Casio ALGEBRA FX 2.0 User manual

Examples

2
examples
kk
kk
k1-Sample ZTest
Set Up
1. On the icon menu, select STAT2.
Execution
2. When using list data (List is selected as the Data parameter), be sure to input data into
the list first.
3. 3(TEST)b(Z)b(1-Smpl) ... 1-Sample ZTest
4. Set calculation parameters.
5. Align the cursor with [Execute]
1(CALC) ... Performs calculation.
6(DRAW) ... Draws graph.

3
examples
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Example Five new members of a football team are timed for the 100-
meter dash, yielding the following times.
A: 12.5 B: 11.6 C: 10.8 D: 12.8 E: 11.4
The average time of current team members is 11.4 seconds,
with a standard deviation of 1.30. Test the null hypothesis that
the times of the five new members are at the same level as
current team members at the 0.05 level of significance.
Procedure
1mSTAT2
2 bc.fwbb.gwba.iw
bc.iwbb.ew
33(TEST) b(Z)b(1-Smpl)
41(LIST) c
1(≠) c
bb.ew
b.dw
1(LIST) bwc
1(1)c
1(None)c
51(CALC)
6(DRAW)
Result Screen
Since P = 0.47003508 > 0.05 (level of significance), we can't reject the null hypothesis and
can conclude the times of the five new numbers is at the same level as current team
members.

4
examples
kk
kk
k2-Sample ZTest
Set Up
1. On the icon menu, select STAT2.
Execution
2. When using list data (List is selected as the Data parameter) , be sure to input data into
the list first.
3. 3(TEST)b(Z)c(2-Smpl) ... 2-Sample ZTest
4. Set calculation parameters.
5. Align the cursor with [Execute]
1(CALC) ... Performs calculation.
6(DRAW) ... Draws graph.

5
examples
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Example A consumer group is testing camp stoves. To test the heating capacity
of a stove they measure the time required to bring 2 qt of water from
50qq
qq
qF to boiling (at sea level). Two competing models are under
consideration. 36 stoves of each model were tested and the following
results were obtained.
Model 1: mean time oo
oo
o1= 11.5min; standard deviation σ1= 2.4min
Model 2: mean time oo
oo
o2= 10 min; standard deviation σ2= 3 min
Is there any difference between the performances of these two
models? (Use a 5% level of significance.)
Procedure
1mSTAT2
23(TEST) b(Z)c(2-Smpl)
32(VAR) c
1(≠) c
c.ew
dw
bb.fw
dgw
baw
dgw
1(None) c
41(CALC)
6(DRAW)
Result Screen
Since P = 0.01914957 < 0.05 (level of significance), we reject the null hypothesis µ1-µ2 =0
and conclude the performances of the two models are different.

6
examples
kk
kk
k1-Prop ZTest
Set Up
1. On the icon menu, select STAT2.
Execution
2. 3(TEST)b(Z)d(1-Prop) ... 1-Prop ZTest
3. Set calculation parameters.
4. Align the cursor with [Execute]
1(CALC) ... Performs calculation.
6(DRAW) ... Draws graph.

7
examples
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Example A team of eye surgeons has developed a new technique for a risky eye
operation to restore the sight of people blinded from a certain disease.
Under the old method it is known that only 30% of the patients who
undergo this operation recover their eyesight. Suppose that surgeons
in various hospitals have performed a total of 250 operations using
the new method and that 93 have been successful (the patients fully
recovered their sight). Can we justify the claim that the new method is
better than the old one?(Use a 1% level of significance.)
Procedure
1mSTAT2
23(TEST) b(Z)d(1-Prop)
33(>) c
.dw
jdw
cfaw
1(None) c
41(CALC)
6(DRAW)
Result Screen
Since P = 6.4915e-3 < 0.01 (level of significance), we can reject the null hypothesis that the
successful probability of new operation is 0.3 and conclude that it can be said the new
method obtains good results more than existing method. That is to say it is more effective by
new methods.

kk
kk
k2-Prop ZTest
Set Up
1. On the icon menu, select STAT2.
Execution
2. 3(TEST)b(Z)e(2-Prop) ... 2-Prop ZTest
3. Set calculation parameters.
4. Align the cursor with [Execute]
1(CALC) ... Performs calculation.
6(DRAW) ... Draws graph.
8
examples

