Post on 25-Jul-2020
transcript
@MEIConference #MEIConf2019
#MEIConf2019
USING CASIO CALCULATORS
FOR STATISTICAL TESTS
#MEIConf2019
The Statistics menu
#MEIConf2019
The Statistics menu
Calculating
summary
statistics
from a set
of data
Calculating probabilities
from distributions.Conducting
hypothesis
tests.
Calculating
confidence
intervals.
#MEIConf2019
Calculating probabilitiesBinomial
#MEIConf2019
Binomial𝑋~𝐵 47,0.2
Find
a) 𝑃 𝑋 = 11
b) 𝑃 𝑋 ≤ 11
c) 𝑃 𝑋 > 11
d) 𝑃 𝑋 < 8
e) 𝑃 3 ≤ 𝑋 ≤ 7
f) 𝑃 10 < 𝑋 < 17
#MEIConf2019
Binomial𝑋~𝐵 47,0.2
Find
a) 𝑃 𝑋 = 11
b) 𝑃 𝑋 ≤ 11
c) 𝑃 𝑋 > 11
d) 𝑃 𝑋 < 8
e) 𝑃 3 ≤ 𝑋 ≤ 7
f) 𝑃 10 < 𝑋 < 17
(a) 0.116
(b) 0.333
(c) 0.217
(d) 0.251
(e) 0.248
(f) 0.326
#MEIConf2019
Calculating ProbabilitiesNormal
#MEIConf2019
Normal𝑌~𝑁 12,25
a) 𝑃(𝑋 ≥ 14.7)
b) 𝑃(7.5 < 𝑋 < 12.3)
c) 𝑃(𝑋 = 13)
d) Find an interval, symmetrical about the mean,
within which Y lies with 95% probability.
#MEIConf2019
Normal𝑌~𝑁 12,25
a) 𝑃(𝑋 ≥ 14.7)
b) 𝑃(7.5 < 𝑋 < 12.3)
c) 𝑃(𝑋 = 13)
d) Find an interval, symmetrical about the mean,
within which Y lies with 95% probability.
(a)0.309
(b)0.340
(c)0
(d)2.2 to 21.8
#MEIConf2019
Hypothesis Testing - BinomialThere is no particular function for this, it is just a
case of using the functions in the dist menu.
#MEIConf2019
Hypothesis Testing - BinomialThere is no particular function for this, it is just a
case of using the functions in the dist menu.
4. It is known that under a treatment for a disease, 10% of patients with
the disease experience side effects within a year.
In a trial of a new treatment, 500 patients with this disease were
randomly selected and the number that experienced side effects within
a year was noted. It was found that 61 of the 500 patients experienced
side effects within one year
Test at the 5% significance level, whether the proportion of patients
experiencing side effects within one year is greater under the new
treatment than the original treatment.
#MEIConf2019
Hypothesis Testing - Binomial10% of patients with the disease experience side effects within a year.
In a trial of a new treatment, 500 patients with this disease were
randomly selected and the number that experienced side effects within
a year was noted. It was found that 61 of the 500 patients experienced
side effects within one year
Test at the 5% significance level, whether the proportion of patients
experiencing side effects within one year is greater under the new
treatment than the original treatment.
#MEIConf2019
Hypothesis Testing - Binomial10% of patients with the disease experience side effects within a year.
In a trial of a new treatment, 500 patients with this disease were
randomly selected and the number that experienced side effects within
a year was noted. It was found that 61 of the 500 patients experienced
side effects within one year
Test at the 5% significance level, whether the proportion of patients
experiencing side effects within one year is greater under the new
treatment than the original treatment.
𝐻0: 𝑝 = 0.1
𝐻1: 𝑝 > 0.1 Test at 5% level
#MEIConf2019
Hypothesis Testing - Binomial10% of patients with the disease experience side effects within a year.
In a trial of a new treatment, 500 patients with this disease were
randomly selected and the number that experienced side effects within
a year was noted. It was found that 61 of the 500 patients experienced
side effects within one year
Test at the 5% significance level, whether the proportion of patients
experiencing side effects within one year is greater under the new
treatment than the original treatment.
𝐻0: 𝑝 = 0.1
𝐻1: 𝑝 > 0.1 Test at 5% level
Under 𝐻0 𝑋 ~𝐵(500,0.1) where X is the number of patients with side
effects.
