Interpreting Opinion Polls Example1: (Confidence Interval for the population proportion): Suppose...

Interpreting Opinion Polls

Example1: (Confidence Interval for the population proportion): Suppose that

the result of sampling yields the following:

p= 0.4 ; n = 36. Use this information to construct a

98% CI for .

Recall from, last week

98% CI for is [0.21 0.59]

Interpreting Opinion Polls

Labour 45 pointsConservatives 39 pointsLib. Dems 15 pointsOther 1 point

Margin of Error :Typically, this means that population proportion of labour voters is between 42% and 48% (with 95% confidence)

Hypothesis Testing

x

Two competing Hypotheses

The Null Hypothesis (H0):The population mean ( =20The Alternative Hypothesis (H1):The population mean ( =25

x

H0: =20 H1: = 25

x

H0: =20 H1: = 25

BenchmarkValue

p-value

BenchmarkValueType 2 Error:

Not rejectingA false H0

Type 1 Error:Rejecting a correct H0

Area in black : Maximum Type2 errorArea in red : Maximum Type1 error

BenchmarkValue

MaximumType 1 ErrorMaximum

Type 2 Error:

BenchmarkValue


Type 2 Error

The Spirit of Hypothesis Testing

H0 is the hypothesis that is challenged

H0 is the established hypothesis

Xo

y

Can we have intersecting indifference curves?

a

a ~ b, a ~ c; so b ~ca contradiction of the axiom more is better

bc

H0 : Yes H1 : No

We work under the assumption that H0 is correct!

x

H0: = 0 H1: = 1

SignificanceLevel (

The Maximum Type 1 Error= SignificanceLevel

The smaller the p-value the more significant is the test

p-value

Fizzy drink cans should have 330 ml of content on average. A consumer group has complained against a manufacturer that its contents are lessthen the stated amount. Use the following data to assess the complaint.(Assume a normal population)

One-Tail and Two-Tail Tests

x < 330 =330

xMEAN = 320 ; = 10 ; n = 16

Step 1: Set up the hypotheses

H0 :

H1 :

Step 2: Select statistic

One-Tail Test

The statistic to be used is xMEAN = 320

Step3 : Identify the distributionof xMEAN

xMEAN ~ Normal16

xMEAN ~ Normal16 Under H0

Step 4: Construct test statistic (z) z= {(xMEAN –} = (320-330)/10/4 = -4

Step 5: Compare with critical value zC

zC = -1.645 > -4

zC = -1.645 for a one-tailed test with significance level () = 0.05

Step 6 : Draw conclusion

The test is significant. Reject H0 at 5% and at 1% ( -4 < -2.33)

Step 7: Interpret result


There is overwhelming evidence to support the claim !

A cereal manufacturer is concerned that the boxes of cereal not be under-filled orover-filled. Each box is supposed to contain 130 grams of cereal. A test is done to see whether the machines are putting an average of 130 grams into a single box. A random sample of 30 boxesis tested. The average weight is found out to be 128 grams and the sample standard deviation 10 grams.

A Two-Tail Test

x < 130 =130 >130

Set up the hypotheses to test whether the average number of grams per box isdifferent from 130. Step 1 H0 :

H1 :

Two-Tail Test

Perform the test at 10% level of significance.


xMEAN ~ 30

xMEAN ~ Normal30 (approx.)

Under H0

Step 2: Select statisticThe statistic to be used is xMEAN = 128

Step 4: Construct test statistic (z) z= {(xMEAN –} = (128-130)/10/ = -1.1


zC = 1.645 for a two-tailed test with significance level () = 0.1

1.645 < 1.1

Explain which hypothesis you are rejecting and why. Do you think the decision to accept/reject the null hypothesis stays unchanged if the significance level is lowered to 5%? Explain carefully.


The test is not significant. Do not reject H0 at 10% and of course not at 5% ( -1.96 < -1.1)


There is no cause for concern, given the data above.

A production process makes components whose strengths are normally distributedwith mean 40kg and unknown variance.

The process is modified and 12 components are selected at random withMean strength 41.125kg and sample standard deviation 1.316591 kg.

Is there any evidence that the modifiedProcess makes stronger components?

Step 1 : Set up the hypothesesH0 : H1 :

One-Tail Test



The statistic to be used is xMEAN = 41.125

Step3 : Identify the distribution

(xMEAN – /s/n has a t-distribution of 11 degrees of freedom

Step 4: Construct test statistic (t) {(xMEAN –}/s/} = (41.125-40)/ 1.316591 /

Step 5: Compare with critical value tC

tC = 1.795884 for a one-tailed test with significance level () = 0.05 and d.o.f . =11t = 2.96 > 1.795884Step 6 : Draw conclusion The test is significant. Reject H0 at 5% ( 2.96 > 1.795884) and at 1% (2.96 > 2.718) 0.006488 is the prob of type 1 error

Step 7: Interpret results

There is overwhelming evidence that the modified process makes stronger components

100 ‘small’ firms and eighty ‘large’ firms were tested for their profitability.