9
examples
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Example The County Clerk wishes to improve voter registration. One method
under consideration is to send reminders in the mail to all citizens in
the county who are eligible to register. As a pilot study to determine if
this method will actually improve voter registration, a random sample
of 1224 potential voters was taken. Then this sample was randomly
divided into two groups.
Group A : There were 612 people in this group. No reminders to
register were sent to them. The number of potential
voters from this group who registered was 280.
Group B : This group also contained 612 people. Reminders were
sent in the mail to each member in the group, and the
number who registered to vote was 335.
The County Clerk claims that the proportion of people to register was
significantly greater in group B. On the basis of this claim the clerk
recommends that the project be funded for the entire population of
Macek County. Use 5% level of significance to test the claim that the
proportion of potential voters who registered was greater in group B,
the group that received reminders.
Procedure
1mSTAT2
23(TEST) b(Z)e(2-Prop)
32(<) c
ciaw
gbcw
ddfw
gbcw
1(None) c
41(CALC)
6(DRAW)
Result Screen
Since P = 8.3277e-4 < 0.05 (level of significance), we can reject the null hypothesis (P1>P2)
and conclude that the County Clerk‘s claim is valid at the 0.05 level of significance.

kk
kk
k1-Sample tTest
Set Up
1. On the icon menu, select STAT2.
Execution
2. When using list data (List is selected as the Data parameter), be sure to input data into
the list first.
3. 3(TEST)c(T)b(1-Smpl) ... 1-Sample tTest
4. Set calculation parameters.
5. Align the cursor with [Execute]
1(CALC) ... Performs calculation.
6(DRAW) ... Draws graph.
10
examples

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Example A company manufactures large rocket engines used to project
satellites into space. The government buys the rockets, and the
contract specifies that these engines are to use an average of 5550 lb
of rocket fuel the first 15 sec operation. The company claims their
engines fit specifications. To test the claim an inspector randomly
selects six such engines from the warehouse. These 6 engines are
fired 15 sec each and the fuel consumption for each engine is meas-
ured. For all six engines, the mean fuel consumption is oo
oo
o= 5750
pounds and the standard deviation is x σ n-1 = 250 lb. Is the claim
justified at the 5% level of significance?
Procedure
1mSTAT2
23(TEST) c(T)b(1-Smpl)
32(VAR) c
1(≠) c
fffaw
fhfaw
cfaw
gw
1(None) c
41(CALC)
6(DRAW)
Result Screen
11
examples
Since P = 0.107344401 > 0.05 (level of significance), we can’t reject the null hypothesis
µ=5550 and conclude that the data does not present sufficient evidence to indicate that the
average fuel consumption from the first 15 seconds of operation is different from µ=5550.

kk
kk
k2-Sample tTest (Pooled On)
Set Up
1. On the icon menu, select STAT2.
Execution
2. When using list data (List is selected as the Data parameter), be sure to input data into
the list first.
3. 3(TEST)c(T)c(2-Smpl) ... 2-Sample tTest
4. Set calculation parameters.
5. Align the cursor with [Execute]
1(CALC) ... Performs calculation.
6(DRAW) ... Draws graph.
12
examples

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Example Two different processes (Type A and Type B) are used to produce
tinplate. The following values show the weights of samples produced
by each process.
Type A: 105 108 86 103 103 107 124 124
Type B: 89 92 84 97 103 107 111 97
Using the level of significance 0.05, test the null hypothesis that the
tinplate produced by the two processes are of the same level.
Procedure
1mSTAT2
2 bafwbaiwigwbadw
badwbahwbcewbcew
e
ijwjcwiewjhw
badwbahwbbbwjhw
33(TEST) c(T)c(2-Smpl)
41(LIST) c
1(≠) c
1(LIST) bwc
1(LIST) cwc
1(1)c
1(1)c
1(On) c
1(None) c
51(CALC)
6(DRAW)
Result Screen
13
examples
Since P = 0.08602732 > 0.05 (level of significance), we can not reject the null hypothesis
and conclude that the tinplate produced by the two processes are of the same level.

kk
kk
k2-Sample tTest (Pooled Off)
Set Up
1. On the icon menu, select STAT2.
Execution
2. When using list data (List is selected as the Data parameter) , be sure to input data into
the list first.
3. 3(TEST)c(T)c(2-Smpl) ... 2-Sample tTest
4. Set calculation parameters.
5. Align the cursor with [Execute]
1(CALC) ... Performs calculation.
6(DRAW) ... Draws graph.
14
examples

