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 1Var Statistics. The result is shown below
Press STAT, then RIGHT ARROW twice to TESTS. Press 1 to select 1:Ztest. If Data is not already highlighted, LEFT ARROW to highlight it, and then press ENTER to select the data mode.



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.


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.

Conclusions based on zvalue:
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 lefttailed
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 Pvalue:
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:Ztest. If STATS is not highlighted, ARROW RIGHT once and press ENTER to select STATS.



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 twotailed 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.


Conclusions based on zvalue:
The test z value shown above is z = 1.89. Using the invNorm
function for =
0.05 and the fact that this is a twotailed 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 Pvalue:
The test p value shown about is p = 0.0593463074. Thus, we cannot reject
:
= 8 since 0.053463074 > 0.05