Hypothesis Testing using Pvalues

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Hypothesis Testing using P-values

• Perform Step 1 and Step 2 as before when using
  rejection points
• Compute the test statistic, z or t.
• Compute the probability of obtaining such an
  extreme test statistic by pure chance, if the
  Null hypothesis were true. This is called the
  p-value of the test.
• Reject or fail to reject the Null Hypothesis by
  determining the p-value is less than or greater
  than a.
• Real-world interpretation

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Interpreting p-values

• All statistical packages give p-values in the
  standard output.
• When we reject Ho we say the test is significant.
• If p-value lt .01, highly significant
  (overwhelming evidence in support of research
  hypothesis)
• If p-value between .01 and .05, significant
  (strong evidence)
• If p-value between .05 and .10, slightly
  significant (weak evidence)
• If p-value gt .10, not significant (no evidence)

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Comparing p-value with alpha Right Tail
If the alternative hypothesis is ?gt20, this is a
right-tailed test with all the a .05 at the
right tail. The p-value (to compare with a) is
the area cut off at the right tail by the
calculated z.
? .05
p-value
z
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Comparing p-value with alpha Left Tail

• If the alternative hypothesis is ? lt 20 and a
  .10,
• this is a left-tailed test with all the .10 at
  the left tail. The p-value (to compare with a) is
  the area cut off at the left tail by the
  calculated z.

?.1
p-value
z (usually negative)
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Example The p-Value for Greater Than
Testing H0 ? ? 50 vs Ha ? gt 50 using rejection
points and p-value. Trash Bag The p-value or the
observed level of significance is the probability
of observing a value of the test statistic
greater than or equal to z when H0 is true. It
measures the weight of the evidence against the
null hypothesis and is also the smallest value of
? for which we can reject H0.
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Hypothesis Test Example ex. 8.7/8.30

• What are we given? n 65 s 2.6424 xbar
  42.954
• For this exercise well consider s to represent
  the true s
• Step 1, establish hypotheses
• H0 ? 42 vs. Ha ? gt 42
• Step 2, specify significance level. a .01
  (given)
• Step 3, compute the test statistic
• z (42.954-42)/(2.6424/sqrt65) 2.91
• Step 4, determine the p-value. Z-table gives P(z
  lt 2.91) 0.9982. So, P(z gt 2.91) 1- 0.9982
  .0018

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Hypothesis Test Example ex. 8.7/8.30

• Step 5, decision reject Ho since p-value (.0018)
  lt ? .01
• Step 6, conclusion within context Conclude there
  is very strong evidence that a typical customer
  is very satisfied since we reject the notion that
  the average rating is 42 with such a small
  p-value

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Hypothesis Test Example ex. 8.9/8.32

• What are we given? n 100 s 32.83 xbar
  86.6
• For this exercise well consider s to represent
  the true s
• Step 1, establish hypotheses
• H0 ? 90 vs. Ha ? lt 90
• Step 2, specify significance level. a .05
  (arbitrary)
• Step 3, compute the test statistic
• z (86.6 - 90)/(32.83/sqrt100) -1.035
• Step 4, determine the p-value. Z-table gives P(z
  lt -1.04) 0.1492.

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Hypothesis Test Example ex. 8.9/8.32

• Step 5, decision fail to reject Ho since p-value
  (.1492) gt ? .05
• Step 6, conclusion within context Conclude there
  is no evidence that the mean audit delay is less
  than 90 days since we fail to reject the notion
  that the mean is 90

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Two Tailed vs. One Tailed Tests
If the alternative is ? ? 20, the test is
two-tailed. a is shared between both tails of the
z-curve. The p-value twice the area cut off at
the tail by the computed z. This p-value is then
compared with a.
a/2 .025
?/2.025
½ p-value
z
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Hypothesis Test Example ex. 8.11/8.45a

• What are we given? n 36 s 0. 1 xbar
  16.05
• For this exercise well consider s to represent
  the true s
• Step 1, establish hypotheses
• H0 ? 16 vs. Ha ? ? 16
• Step 2, specify significance level. a .01
  (given)
• Step 3, compute the test statistic
• z (16.05 - 16)/(0.1/sqrt36) 3.0
• Step 4, determine the p-value. Z-table gives P(z
  lt 3) 0.9987, P(z gt 3) 1 - .9987 0.0013. So
  2-tailed p-value 20.0013 0.0026

