Module 9 Examples
Poisson distribution table
Average 3 calls every 10
seconds during normal day = 0.3 calls @ second
Poison Table
|
|
|
|
Ar |
.3 |
|
0 |
.7408 |
|
1 |
.2222 |
|
2 |
.0333 |
|
3 |
.0033 |
|
4 |
.0003 |
|
5 |
.0000 |
|
6 |
.0000 |
|
7 |
.0000 |
What is the probability that
there will be no more than 1 call in any second?
.2222=22.22%
Calculate:
P(r events) e-
P (1) = .7408 
P (1) = .7408 x .3/1
P (1) = .22224
What is the probability that
there will be more than 3 calls any second?
.0033+.0003=.0036 = .36%
Poisson to Approximate the Binomial
At peak hours 15000 cars per
hour use the tunnel, historical records indicate breaking down in the tunnel is
0.00003. What is the probability that
no more than two car breakdowns in the tunnel during peak hours.
= .45
P(r events) e-
P (0) = e.45
x.45/15000
P (1) = .7408 x .3/1
P (1) = .22224
Sample size 16 average life 170 minutes sample SD = 40 minutes population normal
distribution
What is the 95% confidence
limit for average life
|
t-Distribution
Table |
|
|
Degree of freedom |
Upper-tail area a .025 |
|
15 |
2.131 |
t 0.025 
s/
170
15 0.025 x 40/
170 
2.131x 40/4
170
2.131 x 10
170 
21.31
148.7 – 191.3
95 %
confidence level is between 148.7 and 191.3
What is the 95% confidence
limit for average life
|
t-Distribution
Table |
|
|
Degree of freedom |
Upper-tail area a .05 |
|
15 |
1.753 |
t 0.05 
s/
170
15 0.05 x 40/
170 
1.753x 40/4
170
1.753 x 10
170 
17.53
152.47 – 187.53
90 %
confidence level is between 152.47 and 187.53
Chi-squared Distribution
Variance
of a sample of 25 batteries found life of 900 hours. The variance calculated from past records was 1200
What
is the lower critical value in the test, given a 10% significance level?
|
Chi-squared
Distribution Table |
|
|
Degree of freedom |
Area in Upper-tail .10 |
|
24 |
33.1963 |
Should
the hypothesis be accepted?
Observed
Calculation
X2Obs = (n-1) x Sample variance / Population variance
X2Obs = 24 x
900 / 1200
X2Obs = 21600 /
1200
X2Obs = 18
yes
18 is less than the significance level of 33.1963
A car manufacturer is trying to
find out more about its male customers, by associating the type of car men buy
with their personality type. The table shows the results from a sample of 200
recent customers.
|
Car purchased |
||||||
|
Personality |
Saloon |
Hatchback |
Sports |
Total |
||
|
|
||||||
|
Introvert |
19 |
24 |
|
|
||
|
Extrovert |
22 |
25 |
13 |
|
||
|
Mixed |
39 |
41 |
10 |
|
||
|
|
||||||
|
Total |
80 |
90 |
30 |
200 |
||
7Does
personality type affect car-purchasing behaviour?
Follow the five stages of a significance test.
1.
The null hypothesis is that of independence: the proportions of
different cars purchased are in proportion to the numbers of buyers falling under
each personality type. The significance test is one-tailed since the car
manufacturer wishes to see whether the difference between observed and expected
results is unusually high. It would be of no interest to see whether they were
unusually close together.
2.
The data is the sample of 200 recent car purchasers.
3.
The conventional 55 significance level is used.
4.
The chi-squared distribution is applicable here. The table in the
question is the observed data. Expected results (if the hypothesis is true) are
shown below.
.fe = row total / grand total x column total
|
Personality |
Saloon |
Hatchback |
Sports |
Total |
|
Introvert |
fo=19 fe=20 |
fo=24 fe=22.5 |
fo=7 fe=7.5 |
|
|
Extrovert |
fo=22 fe=24 |
fo=25 fe=27 |
fo=13 fe=9 |
|
|
Mixed |
fo=39 fe=36 |
fo=41 fe=40.5 |
fo=10 fe=13.5 |
|
|
|
||||
|
Total |
80 |
90 |
30 |
200 |
|
Personality |
Saloon |
Hatchback |
Sports |
|
Introvert |
|
|
|
|
Extrovert |
|
|
|
|
Mixed |
|
|
|
= .05+.10+.033+.1667+.1481+1.7778+.25+.0123+.9074
= 3.4456
Degree of Freedom = (r-1) x(c-1)
Degree of Freedom = 4= 2 X2
The critical value of chi-squared for the 5% upper tail is
|
Chi-squared
Distribution Table |
|
|
Degree of freedom |
Area in Upper-tail .05 |
|
4 |
9.48773 |
The observed Chi squared is much smaller than the critical value of chi squared at 9.48773 therefore the hypothesis must be accepted. Personality are of no significant difference in the type of car purchased.
F-Distribution
Hal
has a sample size of 25 and a variance of 900
GO
has a sample size of 16 and a variance of 750
What
is the critical value given a 5% significance level? 2.24
(15, 24)
|
F-Distribution
Table |
|
|
Degree of freedom in
the denominator |
Degree of freedom in
the numerator 24 |
|
15 |
2.24 5% 3.21 1% |
What is the observes F Value
F =
Variance of sample 1 / Variance of sample 2
F =
900 / 750
F =
1.2
Should
the hypothesis be accepted?
Yes accept
the hypothesis the observed F value of 1.2 is less than the 5% Critical value
of 2.24