Known (z-test)
1. Given this Population Data:
Test the claim that
: 
<
80 for these data using 
=
0.10:
|
60 |
70 |
75 |
55 |
80 |
55 |
|
50 |
40 |
80 |
70 |
50 |
95 |
|
120 |
90 |
75 |
85 |
80 |
60 |
|
110 |
65 |
80 |
85 |
85 |
45 |
|
75 |
60 |
90 |
90 |
60 |
95 |
|
110 |
85 |
45 |
90 |
70 |
70 |
Note that 
is
80.
First enter
the data in 
by pressing STAT and selecting 1:Edit by pressing ENTER.
Then run 1-Var Statistics.
The result is shown below
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Press STAT, then RIGHT ARROW twice to TESTS. Press 1 to select 1:Z-test. If Data is not already highlighted, LEFT ARROW to highlight it, and then press ENTER to select the data mode.
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ARROW DOWN once
and enter 80 next to 
:.
ARROW DOWN once to :.
In order to enter 
here, press VARS, type 5 to select 5:Statistics, and then
type 3 to select 3:.
As we ARROW DOWN,
will be replaced by the computed value 19.16097224. Make
sure the defaults of
for List: and 1 for Freq: are also listed.
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ARROW
DOWN to :.
Use the arrow keys to highlight <,
since the claim being tested is <
80. Press ENTER. ARROW
DOWN to select Calculate, and then press ENTER. The
results are shown below.
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Conclusions based on z-value:
The test z value shown above is z = -1.59. Using the invNorm
function for =
0.10, we find that invNorm(0.10) = -1.28.
This is a left-tailed
test, so since -1.59 < -1.28, we are in the rejection region, so reject
:
80, and accept the alternative hypothesis :<
80.
Conclusions based on P-value:
The test p value shown above is p = 0.0561553908. The decision is to reject
: 80, since 0.0561553908 < 0.10; that is, p < .
Test the claim :
= 8
when = 0.6, 
= 8.2, and n = 32 using =
0.05. We use :
8
Press STAT, then RIGHT ARROW twice to TESTS. Press 1 to select 1:Z-test. If STATS is not highlighted, ARROW RIGHT once and press ENTER to select STATS.
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ARROW DOWN once and enter
the value of 8 for .
ARROW DOWN once and enter .6 for .
ARROW DOWN once more and enter 8.2 for .
ARROW DOWN once and enter 36 for n.
Now we must choose which test. Since
this is a two-tailed test, ARROW DOWN once to :
then use the LEFT or RIGHT ARROW to select ,
then press ENTER. ARROW
DOWN one last time to Calculate and then press ENTER. The
results are shown below.
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Conclusions based on z-value:
The test z value shown above is z = 1.89. Using the invNorm
function for =
0.05 and the fact that this is a two-tailed test, we find that invNorm(0.025)
= 1.96. Thus z < -1.96 or z > 1.96 would place our test value in the
rejection region. But it is the case that -1.96 < 1.89 < 1.96.
The decision is that we cannot
reject :
= 8.
Conclusions based
on P-value:
The test p value shown about is p = 0.0593463074. Thus, we cannot reject
:
= 8 since 0.053463074 > 0.05
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