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Further Inference

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Title: Further Inference


1
Chapter 7
  • Further Inference

2
Inference when population standard deviation is
unknown
There is one problem.we do not usually know s so
we cannot calculate sx. We could use the sample
standard deviation, however.
3
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

4
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).

8
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
9
Group Work 90 C.I. For m
Suppose you have collected a sample of
25 observations from a normal distribution, your
sample mean is 8.0 and your sample standard
deviation is 2.1
df 24 t ____
10
The t-procedure for C.I.s is robust. Check for
normality with skewness kurtosis measures (both
must be within (1) and absence of outliers for
very small samples (n lt 15). For 15 lt n lt 40,
t-procedure O.K. as long as skewness kurtosis
within 3 and no outliers. For n gt 40,
t-procedure O.K.
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Hypothesis Test Example 1
  • 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 4, estimate the p-value. Using df 100,
    t-table gives P(T gt 3.390) .0005. So, P(T gt
    t4) lt .0005. Since test is two tailed, p-value
    lt 20.0005 i.e. lt 0.001
  • Step 5, decision reject Ho since p-value (.001)
    lt ? .05
  • Step 6, conclusion within context none given

13
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
3.390
4.0
14
Macro Output for Example 1
15
Hypothesis Test Example 2
  • What are we given? n 30 s 15 x 21 ?
    .05
  • Step 1, establish hypotheses (given)
  • H0 ? 20 vs. Ha ? gt 20
  • Step 2, set significance level. a .05 (given)
  • Step 3, compute the test statistic
  • t (21 - 20)/2.74 0.365
  • Step 4, estimate the p-value. Using df 29,
    t-table gives P(T gt 0.683) .25. So, P(T gt
    0.365) gt .25. Since test is one tailed, this is
    the p-value (gt .25)
  • Step 5, decision fail to reject Ho since p-value
    (gt .25) gt ? .05
  • Step 6, conclusion within context none given

16
SExbar 15/?30 2.74
Using df 29, t-table gives P(T gt 0.683)
.25 P-value P(T gt t) gt .25
0.25
t 0.365
0.683
17
Macro Output for Example 2
18
Hypothesis Test Example 3
  • What are we given? n 30 s 15 x 18.7 ?
    .05
  • Step 1, establish hypotheses (given)
  • H0 ? 20 vs. Ha ? lt 20
  • Step 2, set significance level. a .05 (given)
  • Step 3, compute the test statistic
  • t (18.7 - 20)/2.74 - 0.474
  • Step 4, estimate the p-value. Using df 29,
    t-table gives P(T lt -0.683) .25. So, P(T lt
    -0.474) gt .25. Since test is one tailed, this is
    the p-value (gt .25)
  • Step 5, decision fail to reject Ho since p-value
    (gt .25) gt ? .05
  • Step 6, conclusion within context none given

19
Using df 29, t-table gives P(T lt -0.683)
P-value P(T lt t)
0.25
gt .25
.25
-0.683
-.474
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Confidence Intervals and Hypothesis Tests for
Comparing Two Means
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Difference between Means
  • Do female students perform better than male
    students on average?
  • We can answer this by drawing random samples of
    female students and male students and looking to
    see how far apart the sample means are

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Likely scenario when female and male students
form a single population with one Mean
These two sample means probably ARE NOT very far
apart.
m1
25
If they form two Populations with two Means?
However, if the true means are different we are
more likely to get two sample means which are far
apart.
m1
m2
26
Two Populations, Two Means
However, if the true means are different we are
more likely to get two sample means which are far
apart.
m1
m2
27
Difference between Means
  • If the sample means are far apart (statistically)
    then we conclude that the population means are
    different.
  • We will, first, calculate the difference between
    the sample means and then compute a t-statistic
    (like a z-score) to measure the distance apart in
    terms of standard deviations (called std. error
    here)

28
To test the hypothesis Ho m1 m2 when s1 and s2
are unknown we use the equivalent two-sample
t-statistic
29
To test the hypothesis Ho m1 m2 using the
two-sample t-statistic we recognize that under
Ho, m1 m2 0. So our test statistic
becomes
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Healthy Companies vs. Failed Companies
To test the hypothesis Is the mean current ratio
for healthy companies greater than the current
ratio for failed companies? We need Ho m1 m2
vs. Ha m1 gt m2 which is the same as writing
Ho m1 m2 0 vs. Ha m1 m2 gt 0
34
Excel Output
Back-up hand calculations
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Macro Output
36
Healthy Companies vs. Failed Companies
Suppose we choose a .01. Then we compare our
p-value against a. Decision Rule is reject Ho
(or conclude Ha) if p-value lt a Since p-value lt
.0005 which is lt a .01 we reject Ho (conclude
Ha) The test is highly significant since the
p-value is so small. We conclude that there is
overwhelming evidence that the mean current ratio
among healthy companies is greater than that
among failed companies
37
Group Work
Test the hypothesis, at the 5 significance
level, that the mean salaries for males and
females are different.
38
Excel Output for ex. 7.97
39
Confidence Intervals for the Difference Between
Two Means
40
Difference between Means
  • If the sample means are far apart (statistically)
    then we conclude that the population means are
    different.
  • We will, first, calculate the difference between
    the sample means and then construct a confidence
    interval around that difference.

41
Difference between Means
  • What value will we look for (in the confidence
    interval) to see whether the means are the same?
  • Zero

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Excel Descriptive Statistics and hand
calculations
46
Output from Macro
47
Group Work
Compute a 95 C.I. for the difference mm-mf
48
Sample statistics for ex. 7.97
49
Macro Output for ex. 7.97
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