#MEIConf2019
Hypothesis Testing - Binomial10% of patients with the disease experience side effects within a year.
In a trial of a new treatment, 500 patients with this disease were
randomly selected and the number that experienced side effects within
a year was noted. It was found that 61 of the 500 patients experienced
side effects within one year
Test at the 5% significance level, whether the proportion of patients
experiencing side effects within one year is greater under the new
treatment than the original treatment.
𝐻0: 𝑝 = 0.1
𝐻1: 𝑝 > 0.1 Test at 5% level
Under 𝐻0 𝑋 ~𝐵(500,0.1) where X is the number of patients with side
effects.
We require 𝑃(𝑋 ≥ 61)
#MEIConf2019
Hypothesis Testing - Binomial
𝐻0: 𝑝 = 0.1
𝐻1: 𝑝 > 0.1 Test at 5% level
Under 𝐻0 𝑋 ~𝐵(500,0.1) where X is the number of patients with side
effects.
We require 𝑃(𝑋 ≥ 61)
#MEIConf2019
Hypothesis Testing - Binomial
𝐻0: 𝑝 = 0.1
𝐻1: 𝑝 > 0.1 Test at 5% level
Under 𝐻0 𝑋 ~𝐵(500,0.1) where X is the number of patients with side
effects.
We require 𝑃(𝑋 ≥ 61)
#MEIConf2019
Hypothesis Testing - Binomial
𝐻0: 𝑝 = 0.1
𝐻1: 𝑝 > 0.1 Test at 5% level
Under 𝐻0 𝑋 ~𝐵(500,0.1) where X is the number of patients with side
effects.
𝑃 𝑋 ≥ 61 = 0.0618
#MEIConf2019
Hypothesis Testing - Binomial
𝐻0: 𝑝 = 0.1
𝐻1: 𝑝 > 0.1 Test at 5% level
Under 𝐻0 𝑋 ~𝐵(500,0.1) where X is the number of patients with side
effects.
𝑃 𝑋 ≥ 61 = 0.0618
As 0.0618 > 0.05, we do not reject 𝐻0.
#MEIConf2019
Hypothesis Testing - Binomial
𝐻0: 𝑝 = 0.1
𝐻1: 𝑝 > 0.1 Test at 5% level
Under 𝐻0 𝑋 ~𝐵(500,0.1) where X is the number of patients with side
effects.
𝑃 𝑋 ≥ 61 = 0.0618
As 0.0618 > 0.05, we do not reject 𝐻0, concluding that there is
insufficient evidence at the 5% level that the proportion of patients
experiencing side effects within one year is greater under the new
treatment than the original treatment.
#MEIConf2019
Hypothesis Testing - Binomial5. Neil thinks that a particular 4-sided dice is
biased in favour of 1. He plans to throw the dice
100 times and note the number of times that it
shows a 1. He will then carry out a test at the 2%
significance level.
Find the critical region for the test.
#MEIConf2019
Hypothesis Testing - Binomial5. Neil thinks that a particular 4-sided dice is biased in favour of 1. He
plans to throw the dice 100 times and note the number of times that it
shows a 1. He will then carry out a test at the 2% significance level.
Find the critical region for the test.
𝐻0: 𝑝 = 0.25
𝐻1: 𝑝 > 0.25 Test at 2% level
#MEIConf2019
Hypothesis Testing - Binomial5. Neil thinks that a particular 4-sided dice is biased in favour of 1. He
plans to throw the dice 100 times and note the number of times that it
shows a 1. He will then carry out a test at the 2% significance level.
Find the critical region for the test.
𝐻0: 𝑝 = 0.25
𝐻1: 𝑝 > 0.25 Test at 2% level
Under 𝐻0, the number of ones 𝑋~𝐵(100,0.25)
#MEIConf2019
Hypothesis Testing - Binomial5. Neil thinks that a particular 4-sided dice is biased in favour of 1. He
plans to throw the dice 100 times and note the number of times that it
shows a 1. He will then carry out a test at the 2% significance level.
Find the critical region for the test.
𝐻0: 𝑝 = 0.25
𝐻1: 𝑝 > 0.25 Test at 2% level
Under 𝐻0, the number of ones 𝑋~𝐵(100,0.25)
We require the least value 𝑐 such that 𝑃(𝑋 ≥ 𝑐) ≤ 0.02
#MEIConf2019
Hypothesis Testing - Binomial5. Neil thinks that a particular 4-sided dice is biased in favour of 1. He
plans to throw the dice 100 times and note the number of times that it
shows a 1. He will then carry out a test at the 2% significance level.
