Confidence Intervals and Hypothesis Testing

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Confidence Intervals and Hypothesis Testing

• 7.1/2 Confidence Intervals for a Population Mean
• 7.4 Confidence Intervals for a Population
  Proportion
• 8.1 Null and Alternative Hypotheses and Errors in
  Testing
• 8.2/3 Tests about a Population Mean assuming
  sigma is known use of Z
• 8.4 Testing assuming sigma is unknown

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Motivational Scenario

• A market research agency has been given the task
  to estimate the average number of hours per week
  that young adults spend surfing the web.
• The agency surveys a random sample of 100 young
  adults and obtains a mean of 20 hours and a
  standard deviation of 5 hours
• Can the agency conclude that the true mean number
  of hours per week spent by all young adults
  surfing the web is exactly 20 hours?

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Motivational Scenario contd

• Because the market research agency recognizes
  that the 20 hours was obtained from just one of
  many possible samples of the population they are
  unwilling to say the population mean is exactly
  equal to 20 hours.
• To allow for the variation in the sample estimate
  they may cautiously conjecture that the true mean
  is somewhere between 18 and 22 hours, between 15
  and 25 hours, etc.

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Establishing an Interval for Estimation

• How wide should they make the interval?
• How confident should they be that the named
  interval does indeed contain the true mean?
• On what basis should the choice be made?
• They can use an established fact about how sample
  means vary when random samples are repeatedly
  drawn from any population the central limit
  theorem

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Confidence Intervals for the Mean - Rationale

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Based on this relative frequency idea, if only
one random sample of size n is drawn we can
express 95 confidence that the interval
will contain m. This interval is called
a 95 confidence interval for m because if we
were to repeat the sampling process many times,
95 of the intervals calculated this way would
contain the true mean
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7.1 Confidence Intervals for a Mean When Sigma
is Known
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Example Confidence Interval for Mean assuming
sigma known
Example 7.1 Gas Mileage Case (assuming sigma
known)
95 Confidence Interval
99 Confidence Interval
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Effect of Confidence Coefficient on Interval
For a given sample size, the interval width is
narrower for lower levels of confidence.
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Effect of Sample Size on Confidence Interval
For a given level of confidence the interval
width is narrower for larger samples.
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Confidence Intervals when population standard
deviation is unknown
There is one problem.we do not usually know s so
we cannot calculate We could use the
sample standard deviation, however.
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The t-distribution

• Just as
• has a standard normal (Z) distribution either
  when X is normal or n is large, so does
• follow a t-distribution with n-1 degrees
• of freedom provided X is normally distributed
  

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The t-distribution

• The t-distribution looks almost like the
  standard normal distribution in that it is
  symmetric about zero.
• However, the tails of the t-distribution are
  fatter than that of the standard normal. -
  This is to take into account the use of the
  sample standard deviation (s) instead of the
  population standard deviation (s).

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The t-distribution

• Table values used to construct confidence
  intervals using the t-distribution will be
  different from the standard normal.
• The shape of the t-distribution is different
  depending on the degrees of freedom (df). When
  used to compute confidence intervals for the mean
  (m) the df n -1.
• When the df are very large, the t-distribution is
  close to the standard normal distribution
  (z-distribution).

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Effect of Degrees of Freedom on the
t-distribution
As the number of degrees of freedom increases,
the spread of the t distribution decreases and
the t curve approaches the standard normal curve.
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7.2 Confidence Intervals for a Mean when Sigma
is unknown
If the sampled population is normally distributed
with mean ?, then a (1-a)100 confidence interval
for m is
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95 C.I. for m using the t-distribution
Suppose you have collected a sample of
20 observations from a normal distribution, your
sample mean is 5.5 and your sample standard
deviation is 1.7 df 19 t 2.093
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7.4 Confidence Intervals for a Population
Proportion
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Example Confidence Interval for a Proportion
Example 7.8 Phe-Mycin side effects
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Hypothesis Testing

• Formal way to determine whether or not the data
  support a belief or hypothesis.

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Hypothesis Testing in the Judicial System

• In our judicial system we have the following
  hypotheses
• The accused is innocent The accused is guilty
• We can make two errors
• Convicting the innocent Letting the guilty go
  free

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Hypothesis Testing in the Judicial System

• It is desirable to minimize the chance of
  committing either error. But guarding against
  one usually results in increasing the chance of
  committing the other.
• Society favors guarding against convicting the
  innocent.

