MEI PowerPoint Template · 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

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USING CASIO CALCULATORS

FOR STATISTICAL TESTS

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The Statistics menu

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The Statistics menu

Calculating

summary

statistics

from a set

of data

Calculating probabilities

from distributions.Conducting

hypothesis

tests.

Calculating

confidence

intervals.

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Calculating probabilitiesBinomial

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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

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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

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Calculating ProbabilitiesNormal

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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.

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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

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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.

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Hypothesis Testing - Binomial10% of patients with the disease experience side effects within a year.








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𝐻0: 𝑝 = 0.1

𝐻1: 𝑝 > 0.1 Test at 5% level

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𝐻0: 𝑝 = 0.1


Under 𝐻0 𝑋 ~𝐵(500,0.1) where X is the number of patients with side

effects.

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𝐻0: 𝑝 = 0.1



effects.

We require 𝑃(𝑋 ≥ 61)

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Hypothesis Testing - Binomial

𝐻0: 𝑝 = 0.1



effects.


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𝐻0: 𝑝 = 0.1



effects.


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𝐻0: 𝑝 = 0.1



effects.

𝑃 𝑋 ≥ 61 = 0.0618

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𝐻0: 𝑝 = 0.1



effects.

𝑃 𝑋 ≥ 61 = 0.0618

As 0.0618 > 0.05, we do not reject 𝐻0.

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𝐻0: 𝑝 = 0.1



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



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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.

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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.


𝐻0: 𝑝 = 0.25


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𝐻0: 𝑝 = 0.25


Under 𝐻0, the number of ones 𝑋~𝐵(100,0.25)

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𝐻0: 𝑝 = 0.25



We require the least value 𝑐 such that 𝑃(𝑋 ≥ 𝑐) ≤ 0.02

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𝐻0: 𝑝 = 0.25




How do we find such a value of c?

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𝐻0: 𝑝 = 0.25




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𝐻0: 𝑝 = 0.25




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𝐻0: 𝑝 = 0.25




What’s going on?

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𝐻0: 𝑝 = 0.25




What’s going on?

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𝐻0: 𝑝 = 0.25




What’s going on?

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𝐻0: 𝑝 = 0.25




Reject 𝐻0 if 𝑋 ≥ 35.

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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.

.

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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)



.

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𝐻0: 𝑝 = 0.45





.

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𝐻0: 𝑝 = 0.45





.

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Hypothesis Testing - Binomial.

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Hypothesis Testing - Binomial.

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.(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

.

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Hypothesis Testing - Normal

.

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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.

.

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decreased.

𝐻0: 𝜇 = 67.4

𝐻1: μ < 67.4 Test at 1% level.

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decreased.

𝐻0: 𝜇 = 67.4

𝐻1: μ < 67.4 Test at 1% level

ҧ𝑥 = 65.8 𝑛 = 24 𝜎 = 3.8

.

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decreased.

𝐻0: 𝜇 = 67.4


ҧ𝑥 = 65.8 𝑛 = 24 𝜎 = 3.8

.

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decreased.

𝐻0: 𝜇 = 67.4


ҧ𝑥 = 65.8 𝑛 = 24 𝜎 = 3.8

.

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decreased.

𝐻0: 𝜇 = 67.4


ҧ𝑥 = 65.8 𝑛 = 24 𝜎 = 3.8

𝑃 ത𝑋 < 65.8 = 0.0196

.

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decreased.

𝐻0: 𝜇 = 67.4


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.

.

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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.

.

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𝐻0: 𝜇 = 5

𝐻1: μ ≠ 5 Test at 5% level

ҧ𝑥 = 4.5 𝑛 = 100 𝜎 = 2

.

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𝐻0: 𝜇 = 5


ҧ𝑥 = 4.5 𝑛 = 100 𝜎 = 2

.

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𝐻0: 𝜇 = 5


ҧ𝑥 = 4.5 𝑛 = 100 𝜎 = 2

𝑃 ത𝑋 < 4.5 = 0.0062

.

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𝐻0: 𝜇 = 5


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.

.

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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..

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increased.

𝐻0: 𝜇 = 132.5

𝐻1: μ > 132.5 Test at 5% level

𝑛 = 50 𝜎 = 28.2

How should we approach this?

.

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increased.

𝐻0: 𝜇 = 132.5


𝑛 = 50 𝜎 = 28.2


.

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increased.

𝐻0: 𝜇 = 132.5


𝑛 = 50 𝜎 = 28.2


.

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increased.

𝐻0: 𝜇 = 132.5


𝑛 = 50 𝜎 = 28.2

Reject 𝐻0 if ത𝑋 > 139.06

.

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Hypothesis Testing - Poisson

.

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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?

.

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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.

.

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decreased?

𝐻0: 𝜆 = 12𝐻1: 𝜆 < 12



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.

.

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decreased?

𝐻0: 𝜆 = 12𝐻1: 𝜆 < 12



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

.

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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

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Number of

throws

Frequency

1 2 3 4 5 6 7

32 25 22 14 3 3 1

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Number of

throws

Frequency

1 2 3 4 5 6 7

32 25 22 14 3 3 1

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Number of

throws

Frequency

1 2 3 4 5 6 7

32 25 22 14 3 3 1

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What about 7?

Number of

throws

Frequency

1 2 3 4 5 6 7

32 25 22 14 3 3 1

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What about 7?

Number of

throws

Frequency

1 2 3 4 5 6 7

32 25 22 14 3 3 1

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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

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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

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Goodness of fit

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Goodness of fit

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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.

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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

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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.

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Difference in meansDesign 1: 𝑛 = 12 ҧ𝑥 = 9.85 𝑠𝑥 = 1.41

Design 2: 𝑛 = 10 ത𝑦 = 8.76 𝑠𝑦= 1.33



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Design 2: 𝑛 = 10 ത𝑦 = 8.76 𝑠𝑦= 1.33



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Design 2: 𝑛 = 10 ത𝑦 = 8.76 𝑠𝑦= 1.33



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MEI PowerPoint Template · 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

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