Find the critical region for the test.
𝐻0: 𝑝 = 0.25
𝐻1: 𝑝 > 0.25 Test at 2% level
Under 𝐻0, the number of ones 𝑋~𝐵(100,0.25)
We require the least value 𝑐 such that 𝑃(𝑋 ≥ 𝑐) ≤ 0.02
How do we find such a value of c?
#MEIConf2019
Hypothesis Testing - Binomial5. Neil thinks that a particular 4-sided dice is biased in favour of 1. He
plans to throw the dice 100 times and note the number of times that it
shows a 1. He will then carry out a test at the 2% significance level.
Find the critical region for the test.
𝐻0: 𝑝 = 0.25
𝐻1: 𝑝 > 0.25 Test at 2% level
Under 𝐻0, the number of ones 𝑋~𝐵(100,0.25)
We require the least value 𝑐 such that 𝑃(𝑋 ≥ 𝑐) ≤ 0.02
#MEIConf2019
Hypothesis Testing - Binomial5. Neil thinks that a particular 4-sided dice is biased in favour of 1. He
plans to throw the dice 100 times and note the number of times that it
shows a 1. He will then carry out a test at the 2% significance level.
Find the critical region for the test.
𝐻0: 𝑝 = 0.25
𝐻1: 𝑝 > 0.25 Test at 2% level
Under 𝐻0, the number of ones 𝑋~𝐵(100,0.25)
We require the least value 𝑐 such that 𝑃(𝑋 ≥ 𝑐) ≤ 0.02
#MEIConf2019
Hypothesis Testing - Binomial5. Neil thinks that a particular 4-sided dice is biased in favour of 1. He
plans to throw the dice 100 times and note the number of times that it
shows a 1. He will then carry out a test at the 2% significance level.
Find the critical region for the test.
𝐻0: 𝑝 = 0.25
𝐻1: 𝑝 > 0.25 Test at 2% level
Under 𝐻0, the number of ones 𝑋~𝐵(100,0.25)
We require the least value 𝑐 such that 𝑃(𝑋 ≥ 𝑐) ≤ 0.02
What’s going on?
#MEIConf2019
Hypothesis Testing - Binomial5. Neil thinks that a particular 4-sided dice is biased in favour of 1. He
plans to throw the dice 100 times and note the number of times that it
shows a 1. He will then carry out a test at the 2% significance level.
Find the critical region for the test.
𝐻0: 𝑝 = 0.25
𝐻1: 𝑝 > 0.25 Test at 2% level
Under 𝐻0, the number of ones 𝑋~𝐵(100,0.25)
We require the least value 𝑐 such that 𝑃(𝑋 ≥ 𝑐) ≤ 0.02
What’s going on?
#MEIConf2019
Hypothesis Testing - Binomial5. Neil thinks that a particular 4-sided dice is biased in favour of 1. He
plans to throw the dice 100 times and note the number of times that it
shows a 1. He will then carry out a test at the 2% significance level.
Find the critical region for the test.
𝐻0: 𝑝 = 0.25
𝐻1: 𝑝 > 0.25 Test at 2% level
Under 𝐻0, the number of ones 𝑋~𝐵(100,0.25)
We require the least value 𝑐 such that 𝑃(𝑋 ≥ 𝑐) ≤ 0.02
What’s going on?
#MEIConf2019
Hypothesis Testing - Binomial5. Neil thinks that a particular 4-sided dice is biased in favour of 1. He
plans to throw the dice 100 times and note the number of times that it
shows a 1. He will then carry out a test at the 2% significance level.
Find the critical region for the test.
𝐻0: 𝑝 = 0.25
𝐻1: 𝑝 > 0.25 Test at 2% level
Under 𝐻0, the number of ones 𝑋~𝐵(100,0.25)
We require the least value 𝑐 such that 𝑃(𝑋 ≥ 𝑐) ≤ 0.02
Reject 𝐻0 if 𝑋 ≥ 35.