15
examples
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Example Two competing headache remedies claim to give fast-acting relief. An
experiment was performed to compare the mean lengths of time
required for bodily absorption of brand A and brand B headache
remedies. 12 people were randomly selected and given an oral dosage
of brand A. Another 12 were randomly selected and given an equal
dosage of brand B. The length of time in minutes for the drugs to
reach a specified level in the blood was recorded. The means, stand-
ard deviations, and sizes of the two samples follow;
Brand A: oo
oo
o1 = 19.8; x1 σ n-1 = 8.6; n1 = 12
Brand B: oo
oo
o2 = 19 ; x2 σ n-1 = 6 ; n2 = 12
Past experience with the drug composition of the two remedies
permits researchers to assume the standard deviations of the two time
distributions are approximately equal. Let us use a 5% level of
significance to test the null hypothesis that there is no difference in
the mean time required for bodily absortion.
Procedure
1mSTAT2
23(TEST) c(T)c(2-Smpl)
32(VAR) c
1(≠) c
bj.iw
i.gw
bcw
bjw
gw
bcw
2(Off) c
1(None) c
41(CALC)
6(DRAW)
Result Screen
Since P = 0.79431564 > 0.05 (level of significance), the difference is in acceptance region.
We conclude that the data does not contain sufficient evidence to reject H0. Therefore, at the
5% level of significance we conclude that there is no difference in the mean times.

16
examples
kk
kk
kLinearReg tTest
Set Up
1. On the icon menu, select STAT2.
Execution
2. Input data into the list.
3. 3(TEST)c(T)d(LinReg) ... LinearReg tTest
4. Set calculation parameters.
5. Align the cursor with [Execute]
1(CALC) ... Performs calculation.

17
examples
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Example A survey conducted by a city on its real estate investments revealed
the following data on the relationship between area and price.
Does the data indicate that the value β 1, that is, the slope of the popula-
tion regression line, is not zero, which would mean that x(Area) can be
use as a predictor of y (Sales Price)? Use a 5% level of significance.
Procedure
1mSTAT2
2jwbfwbawbbwbaw
e
dgwiaweewffwdfw
33(TEST) c(T)d(LinReg)
41(≠) c
1(LIST) bwc
1(LIST) cwc
1(1)c
1(None) c
51(CALC)
Result Screen
Since P = 6.4531e-3 < 0.05 (level of significance), we can reject the null hypothesis β 1 ≠ 0
and conclude that there is a linear relationship y= β 0 + β 1x + εbetween area(x) and price(y).
x Area (square feet) 9 15101110
y Sales Price (×$1000) 36 80 44 55 35

18
examples
kk
kk
k χ 2Test
Set Up
1. On the icon menu, select STAT2.
Execution
2. 3(TEST)d( χ
2)...
χ
2Test
3. Input data into the Matrix.
4. Set calculation parameters.
5. Align the cursor with [Execute]
1(CALC) ... Performs calculation.
6(DRAW) ... Draws graph.

19
examples
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Example A certain club collected data on the attendance at meetings by married
status, and obtained the data shown below.
Married Divorced Widowed Single Row Totals
Often Absent 34 16 14 36 100
Seldom Absent 64 34 20 82 200
Never Absent 52 50 16 82 200
Column Totals 150 100 50 200 500
Test the null hypothesis that two phenomena are independent using
the level of significance of 0.05.
Procedure
1mSTAT2
23(TEST) d( χ2)
32('MAT)
1(DIM)
dw
eww
dewbgwbewdgw
gewdewcawicw
fcwfawbgwicw
ii
41(MAT) v(A) wc
1(MAT) l(B) wc
1(None) c
51(CALC)
6(DRAW)
Result Screen
Since P = 0.17546141 > 0.05 (level of significance), we can not reject the null hypothesis
and conclude that there the two phenomena are independent of each other.

kk
kk
k2-Sample FTest
Set Up
1. On the icon menu, select STAT2.
Execution
2. When using list data (List is selected as the Data parameter), be sure to input data into
the list first.
3. 3(TEST)e(F)... 2-Sample FTest
4. Set calculation parameters.
5. Align the cursor with [Execute]
1(CALC) ... Performs calculation.
6(DRAW) ... Draws graph.
20
examples