Large firms made a profit of 25p for every £1 of turnover with a standard deviation of 10p.Small firms made a profit of 20p for every £1 of turnover with a standard deviation of 5p.Test the hypothesis that size is related toprofitability

x = large firm; y = small firm

xMEAN = 25; sX = 10 nX =80yMEAN = 20; sy = 5 ; nY =100

Step 1 H0 : x

- Y = 0H1 : x

- Y Two-Tail Test


Step3 : Identify the distributionof xMEAN -yMEAN

xMEAN -yMEAN ~ Normalx

YX 80 + Y 100

~ NormalsX 80 + sY 100(approx.) Under H0

Step 2: Select statisticThe statistic to be used is xMEAN -yMEAN = 25-20 = 5

~ Normal10080 + 25100(approx.)

Step 4: Construct test statistic (z) z= {(xMEAN -yMEAN –(x

–Y} = (5-0)/ = 4.46



1.96 < 4.46


The test is significant. Reject H0 at 5% ( 1.96 < 4.46) or even at 1%


There is overwhelming evidence that large firms are more profitable

x

BenchmarkValue

p-value

Prob (Type 1 error) = p- value = 4.10158E-06

Sample size ‘large’Hypothesis Testing with xMEAN

known unknown

Perform Z-testReplace s for and perform Z-test

Sample size ‘small’

Normal Population

known unknown

Non-normal Population

Parametric TestNot possiblePerform Z-test Perform t-test.

Replace s for

Learning ObjectivesLearning Objectives

• Understand the logic of hypothesis testing, and know how to establish null and alternate hypotheses.

• Understand Type I and Type II errors.• Use large samples to test hypotheses about a

single population mean and about the differences of two population means

• Use large samples to test hypotheses about a single population proportion and about the differences of two population proportions

• Test hypotheses about a single population mean using small samples when is unknown and the population is normally distributed.

Hypothesis Testing

A process of testing hypotheses about parameters by setting up null and alternative hypotheses, gathering sample data, computing statistics from the samples, and using statistical techniques to reach conclusions about the hypotheses.

Null and Alternative HypothesesNull and Alternative Hypotheses

• The Null and Alternative Hypotheses are mutually exclusive. Only one of them can be true.

• The Null Hypothesis is assumed to be true.

• The burden of proof falls on the Alternative Hypothesis.

The Spirit of Hypothesis Testing

H0 is the hypothesis that is challenged

H0 is the established hypothesis

Xo

y

Can we have intersecting indifference curves?

a

a ~ b, a ~ c; so b ~ca contradiction of the axiom more is better

bc

H0 : Yes H1 : No

We work under the assumption that H0 is correct!

Null and Alternative Hypotheses: ExampleNull and Alternative

Hypotheses: Example

• A soft drink company is filling 12 oz. cans with cola.• The company hopes that the cans are averaging 12

ounces.

H oz

H oz

o

a

:

:

12

12

Rejection and Non-rejection Regions

Rejection and Non-rejection Regions

=12 oz

Nonrejection Region

Rejection Region

Critical Value

Rejection Region

Critical Value

Type I and Type II ErrorsType I and Type II Errors

• Type I Error– Rejecting a true null hypothesis – The probability of committing a Type I error

is called , the level of significance.• Type II Error

– Failing to reject a false null hypothesis– The probability of committing a Type II

error is called .– Power is the probability of rejecting a false

null hypothesis, and equal to 1-

x

H0: =20 H1: = 25

Two competing Hypotheses

x

H0: =20 H1: = 25

BenchmarkValue

p-value

BenchmarkValueType 2 Error:

Not rejectingA false H0

Type 1 Error:Rejecting a correct H0

Area in black : Maximum Type2 errorArea in red : Maximum Type1 error

BenchmarkValue


Type 2 Error:


p-value

SignificanceLevel (

The Maximum Type 1 Error= SignificanceLevel

Decision Table for Hypothesis Testing

Decision Table for Hypothesis Testing

(

( )

Null True Null False

Fail toreject null

CorrectDecision

Type II error

Reject null Type I error

Correct Decision (Power)


p-value

Fizzy drink cans should have 330 ml of content on average. A consumer group has complained against a manufacturer that its contents are lessthen the stated amount. Use the following data to assess the complaint.(Assume a normal population)

A One-Tail Test

x < 330 =330

xMEAN = 320 ; = 10 ; n = 16


H0 :

H1 :


One-Tail Test

The statistic to be used is xMEAN = 320


xMEAN ~ Normal16

xMEAN ~ Normal16 Under H0

Step 4: Construct test statistic (z) z= {(xMEAN –} = (320-330)/10/4 = -4


zC = -1.645 > -4



The test is significant. Reject H0 at 5% and at 1% ( -4 < -2.33)



There is overwhelming evidence to support the claim !