#MEIConf2019
Hypothesis Testing - Binomial6. Past records show that the probability that a particular seed
germinates is 0.45. A new method of storing seeds is to be trialled
and Susan wishes to know if this changes the proportion that
germinate. She intends to take a sample of 30 seeds.
(a) State the hypotheses that Susan should use
(b) Find the critical region for a test of the hypothesis that the proportion
is unchanged at the 5% level.
(c) State the actual (true) significance level for this test.
.
#MEIConf2019
Hypothesis Testing - Binomial6. Past records show that the probability that a particular seed
germinates is 0.45. A new method of storing seeds is to be trialled
and Susan wishes to know if this changes the proportion that
germinate. She intends to take a sample of 30 seeds.
(a) State the hypotheses that Susan should use
𝐻0: 𝑝 = 0.45
𝐻1: 𝑝 ≠ 0.45 Test at 5% level – 2 tailed test
Under 𝐻0, the number that germinate 𝑋~𝐵(30,0.45)
(b) Find the critical region for a test of the hypothesis that the proportion
is unchanged at the 5% level.
.
#MEIConf2019
Hypothesis Testing - Binomial6. Past records show that the probability that a particular seed
germinates is 0.45. A new method of storing seeds is to be trialled
and Susan wishes to know if this changes the proportion that
germinate. She intends to take a sample of 30 seeds.
(a) State the hypotheses that Susan should use
𝐻0: 𝑝 = 0.45
𝐻1: 𝑝 ≠ 0.45 Test at 5% level – 2 tailed test
Under 𝐻0, the number that germinate 𝑋~𝐵(30,0.45)
(b) Find the critical region for a test of the hypothesis that the proportion
is unchanged at the 5% level.
.
#MEIConf2019
Hypothesis Testing - Binomial6. Past records show that the probability that a particular seed
germinates is 0.45. A new method of storing seeds is to be trialled
and Susan wishes to know if this changes the proportion that
germinate. She intends to take a sample of 30 seeds.
(a) State the hypotheses that Susan should use
𝐻0: 𝑝 = 0.45
𝐻1: 𝑝 ≠ 0.45 Test at 5% level – 2 tailed test
Under 𝐻0, the number that germinate 𝑋~𝐵(30,0.45)
(b) Find the critical region for a test of the hypothesis that the proportion
is unchanged at the 5% level.
.
#MEIConf2019
Hypothesis Testing - Binomial.
#MEIConf2019
Hypothesis Testing - Binomial.
#MEIConf2019
Hypothesis Testing - Binomial
.(b) Find the critical region for a test of the hypothesis that the
proportion is unchanged at the 5% level.
Reject 𝐻0 if 𝑋 ≤ 7 or 𝑋 ≥ 20
(c) State the actual (true) significance level for this test.
0.0121 + 0.0138 = 0.0259
.
#MEIConf2019
Hypothesis Testing - Normal
.
#MEIConf2019
Hypothesis Testing - Normal7. The masses of adult students are known to be Normally distributed
with mean 67.4kg and standard deviation 3.8kg.
A sample of size 24 is taken and the mean found to be 65.8kg.
Assuming that the standard deviation is unchanged test, at the 1% level
of significance, whether the mean mass of adult students has
decreased.
.
#MEIConf2019
Hypothesis Testing - Normal7. The masses of adult students are known to be Normally distributed
with mean 67.4kg and standard deviation 3.8kg.
A sample of size 24 is taken and the mean found to be 65.8kg.
Assuming that the standard deviation is unchanged test, at the 1% level
of significance, whether the mean mass of adult students has
decreased.
𝐻0: 𝜇 = 67.4
𝐻1: μ < 67.4 Test at 1% level.
#MEIConf2019
Hypothesis Testing - Normal7. The masses of adult students are known to be Normally distributed
with mean 67.4kg and standard deviation 3.8kg.
A sample of size 24 is taken and the mean found to be 65.8kg.
Assuming that the standard deviation is unchanged test, at the 1% level
of significance, whether the mean mass of adult students has
decreased.
𝐻0: 𝜇 = 67.4
𝐻1: μ < 67.4 Test at 1% level
ҧ𝑥 = 65.8 𝑛 = 24 𝜎 = 3.8
.
#MEIConf2019
Hypothesis Testing - Normal7. The masses of adult students are known to be Normally distributed
with mean 67.4kg and standard deviation 3.8kg.
A sample of size 24 is taken and the mean found to be 65.8kg.
