Teaching S1/S2 statistics using graphing technology

Teaching S1/S2
statistics using
graphing technology

CALCULATOR HINTS FOR S1 & 2 STATISTICS
- STAT MENU (2) on Casio
.
It is advised that mean and standard deviation are obtained
directly from a calculator.
1. Numerical measures
Mean and standard deviation
No frequencies
List 1: input x 1 2 3 4 5
F2 (CALC)
3

F6 SET
1 Var X List: List 1
F1 1 Var Freq : 1
EXIT
F1 (1 VAR)

Frequencies
List 1: input x 1 2 3 4 5
List 2: input frequencies 2 5 8 4 2
For grouped data, the mid values should be
entered into List 1. This can be done by entering, for example,
(34.5 + 39.5)/2 directly into List 1.
F2 (CALC)
F6 SET
1 Var X List : List 1
1 Var Freq : List 2
EXIT
F1 (1 VAR) 5

1. Find the mode.
2. Find the mean.
3. Find the median.

It is advised that calculator is used to obtain probabilities.
Binomial Distribution
(a) Probabilities of type P(X = x):
eg B( 15, 0.2) P( X = 3)
F5 DIST
F5 BINOMIAL
F1 Bpd
Data Variable
x : 3 EXE
Numtrial : 15 EXE
p : 0.2 EXE
Execute F1 (calc)

(b) Probabilities of type P(X ≤ x):
eg B( 15, 0.2) P( X ≤ 3)
F2 Bcd
x : 3 EXE
Numtrial : 15 EXE
p : 0.2 EXE
Execute F1 (calc)

AQA Jan 2007 Statistics 1B

It is advised that students should be aware of the method involved as
questions may involve being asked to show a result.
Discrete Probability Distributions: Expectation and variance
List 1: input x 0 1 2 3
List 2: input probabilities 0.1 0.3 0.4 0.2
CALC
SET
EXIT
1 VAR
X x 2
= E(X) = E(X) =E(X²) σ given
Check n = 1
x 12

Example
For the following probability distribution,
x 0 1 2 3 4 5
P (X = x) 0.08 0.30 0.34 0.15 0.10 0.03
calculate (a) E(X) ;
(b) E( X²) ;
(c) the variance of X;
(d) the standard deviation of X.
13

(a) E(X) =1.98
(b) E(X²) = 5.36
(c) Var = 5.36 – 1.98² = 1.44
(d) sd = 1.20
14

It is advised that z values and methods are shown and the calculator is
used for checking results.
Normal Distribution: Calculation of probabilities
(a) Probabilities of types P(X ≤ x), P(X < x), P(X > x) and P(X ≥ x) can be
found directly
eg X~ Normal mean 135 st deviation 15 P(X ≤ 127) or P(X < 127)
F5 DIST F1 Norm F2 Ncd
Select a suitable Lower
value – enter info

eg X~ Normal mean 135 st deviation 15 P(X ≥ 118) or P(X >118)
Select a suitable Upper
value - enter info
eg X~ Normal mean 135 st deviation 15 P( 119 < X < 128)

(b) Problems involving inverse normal probabilities :
F5 DIST F1 NORM F3 InvN
eg X~ Normal mean 135 st deviation 15 Find value of x such that
P(X< x) = 0.15
F3 Inv N
Enter data – Area Left
z scores corresponding to area can also be obtained (as Inv Normal tables)

eg X~ Normal mean 135 st deviation 15 Find value of x such that
P(X > x) = 0.30
Area right this time

AQA Jan 2011 Statistics 1B

(iii) 0
(i)
(ii)

Correlation and regression
It is advised that the calculator is used, if possible, to find the values
of r, a and b.
List 1: input x 1 2 3 4 5
List 2: input y 21 28 34 45 51
CALC
F3 REG F1 X F2 a + bx
This gives a,b,r (and r² )
for regression equation y = a + bx
y = 12.7 + 7.7x

This can also be done from the GRAPH, Scatter Function
F1 graph
Calc X a +bx
F1 graph 1 (default scatter) F1 calc will produce results
Draw
Residuals
calculation can be
set up
Shift Setup
Resid List

Example
The following data were collected by a UK Gas supply company. It shows, for a house
in London, the weekly gas consumption, y, in thousands of cubic feet, and the
average outside temperature, x° C.
week 1 2 3 4 5 6 7 8 9 10
x 7.0 6.2 3.5 3.0 4.0 8.2 2.7 -1.0 7.0 7.8
y 3.5 4.2 5.6 6.0 5.0 4.1 6.2 7.5 4.5 4.2
(a) Calculate the regression line of gas consumption on average outside temperature.
(b) Evaluate the residuals for the points where x = 3.0° and x = 7.8°
(c) Find an estimate for the weekly gas consumption when the outside
temperature is 5°C.