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Procedure for Guarding Against Convicting Innocent

• Assume accused is innocent.
• Gather evidence to prove guilt
• Convict only if evidence is strong enough

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Equivalent Procedure in Statistics

• To test the belief m gt 20 (alternative
  hypothesis, Ha)
• Assume m not gt 20 (Null hypothesis, Ho m ? 20)
• Gather random sample from population compute
  sample mean, x
• Conclude m gt 20 (Ha) only if evidence is strong
  enough, i.e. if x is so many standard deviations
  away from 20, the probability of this occurring
  by pure random chance is very small

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8.1 Null and Alternative Hypotheses
The null hypothesis, denoted H0, is a statement
of the basic proposition being tested. The
statement generally represents the status quo and
is not rejected unless there is convincing sample
evidence that it is false.
The alternative or research hypothesis, denoted
Ha, is an alternative (to the null hypothesis)
statement that will be accepted only if there is
convincing sample evidence that it is true.
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Types of Hypothesis
One-Sided, Greater Than H0 ? ? 50 Ha ? gt
50 (Trash Bag)
One-Sided, Less Than H0 ? ? 19.5 Ha ? lt
19.5 (Accounts Receivable)
Two-Sided, Not Equal To H0 ? 4.5 Ha ? ?
4.5 (Camshaft)
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Hypothesis Testing Definitions

• Type I error concluding Ha when Ho is true
  (convicting the innocent)
• Type II error concluding Ho when in fact it is
  false (letting the guilty go free)
• ? Prob (Type I error) significance level
• b Prob (Type II error)

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Formal Hypothesis Testing

• Establish the Null Hypothesis and the Alternative
  Hypothesis
• H0 ? 20.00 Ha ? ? 20.00 (two tailed)
• OR
• H0 ? 20.00 Ha ? gt 20.00 (right tailed)
• OR
• H0 ? 20.00 Ha ? lt 20.00 (left tailed)
• Ho must always have an equal sign and Ha must be
  what you want to prove

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Formal Hypothesis Testing for m

• Specify ?, the probability of a Type I error.
• For example, ?.05.
• You set your standard for how extreme the
  sample results must be (in support of the
  alternative hypothesis in order for you to reject
  the null.
• Here, the sample results must be strong
  enough in favor of Ha that you would falsely
  reject the null only 5 of the time.

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Formal Hypothesis Testing for m

• Compute the test statistic, z.
• a) Determine the rejection point for z, za
• b) Compute the probability of obtaining such an
  extreme test statistic z 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 a) whether z gt za or b) the p-value
  is less than or greater than a.
• Interpret statistical result in (real-world)
  managerial terms

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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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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 4a, determine the rejection point, z..01
  2.326
• Step 4b, 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 a) test
  statistic, z (2.91) gt rejection point (2.326) or
  since b) 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 4a, determine the rejection point, -z.05
  -1.645
• Step 4b, 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 a) test
  statistic, z (-1.035) not beyond rejection point
  (-1.645) or since b) 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 4a, determine the rejection points, z.005
  2.576
• Step 4b, 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 since a) test
  statistic, z 3.0) is beyond one of the rejection
  points (2.576) or since b) 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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8.4 Tests about a Population Mean Sigma Unknown
If the sampled population is normal, we can
reject H0 ? ?0 at the ? level of significance
(probability of Type I error equal to ?) if and
only if the appropriate rejection point condition
holds or, equivalently, if the corresponding
p-value is less than ?.
Reject H0 if
p-Value
Alternative
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Example 1 Test about a Mean 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 4a, determine rejection point t.05,14
  1.761
• Step 4b, estimate p-value df14, P(t lt 4.14)
  .0005. So P(t lt 4.97) lt .0005

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Example 1 Test about a Mean Sigma Unknown.
Credit Card Case (pg 322)
Step 5, decision reject Ho since (a) test
statistic, t (-4.97) lt rejection point (-1.761)
or (b) 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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Hypothesis Test 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 4a, determine the rejection points, t.10,29
  1.311
• 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

SExbar 15/?30 2.74
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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 (a)
  test statistic, t (0.365) lt rejection point
  (1.311) or (b) 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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Hypothesis Test 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 4a, determine the rejection points,
  t.025,399 1.96
• 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 (a) test
  statistic, t (4.0) gt rejection point (1.96) or
  (b) 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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Hypothesis Test 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 4a, determine the rejection point, t.05,24
  1.711
• 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 (a) test
  statistic, t (1.63) is not as extreme as
  rejection point (1.711) or (b) 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