Assuming that the standard deviation is unchanged test, at the 1% level
of significance, whether the mean mass of adult students has
decreased.
𝐻0: 𝜇 = 67.4
𝐻1: μ < 67.4 Test at 1% level
ҧ𝑥 = 65.8 𝑛 = 24 𝜎 = 3.8
.
#MEIConf2019
Hypothesis Testing - Normal7. The masses of adult students are known to be Normally distributed
with mean 67.4kg and standard deviation 3.8kg.
A sample of size 24 is taken and the mean found to be 65.8kg.
Assuming that the standard deviation is unchanged test, at the 1% level
of significance, whether the mean mass of adult students has
decreased.
𝐻0: 𝜇 = 67.4
𝐻1: μ < 67.4 Test at 1% level
ҧ𝑥 = 65.8 𝑛 = 24 𝜎 = 3.8
.
#MEIConf2019
Hypothesis Testing - Normal7. The masses of adult students are known to be Normally distributed
with mean 67.4kg and standard deviation 3.8kg.
A sample of size 24 is taken and the mean found to be 65.8kg.
Assuming that the standard deviation is unchanged test, at the 1% level
of significance, whether the mean mass of adult students has
decreased.
𝐻0: 𝜇 = 67.4
𝐻1: μ < 67.4 Test at 1% level
ҧ𝑥 = 65.8 𝑛 = 24 𝜎 = 3.8
𝑃 ത𝑋 < 65.8 = 0.0196
.
#MEIConf2019
Hypothesis Testing - Normal7. The masses of adult students are known to be Normally distributed
with mean 67.4kg and standard deviation 3.8kg.
A sample of size 24 is taken and the mean found to be 65.8kg.
Assuming that the standard deviation is unchanged test, at the 1% level
of significance, whether the mean mass of adult students has
decreased.
𝐻0: 𝜇 = 67.4
𝐻1: μ < 67.4 Test at 1% level
ҧ𝑥 = 65.8 𝑛 = 24 𝜎 = 3.8
𝑃 ത𝑋 < 65.8 = 0.0196
0.0196 > 0.01 so we do not reject 𝐻0, concluding that there is insufficient
evidence at the 1% level that the mean mass has decreased.
.
#MEIConf2019
Hypothesis Testing - Normal8.The masses of ball bearings produced at a certain factory are thought
to be normally distributed with a mean of 5g and a standard deviation
2g. The quality control inspector takes a sample of 100 to check whether
the mean has changed (she assumes that the standard deviation is
unchanged). Test at the 5% the hypothesis that the mean has changed
if her sample has a mean of 4.5g.
.
#MEIConf2019
Hypothesis Testing - Normal8.The masses of ball bearings produced at a certain factory are thought
to be normally distributed with a mean of 5g and a standard deviation
2g. The quality control inspector takes a sample of 100 to check whether
the mean has changed (she assumes that the standard deviation is
unchanged). Test at the 5% the hypothesis that the mean has changed
if her sample has a mean of 4.5g.
𝐻0: 𝜇 = 5
𝐻1: μ ≠ 5 Test at 5% level
ҧ𝑥 = 4.5 𝑛 = 100 𝜎 = 2
.
#MEIConf2019
Hypothesis Testing - Normal8.The masses of ball bearings produced at a certain factory are thought
to be normally distributed with a mean of 5g and a standard deviation
2g. The quality control inspector takes a sample of 100 to check whether
the mean has changed (she assumes that the standard deviation is
unchanged). Test at the 5% the hypothesis that the mean has changed
if her sample has a mean of 4.5g.
𝐻0: 𝜇 = 5
𝐻1: μ ≠ 5 Test at 5% level
ҧ𝑥 = 4.5 𝑛 = 100 𝜎 = 2
.
#MEIConf2019
Hypothesis Testing - Normal8.The masses of ball bearings produced at a certain factory are thought
to be normally distributed with a mean of 5g and a standard deviation
2g. The quality control inspector takes a sample of 100 to check whether
the mean has changed (she assumes that the standard deviation is
unchanged). Test at the 5% the hypothesis that the mean has changed
if her sample has a mean of 4.5g.
𝐻0: 𝜇 = 5
𝐻1: μ ≠ 5 Test at 5% level
ҧ𝑥 = 4.5 𝑛 = 100 𝜎 = 2
𝑃 ത𝑋 < 4.5 = 0.0062
.