(a) Calculate the regression line of gas consumption on average outside temperature.
(b) Evaluate the residuals for the points where x = 3.0° and x = 7.8 5°C
(c) Find an estimate for the weekly gas consumption when the outside temp is 5°C
(a)
equation is y = - 0.404x + 7.035
(b)
Residual x =3
Residual x =7.8
(c) Predict y when x = 5 y = - 0.404x 5 + 7.035 = 5.02

It is advised that calculator/tables are used.
Discrete Probability Distributions: Poisson distribution
All probabilities: MENU STAT
Example
Probabilities of type P(X = x):
F5 (DIST) F6 F1 (POISN) F1 (Ppd)
eg Poisson distribution with λ = 2.8 P( X = 4)

(b) Probabilities of type P(X ≤ x):
F5 (DIST) F6 F1 (POISN) F2 (Pcd)
Example Poisson distribution with λ = 2.8 P( X ≤ 3)

It is advised that the z values used and the method are shown, with the
calculator used for checking.
Confidence intervals for μ (known variance or large sample using z)
eg A random sample of 12 packets of crisps with nominal weight 35g is
obtained. The weights of such packets can be modelled by a normal
distribution with standard deviation 3.5g. The weights of each of the
packets in the sample are:
34.6 35.9 37.4 36.2 37.5 34.9 35.8 38.2 37.8 37.2 35.1 35.9
Calculate a 95% confidence interval for the mean weight.
F1 z
F1
Data in list 1
95%
( 34.39 , 36.36 )

eg A random sample of 112 rods, that had mean 4.76 mm and
standard deviation 1.46 mm, was obtained from a production
process on a particular Friday. Calculate a 90% confidence interval
for the mean length of rods produced on this Friday.
Variable as
mean and
stdev given
Interval ( 4.53, 4.99)
28

It is advised that the t values used and the method are shown, with the
calculator used for checking.
Confidence intervals for μ (random sample from a normal distribution
with unknown standard deviation using t- distribution.)
eg A company manufactures components for racing cars. A random sample
of these components is taken from the production line during an afternoon
shift. The lengths, in mm, of the components in this sample were:
135.6 134.8 135.1 136.4 135.2 135.7 135.9 136.1 135.8
134.8 136.2
Assuming the lengths may be modelled by a normal distribution, calculate
a 90% confidence interval for the mean length.

F2 t F1 1 sample
Data in List 1
90%
so the interval is (135.29 , 135.90 )

Summations given
Standard deviation not known use t
First find the mean and an estimate for the standard deviation
31

.
It is advised that the method used for obtaining the test statistic is shown
with the calculator used for checking.
Hypothesis test on population mean based on sample from a normal
distribution with known or unknown standard deviation using
z or t- distribution
Example List of data given – z test H0 μ = 35 H1 μ > 35 5% level
A random sample of 10 packets of crisps with nominal weight 35g is obtained.
These weights can be modelled by a normal distribution with standard
deviation known to be 1.5 g. The weights of each of the packets in the sample
are : 34.6 35.9 37.4 36.2 34.9 35.8 38.2 37.8 35.1 35.9
F3 test F1 z F1 I sample

H 1 is μ > 35
z = 2.488 (test statistic)
p = 0.00643 0.0643 < 0.05 sig level
X
= 36.18 sx not needed
n = 10
Sig evidence to reject Ho and conclude that there is significant
evidence to suggest that the mean weight of crisps is higher
than 35g

Example List of data given – t test H0 μ = 35 H1 μ 35 5% level
A random sample of 10 packets of crisps with nominal weight 35g is
obtained. These weights can be modelled by a normal distribution
The weights of each of the packets in the sample are :
34.6 35.9 37.4 36.2 34.9 35.8 38.2 37.8 35.1 35.9
F3 test F2 t F1 I sample

H 1 is ≠ 35
t = 3.0138 (test statistic)
p = 0.014623
X
0.0146 < 0.05 sig level
= 36.18 Sig evidence to reject Ho and conclude
S x = 1.238 that the mean weight of crisps is not
n = 10 35g ( appears to be higher )
.

It is advised that the method for obtaining expected frequencies is shown
and also the method used for obtaining the test statistic.
Application of Contingency tables: use of
Example
A random sample of people was asked their opinion on a road scheme.
The people lived in the road to be involved in the scheme, one road away
from the scheme or 2 or more roads away from the scheme.
The results are summarised in the following table.
( O
i
Ei
E
i
2
)
2
Test, using the distribution and the 1% level of significance,
whether opinion is independent of where a person lives .

F3 test F3 Chi F2 2 way
F2 to insert table Set dimensions EXE
Enter data and EXIT Exe test stat and p value

F6 Go to Mat B
Mat B gives the
expected values
Ho No association
H1 Association
p value < 0.01 ( sig level)
Reject Ho
Significant evidence that opinion is associated with whether respondent
lives in the road involved, one road away from the scheme or 2 or more
roads away from the scheme

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