
  178.
• If the account are approximately normally
  distributed with s 65, can we conclude that
  the new system will be cost effective?

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• Solution
• The population of interest is the credit accounts
  at the store.
• We want to show that the mean account for all
  customers is greater than 170.

H1 m gt 170

• The null hypothesis must specify a single value
  of the parameter m

H0 m 170
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• Is a sample mean of 178 sufficiently greater
  than 170 to infer that the population mean is
  greater than 170?

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• Example 10.1 - continued
• H0 m 170
• H1 m gt 170
• Test statistic
• Rejection region z gt z.05 1.645.
• Conclusion Since 2.46 gt 1.645, reject the null
  hypothesis in favor of the alternative
  hypothesis.
• 2.46 is test statistic value, p-value is
  probability associated w/ the test statistic
  value (see below).
• 1.645 is the critical value. Significance level
  of the test is the probability associated with
  this value.

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The probability of observing a test statistic at
least as extreme as 178, given that the null
hypothesis is true is
The p-value
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• The p-value and rejection region methods
• The p-value can be used when making decisions
  based on rejection region methods as follows
• Define the hypotheses to test, and the required
  significance level a.
• Perform the sampling procedure, calculate the
  test statistic and the p-value associated with
  it.
• Compare the p-value to a. Reject the null
  hypothesis only if p lta otherwise, do not reject
  the null hypothesis.

P-value 0.0069
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• Example 10.2
• A government inspector samples 25 bottles of
  catsup labeled Net weight 16 ounces, and
  records their weights.
• From previous experience it is known that the
  weights are normally distributed with a standard
  deviation of 0.4 ounces.
• Can the inspector conclude that the product
  label is unacceptable?

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• Solution
• We need to draw a conclusion about the mean
  weights of all the catsup bottles.
• We investigate whether the mean weight is less
  than 16 ounces (bottle label is unacceptable).

H0 m 16
Then
H1 m lt 16

• Select a significance level
• a 0.05

• Define the rejection region
• z lt - za -1.645

One tail test
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we want this
mistake to happen not more than 5 of the time.
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A sample mean far below 16, should be a rare
event if m 16.
-za -1.645
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Since the value of the test statistic does not
fall in the rejection region, we do not reject
the null hypothesis in favor of the alternative
hypothesis.
There is insufficient evidence to infer that the
mean is less than 16 ounces.
The p-value P(Z lt - 1.25) .1056 gt .05
0
-za -1.645
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• Example 10.3
• The amount of time required to complete a
  critical part of a production process on an
  assembly line is normally distributed. The mean
  was believed to be 130 seconds.
• To test if this belief is correct, a sample of
  100 randomly selected assemblies was drawn, and
  the processing time recorded. The sample mean was
  126.8 seconds.
• If the process time is really normal with a
  standard deviation of 15 seconds, can we conclude
  that the belief regarding the mean is incorrect?

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• Solution
• Is the mean different than 130?

H0 m 130
Then

• Define the rejection region
• z lt - za/2 or z gt za/2

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we want this mistake to happen not more than 5
of the time.
130
A sample mean far below 130 or far above 130,
should be a rare event if m 130.
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Since the value of the test statistic falls in
the rejection region, we reject the null
hypothesis in favor of the alternative
hypothesis.
There is sufficient evidence to infer that the
mean is not 130.
The p-value P(Z lt - 2.13)P(Z gt 2.13)
2(.0166) .0332 lt .05
a/2 0.025
a/2 0.025
0
-2.13
2.13
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Testing hypotheses and intervals estimators

• Interval estimators can be used to test
  hypotheses.
• Calculate the 1 - a confidence level interval
  estimator, then
• if the hypothesized parameter value falls within
  the interval, do not reject the null hypothesis,
  while
• if the hypothesized parameter value falls outside
  the interval, conclude that the null hypothesis
  can be rejected (m is not equal to the
  hypothesized value).

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• Drawbacks
• Two-tail interval estimators may not provide the
  right answer to the question posed in one-tail
  hypothesis tests.
• The interval estimator does not yield a p-value.

There are cases where only tests produce the
information needed to make decisions.
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Calculating the Probability of a Type II Error

• To properly interpret the results of a test of
  hypothesis, we need to
• specify an appropriate significance level or
  judge the p-value of a test
• understand the relationship between Type I and
  Type II errors.
• How do we compute a type II error?

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• Calculation of a type II error requires that
• the rejection region be expressed directly, in
  terms of the parameter hypothesized (not
  standardized).
• the alternative value (under H1) be specified.

H0 m m0 H1 m m1 (m0 is not equal to m1)
m m0
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• Revisiting example 10.1
• The rejection region was with
  a .05.
• A type II error occurs when a false H0 is not
  rejected.

Do not reject H0
m0 170
175.34
but H0 is false
m1 180
175.34
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• Effects on b of changing a
• Decreasing the significance level a, increases
  thethe value of b, and vice versa.

a1
gt a2
b1
lt b2
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Inference About the Description of a Single
Population

• Chapter 11

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11.1 Introduction

• In this chapter we utilize the approach developed
  before for making statistical inference about
  populations.
• Identify the parameter to be estimated or tested
  .
• Specify the parameters estimator and its
  sampling distribution.
• Construct an interval estimator or perform a test.

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• We will develop techniques to estimate and test
  three population parameters.
• The expected value m
• The variance s2
• The population proportion p (for qualitative
  data)
• Examples
• A bank conducts a survey to estimate the number
  of times customer will actually use ATM machines.
• A random sample of processing times is taken to
  test the mean production time and the variance of
  production time on a production line.

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11.2 Inference About a Population Mean When the
Population Standard Deviation Is Unknown
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Z
Z
t
t
Z
t
t
Z
t
t
Z
t
t
Z
t
t
t
t
t
s
s
s
s
s
s
s
s
s
s
When the sampled population is normally
distributed, the statistic t is Student t
distributed.
The degrees of freedom, a function of the
sample size determines how spread
the distribution is (compared to the normal
distribution)
The t distribution is mound-shaped, and
symmetrical around zero.
d.f. n2
d.f. n1
n1 lt n2
0
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Probability calculations for the t distribution

• The t table provides critical value for various
  probabilities of interest.
• The form of the probabilities that appear in
  table 4 Appendix B are
• P(t gt tA, d.f.) A
• For a given degree of freedom, and for a
  predetermined right hand tail probability A, the
  entry in the table is the corresponding tA.
• These values are used in computing interval
  estimates and performing hypotheses tests.

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tA
t.100
t.05
t.025
t.01
t.005
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Testing the population mean when the population
standard deviation is unknown

• If the population is normally distributed, the
  test statistic for m when s is unknown is t.
• This statistic is Student t distributed with n-1
  degrees of freedom.