#MEIConf2019
Hypothesis Testing - Normal8.The masses of ball bearings produced at a certain factory are thought
to be normally distributed with a mean of 5g and a standard deviation
2g. The quality control inspector takes a sample of 100 to check whether
the mean has changed (she assumes that the standard deviation is
unchanged). Test at the 5% the hypothesis that the mean has changed
if her sample has a mean of 4.5g.
𝐻0: 𝜇 = 5
𝐻1: μ ≠ 5 Test at 5% level
ҧ𝑥 = 4.5 𝑛 = 100 𝜎 = 2
As 𝑃 ത𝑋 < 4.5 = 0.0062 we reject 𝐻0 concluding there is evidence at the
5% that the mean mass of then ball bearings has changed.
.
#MEIConf2019
Hypothesis Testing - Normal9.Last year the time spent online, in minutes, by students at a certain
university had a mean 132.5 and standard deviation 28.2.
This year is it is suspected that the mean time has increased. A sample
of size 50 is to taken.
Assuming that the times are normally distributed and that the standard
deviation remains unchanged, find the critical region for a test at the 5%
significance level that the mean time spent by users online has
increased..
#MEIConf2019
Hypothesis Testing - Normal9.Last year the time spent online, in minutes, by students at a certain
university had a mean 132.5 and standard deviation 28.2.
This year is it is suspected that the mean time has increased. A sample
of size 50 is to taken.
Assuming that the times are normally distributed and that the standard
deviation remains unchanged, find the critical region for a test at the 5%
significance level that the mean time spent by users online has
increased.
𝐻0: 𝜇 = 132.5
𝐻1: μ > 132.5 Test at 5% level
𝑛 = 50 𝜎 = 28.2
How should we approach this?
.
#MEIConf2019
Hypothesis Testing - Normal9.Last year the time spent online, in minutes, by students at a certain
university had a mean 132.5 and standard deviation 28.2.
This year is it is suspected that the mean time has increased. A sample
of size 50 is to taken.
Assuming that the times are normally distributed and that the standard
deviation remains unchanged, find the critical region for a test at the 5%
significance level that the mean time spent by users online has
increased.
𝐻0: 𝜇 = 132.5
𝐻1: μ > 132.5 Test at 5% level
𝑛 = 50 𝜎 = 28.2
How should we approach this?
.
#MEIConf2019
Hypothesis Testing - Normal9.Last year the time spent online, in minutes, by students at a certain
university had a mean 132.5 and standard deviation 28.2.
This year is it is suspected that the mean time has increased. A sample
of size 50 is to taken.
Assuming that the times are normally distributed and that the standard
deviation remains unchanged, find the critical region for a test at the 5%
significance level that the mean time spent by users online has
increased.
𝐻0: 𝜇 = 132.5
𝐻1: μ > 132.5 Test at 5% level
𝑛 = 50 𝜎 = 28.2
How should we approach this?
.
#MEIConf2019
Hypothesis Testing - Normal9.Last year the time spent online, in minutes, by students at a certain
university had a mean 132.5 and standard deviation 28.2.
This year is it is suspected that the mean time has increased. A sample
of size 50 is to taken.
Assuming that the times are normally distributed and that the standard
deviation remains unchanged, find the critical region for a test at the 5%
significance level that the mean time spent by users online has
increased.
𝐻0: 𝜇 = 132.5
𝐻1: μ > 132.5 Test at 5% level
𝑛 = 50 𝜎 = 28.2
Reject 𝐻0 if ത𝑋 > 139.06
.
#MEIConf2019
Hypothesis Testing - Poisson
.
#MEIConf2019
Hypothesis Testing - Poisson10. An estate agent is concerned that the housing market appears to
be slowing down. His records in recent years show that, on average, he
sells 12 houses during the month of June each year. In June this year,
he sells only 6 houses. Does this provide evidence, at the 5% level,
that the mean number of houses he sells during the month of June has
decreased?
.
#MEIConf2019
Hypothesis Testing - Poisson10. An estate agent is concerned that the housing market appears to
be slowing down. His records in recent years show that, on average, he
sells 12 houses during the month of June each year. In June this year,
he sells only 6 houses. Does this provide evidence, at the 5% level,
that the mean number of houses he sells during the month of June has
decreased?
