Tests of Hypothesis Tests of Means and Variances

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Tests of HypothesisTests of Means and Variances

• Probability and Statistics for Scientists and
  Engineers

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Example

• A company produces and markets coffee in cans
  which are advertised as containing one pound of
  coffee. What this means is that the true mean
  weight of coffee per can is 1 pound. If the true
  mean weight of coffee per can exceeds 1 pound,
  the companys profit will suffer. On the other
• hand, if the true mean weight is very much less
  than 1 pound, consumers will complain and sales
  may decrease. To monitor the process, 25 cans of
  coffee are randomly selected during each days
  production. The process will be adjusted if there
  
• is evidence to indicate that the true mean amount
  of coffee is not 1 pound.

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Example Continued

• A decision rule is desired so that the
  probability of adjusting the process when the
  true mean weight of coffee per can is equal to 1
  pound is 1. Assume that weight of coffee per can
  has a normal distribution with unknown mean and
  standard deviation.

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Example - solution

• The decision rule is
• Action 1 - adjust process if
• or if

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Example solution Continued

• The decision rule is
• Action 2 - do not adjust process if

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Example solution Continued

• Suppose that for a given day
• and
• s 0.012
• Then t

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Example - solution

• so that - 2.797 lt 2.5 lt 2.797
• and Action 2 no adjustment, is taken. We
  conclude that the true mean weight of coffee per
  can is 1 pound. We have thus tested the
  statistical hypothesis that ? 1 pound versus
  the alternative hypothesis that ? does not equal
  1 pound at the
• 1 level of significance.

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Test of Means

• Let X1, , Xn, be a random sample of size n, from
  a normal distribution with mean ? and standard
  deviation ?, both unknown.
• To test the Null Hypothesis
• H0 ? ?0 , a given or specified value
• against the appropriate Alternative Hypothesis
• 1. HA ? lt ?0 ,
• or
• 2. HA ? gt ?0 ,
• or
• 3. HA ? ? ?0 ,

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Test of Means Solution

• at the 100 . ? level of significance. Calculate
  the value of the test statistic
• Reject H0 if
• 1. t lt -t?, n-1 ,
• 2. t gt t?, n-1 ,
• 3. t lt -t?/2, n-1 , or if t gt t?/2, n-1 ,
• depending on the Alternative Hypothesis.

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Test on Two Means

• Let X11, X12, , X1n be a random sample of size
  n1 from
• N(m1, s1) and X21, X22, , X2n be a random
  sample of size n2 from N(m2, s2), where m1,s1, m2
  and s2 are all unknown.
• To test against the appropriate alternative
  hypothesis
• H0 µ1 - µ2 do, where do ? 0 (usually do0)

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2
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Test on Two Means

• 1. H1 µ1 - µ2 lt do, where do ? 0,
• or
• 2. H1 µ1 - µ2 gt do, where do ? 0,
• or
• 3. H1 µ1 - µ2 ? do, where do ? 0,
• at the a ? 100 level of significance, calculate
  the value of the test statistic.

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Test on Two Means Continued

• Calculate the value of the test statistic

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Test on Two Means Continued

• Reject H0 if
• 1. t' lt -ta,?
• or 2. t' gt ta,?
• or 3. t' lt -ta/2, ? or t' gt
  ta/2, ? ,
• depending on the alternative hypothesis, where

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Example - Test on Two Means

• An experiment was performed to compare the
  abrasive wear of two different laminated
  materials. Twelve pieces of material 1 were
  tested, by exposing each piece to a machine
  measuring wear. Ten pieces of material 2 were
  similarly tested. In each case, the depth of
  wear was observed. The samples of material 1
  gave an average (coded) wear of 85 units with a
  standard deviation of 4, while samples of
  material 2 gave an average of 81 and a standard
  deviation of 5. Test the hypothesis that the two
  types of material exhibit the same mean abrasive
  wear at the 0.10 level of significance. Assume
  the populations to be approximately normal.

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Example

• Test
• H0 m1 m2 or m1 - m2 0.
• Vs.
• H1 m1 ? m2 or m1 - m2 ? 0.
• With a 10 level of significance, i.e., a 0.10
• Then

'
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Example

• where
• and
• The calculate Critical Region is t' lt -1.725
  and t' gt 1.725,

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Example

• Since t' 2.07, we can reject H0 and conclude
  that the two materials do not exhibit the same
  abrasive wear.

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Test of Variances

• Let X1, , Xn, be a random sample of size n, from
  a normal distribution with mean ? and standard
  deviation ?, both unknown.
• To test the Null Hypothesis
• H0 ?2 ?o2 , a specified value
• against the appropriate Alternative Hypothesis
• 1. HA ?2 lt ?o2,
• or
• 2. HA ?2 gt ?o2,
• or
• 3. HA ?2 ? ?o2,

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Test of Variances

• at the ? . 100 level of significance. Calculate
  the value of the test statistic
• Reject H0 if
• 1. ?2 lt ?21-?, n-1 ,
• 2. ?2 gt ?2 ?, n-1 ,
• 3. ?2 lt ?21- ? /2, n-1 , or if ?2 gt ?2 ?/2, n-1
  ,
• depending on the Alternative Hypothesis.

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Test on Two Variances

• Let X11, X12, , X1n be a random sample of size
  n1 from
• N(m1, s1) and X21, X22, , X2n be a random
  sample of size n2
• from N(m2, s2), where m1, s1, m2 and s2 are all
  unknown.
• To test
• H0
• against the appropriate alternative hypothesis

1
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Test on Two Variances

• 1. H1
• or
• 2. H1
• or
• 3. H1
• at the a . 100 level of significance, calculate
  the value of the test statistic.

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Test on Two Variances

• Reject H0 if
• or
• or
• depending on the alternative hypothesis,
• and where
• and

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Example - Test on Variances

• An experiment was performed to compare the
  abrasive wear of two different laminated
  materials. Twelve pieces of material 1 were
  tested, by exposing each piece to a machine
  measuring wear. Ten pieces of material 2 were
  similarly tested. In each case, the depth of
  wear was observed. The samples of material 1
  gave an average (coded) wear of 85 units with a
  standard deviation of 4, while samples of
  material 2 gave an average of 81 and a standard
  deviation of 5. Test the hypothesis that the two
  types of material exhibit the same variation in
  abrasive wear at the 0.10 level of significance.

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Example - Test of Variances

• H0 s12 s22
• H1 s12 ? s22
• With a 10 level of significance, i.e., a 0.10
• Critical region From the graph we see that
  F0.05(11,9) 3.11

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Example - Test of Variances

• Therefore, the null hypothesis is rejected when F
  lt 0.34 or F gt 3.11.
• Decision
• Do not reject H0. Conclude that there is
  insufficient evidence that the variances differ.