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Hypothesis Test Example ex. 8.11/8.45

• Step 5, decision reject Ho p-value (.0026) lt ?
  .01
• Step 6, conclusion within context Conclude there
  is very strong evidence that the filler needs
  readjusting since we reject the notion that the
  mean is 16

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Estimating P-value with t-tables Sigma Unknown.
Credit Card Case (pg 322)

• What are we given? n 15 s 1.538 xbar
  16.827 ? .05
• Step 1, establish hypotheses
• H0 ? ? 18.8 vs Ha ? lt 18.8
• Step 2, set significance level. a .05 (given)
• Step 3, compute the test statistic
• Step 4b, estimate p-value df14, P(t lt 4.14)
  .0005. So P(t lt 4.97) lt .0005

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Estimating P-value with t-tables Sigma Unknown.
Credit Card Case (pg 322)
Step 5, decision reject Ho since p-value
(lt.0005) lt a .05 Step 6, conclusion within
context there is overwhelming evidence that the
current mean credit card interest rate is less
than 18.8.
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MegaStat Output for Example 1
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Estimating p-value, Example 2

• What are we given? n 30 s 15 x 21 ?
  .10
• Step 1, establish hypotheses
• H0 ? 20 vs. Ha ? gt 20
• Step 2, set significance level. a .10 (given)
• Step 3, compute the test statistic
• t (21 - 20)/2.74 0.365
• Step 4b, estimate the p-value. Using df 29,
  t-table gives P(T gt 1.311) .10. So, P(T gt t
  .365) gt .10 Best estimate of p-value is gt 0.10

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SExbar 15/?30 2.74
Using df 29, t-table gives P(T gt 1.311)
.10 P-value P(T gt t) gt .10
0.10
t 0.365
1.311
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Hypothesis Test Example 2

• Step 5, decision fail to reject Ho since p-value
  (gt.10) gt ? .10
• Step 6, conclusion within context no context
  given but we can say there is insufficient
  evidence that the population mean is greater than
  20. Notice we do NOT say we have evidence that m
  is less than or equal 20. In other words we can
  prove Ha but not Ho

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MegaStat Output for Example 2
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Estimating P-value, Example 3

• What are we given? n 400 s 15 x 23 ?
  .05
• Step 1, establish hypotheses
• H0 ? 20 vs. Ha ? ? 20
• Step 2, set significance level. a .05 (given)
• Step 3, compute the test statistic
• t (23 20)/0.75 4.0
• Step 4b, estimate the p-value. Using df 8,
  t-table gives P(T gt 3.291) .0005. So, P(T gt t
  4) lt .0005. Since test is two tailed, p-value
  lt 20.0005 i.e. lt 0.001

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Since we have a 2-tailed test, p-value 2 x P(T
gt t). 2 x .0005 .001, so p-value lt .001 lt a
(.05). Reject H0 since p-value lt a
.0005
½ p-value lt .0005
3.291
4.0
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Hypothesis Test Example 3

• Step 5, decision reject Ho since p-value (lt.001)
  lt ? .05
• Step 6, conclusion within context no context
  given but we can say there is very strong
  evidence that the population mean is not equal to
  20 and is most likely gt 20 since we rejected Ho
  at the upper tail.

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MegaStat Output for Example 3
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Estimating P-value, Example 4

• What are we given? n 25 s 4 x 18.7 ?
  .05
• Step 1, establish hypotheses
• H0 ? 20 vs. Ha ? lt 20
• Step 2, set significance level. a .05 (given)
• Step 3, compute the test statistic
• t (18.7 20)/0.80 1.63
• Step 4b, estimate the p-value. Using df 24,
  t-table gives P(T lt 1.711) .05 and P(T lt
  1.318) .10 Since 1.711 lt (t 1.63) lt
  1.318 , p-value is between 0.05 and 0.10

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Using df 24, t-table gives P(T lt 1.711) 0.05
and P(T lt 1.318) 0.10 . Since t 1.63 which
lies between 1.711 and 1.318 then 0.05 lt
P-value lt 0.10
.10
.05
-1.711 -1.63
-1.318
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Hypothesis Test Example 4

• Step 5, decision F.T.R. Ho since p-value
  (between .05 and .10) gt ? .05
• Step 6, conclusion within context no context
  given but we can say there is only weak evidence
  that the population mean is less than 20 and the
  finding is insignificant at the 5 level

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MegaStat Output for Example 4