𝐻0: 𝜆 = 12𝐻1: 𝜆 < 12
Under 𝐻0, 𝑋~𝑃𝑜 12 .
Where X is the number of houses sold in June.
We require 𝑃 𝑋 ≤ 6 . If this is greater than 5% will assume 𝐻0 is
correct. If it is equal to or less than 5%, we reject 𝐻0.
.
#MEIConf2019
Hypothesis Testing - Poisson10. An estate agent is concerned that the housing market appears to
be slowing down. His records in recent years show that, on average, he
sells 12 houses during the month of June each year. In June this year,
he sells only 6 houses. Does this provide evidence, at the 5% level,
that the mean number of houses he sells during the month of June has
decreased?
𝐻0: 𝜆 = 12𝐻1: 𝜆 < 12
Under 𝐻0, 𝑋~𝑃𝑜 12 .
Where X is the number of houses sold in June.
We require 𝑃 𝑋 ≤ 6 . If this is greater than 5% will assume 𝐻0 is
correct. If it is equal to or less than 5%, we reject 𝐻0.
.
#MEIConf2019
Hypothesis Testing - Poisson10. An estate agent is concerned that the housing market appears to
be slowing down. His records in recent years show that, on average, he
sells 12 houses during the month of June each year. In June this year,
he sells only 6 houses. Does this provide evidence, at the 5% level,
that the mean number of houses he sells during the month of June has
decreased?
𝐻0: 𝜆 = 12𝐻1: 𝜆 < 12
Under 𝐻0, 𝑋~𝑃𝑜 12 .
Where X is the number of houses sold in June.
As 𝑃 𝑋 ≤ 6 < 0.05, we reject 𝐻0 concluding that there is evidence at
the 5% to suggest that the mean number of houses he sells during the
month of June has decreased
.
#MEIConf2019
Goodness of fit11.Elise rolls a six-sided dice until she gets a six. She repeats this experiment 100 times.
The number of rolls needed to get a six each time is recorded in the table below:
Elise believes that the random variable ‘the number of throws needed to get a six’ follows
geometric distribution with parameter 0.4.
Test at the 5% level whether Elise’s belief is correct, stating your hypotheses clearly
Number of
throws
Frequency
1 2 3 4 5 6 7
32 25 22 14 3 3 1
#MEIConf2019
Goodness of fit11.Elise rolls a six-sided dice until she gets a six. She repeats this experiment 100 times.
The number of rolls needed to get a six each time is recorded in the table below:
Elise believes that the random variable ‘the number of throws needed to get a six’ follows
geometric distribution with parameter 0.4.
Test at the 5% level whether Elise’s belief is correct, stating your hypotheses clearly
Number of
throws
Frequency
1 2 3 4 5 6 7
32 25 22 14 3 3 1
#MEIConf2019
Goodness of fit11.Elise rolls a six-sided dice until she gets a six. She repeats this experiment 100 times.
The number of rolls needed to get a six each time is recorded in the table below:
Elise believes that the random variable ‘the number of throws needed to get a six’ follows
geometric distribution with parameter 0.4.
Test at the 5% level whether Elise’s belief is correct, stating your hypotheses clearly
Number of
throws
Frequency
1 2 3 4 5 6 7
32 25 22 14 3 3 1
#MEIConf2019
Goodness of fit11.Elise rolls a six-sided dice until she gets a six. She repeats this experiment 100 times.
The number of rolls needed to get a six each time is recorded in the table below:
Elise believes that the random variable ‘the number of throws needed to get a six’ follows
geometric distribution with parameter 0.4.
Test at the 5% level whether Elise’s belief is correct, stating your hypotheses clearly
Number of
throws
Frequency
1 2 3 4 5 6 7
32 25 22 14 3 3 1
#MEIConf2019
Goodness of fit11.Elise rolls a six-sided dice until she gets a six. She repeats this experiment 100 times.
The number of rolls needed to get a six each time is recorded in the table below:
Elise believes that the random variable ‘the number of throws needed to get a six’ follows
geometric distribution with parameter 0.4.
Test at the 5% level whether Elise’s belief is correct, stating your hypotheses clearly
What about 7?
Number of
throws
Frequency
1 2 3 4 5 6 7
32 25 22 14 3 3 1
#MEIConf2019
Goodness of fit11.Elise rolls a six-sided dice until she gets a six. She repeats this experiment 100 times.
The number of rolls needed to get a six each time is recorded in the table below:
Elise believes that the random variable ‘the number of throws needed to get a six’ follows
geometric distribution with parameter 0.4.
Test at the 5% level whether Elise’s belief is correct, stating your hypotheses clearly
What about 7?
Number of
throws
Frequency
1 2 3 4 5 6 7
32 25 22 14 3 3 1
#MEIConf2019
Goodness of fit: Number of throws
Observed
1 2 3 4 5 ≥6
32 25 22 14 3 4
Expected 40 24 14.4 8.64 5.18 7.78
#MEIConf2019
Goodness of fit: Number of throws
Observed
1 2 3 4 5 ≥6
32 25 22 14 3 4
Expected 40 24 14.4 8.64 5.18 7.78
#MEIConf2019
Goodness of fit
#MEIConf2019
Goodness of fit
#MEIConf2019
Goodness of fitElise believes that the random variable ‘the number of throws needed
to get a six’ follows geometric distribution with parameter 0.4.
Test at the 5% level whether Elise’s belief is correct, stating your
hypotheses clearly.
𝐻0: the data follows a geometric distribution with parameter 0.4.
𝐻1: the data does not follow a geometric distribution with parameter 0.4.
Under 𝐻0 the number of throws needed to get a six 𝑋~𝐺𝑒𝑜(0.4)
The number of degrees of freedom is 6 – 1 = 5. χ2 = 11.73 from calculator.
𝑃 χ2 > 11.73 = 0.0387 from calculator.
So reject 𝐻0, concluding that the Geometric with mean 0.4 is not a good fit for this
data.
#MEIConf2019
Difference in meansത𝑋1 − ത𝑋2 ~ N 𝜇1 − 𝜇2,
𝜎2
𝑛1+𝜎2
𝑛2= N 𝜇1 − 𝜇2, 𝜎
21
𝑛1+
1
𝑛2
ത𝑋1 − ത𝑋2 ~ 𝑡𝑛1+𝑛2−2 𝜇1 − 𝜇2, 𝑠2
1
𝑛1+
1
𝑛2
• There are two estimates of the common variance based on the two samples. These are
𝑠12 =
σ 𝑥1− ҧ𝑥12
𝑛1−1and 𝑠2
2 =σ 𝑥2− ҧ𝑥2
2
𝑛2−1
• These estimates are combined as a weighted average:
𝑠2 =𝑛1 − 1 𝑠1
2 + 𝑛2 − 1 𝑠22
𝑛1 − 1 + 𝑛2 − 1
• And this estimate has n1 + n2 – 2 degrees of freedom
#MEIConf2019
Difference in meansTwo different designs for a large open plan office are being compared
in respect of the amount of light available in the locations where
employees will be working. The amount of light is measured by
photoelectric cells at 12 randomly selected locations for one design
and, independently, at 10 randomly selected locations for the other
design. The data, in a standard unit of light, are summarised below:
Design 1: 𝑛 = 12 ҧ𝑥 = 9.85 𝑠𝑥 = 1.41
Design 2: 𝑛 = 10 ത𝑦 = 8.76 𝑠𝑦= 1.33
Test at the 5% level whether the mean amount of light delivered is the
same in the two designs.
#MEIConf2019
Difference in meansDesign 1: 𝑛 = 12 ҧ𝑥 = 9.85 𝑠𝑥 = 1.41
Design 2: 𝑛 = 10 ത𝑦 = 8.76 𝑠𝑦= 1.33
Test at the 5% level whether the mean amount of light delivered is the
same in the two designs.
#MEIConf2019
Difference in meansDesign 1: 𝑛 = 12 ҧ𝑥 = 9.85 𝑠𝑥 = 1.41
Design 2: 𝑛 = 10 ത𝑦 = 8.76 𝑠𝑦= 1.33
Test at the 5% level whether the mean amount of light delivered is the
same in the two designs.
#MEIConf2019
Difference in meansDesign 1: 𝑛 = 12 ҧ𝑥 = 9.85 𝑠𝑥 = 1.41
Design 2: 𝑛 = 10 ത𝑦 = 8.76 𝑠𝑦= 1.33
Test at the 5% level whether the mean amount of light delivered is the
same in the two designs.
#MEIConf2019
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