A cereal manufacturer is concerned that the boxes of cereal not be under-filled orover-filled. Each box is supposed to contain 130 grams of cereal. A test is done to see whether the machines are putting an average of 130 grams into a single box. A random sample of 30 boxesis tested. The average weight is found out to be 128 grams and the sample standard deviation 10 grams.

A Two-Tail Test

x < 130 =130 >130

Set up the hypotheses to test whether the average number of grams per box isdifferent from 130. Step 1 H0 :

H1 :

Two-Tail Test



xMEAN ~ 30

xMEAN ~ Normal30 (approx.)

Under H0

Step 2: Select statisticThe statistic to be used is xMEAN = 128

Step 4: Construct test statistic (z) z= {(xMEAN –} = (128-130)/10/ = -1.1



1.645 < 1.1


The test is not significant. Do not reject H0 at 10% and of course not at 5% ( -1.96 < -1.1)


There is no cause for concern, given the data above.

100 ‘small’ firms and eighty ‘large’ firms were tested for their profitability.

Large firms made a profit of 25p for every £1 of turnover with a standard deviation of 10p.Small firms made a profit of 20p for every £1 of turnover with a standard deviation of 5p.Test the hypothesis that size is related toprofitability

x = large firm; y = small firm

xMEAN = 25; sX = 10 nX =80yMEAN = 20; sy = 5 ; nY =100

Step 1 H0 : x

- Y = 0H1 : x

- Y Two-Tail Test


Step3 : Identify the distributionof xMEAN -yMEAN

xMEAN -yMEAN ~ Normalx

YX 80 + Y 100

~ NormalsX 80 + sY 100(approx.) Under H0

Step 2: Select statisticThe statistic to be used is xMEAN -yMEAN = 25-20 = 5

~ Normal10080 + 25100(approx.)

Step 4: Construct test statistic (z) z= {(xMEAN -yMEAN –(x

–Y} = (5-0)/ = 4.46



1.96 < 4.46


The test is significant. Reject H0 at 5% ( 1.96 < 4.46) or even at 1%


There is overwhelming evidence that large firms are more profitable

Example: Thirty-five percent of the consumers were dissatisfied with the Product when a new CEO was elected.

Test of Proportion

Five months later, 108 out of 360 were found out to be dissatisfied.

Can we conclude (at 5%) that the newly elected CEO has significantly reduced the level of dissatisfaction amongst consumers?


The statistic to be used is p = 108/360 = 0.3Step3 : Identify the distributionof p

First we need to perform the validity check

H0 : H1 :


Here n = 360 and = 0.35 under H0. So the validity check is satisfied.

This means that p ~ Normal(360 where 2 = X 0.65 = 0.2275Step 4: Construct test statistic (z) z= {(p –0.350.2275)/360} = (0.3-0.35)/0.477/18.97367 -19.9


zC = -19.9 < -1.645 Step 6 : Draw conclusion

The test is significant. Reject H0


Based on the analysis of data, we conclude that there is strong evidence that consumer dissatisfaction has been significantly reduced.

Example: 70 out of 90 large firms (L) and100 of 110 small firms (S) were voted eco-friendly. Test the hypothesis that large firms are less eco friendly than small ones..

Step 1: Set up the hypothesesH0 : S L H1 : SL

Test of Difference in Proportions


The statistic to be used is pS–pL

where pL = 7/9 = 0.78; pS = 10/11 = 0.9

Step3 : Identify the distributionof pS –pL

First we need to perform the validity check

Here nS = 110 and nL = 90 but S and L

are unknown. So we perform the validity check by using the values of pS and pL for S and L respectively. The validity check is satisfied.This means that (pS - pL) ~ Normal(S L S

nS LnL

Under H0, (L S) = 0

Also, S = S S)

and L = L L )

Usep = (nS pS + nL pL) / (nS + nL ) for both S and L . Thusp = [(100/110) *110 + (70/90)*90]/200 = 0.85

Hence, S ) = 0.1275

Similarly, L ) = 0.1275

(pS - pL) ~ Normal( 0.1275110 0.127590 (pS - pL) ~ Normal(

Step 4: Construct test statistic (z) z= {(pS - pL) –0}/(0.059) = (0.9-0.78)/0.059 2


zC = 1.645 < 2 Step 6 : Draw conclusion

The test is significant. Reject H0 at 5% but not at 1% ( 2 < 2.33)


Based on the analysis of data, we conclude that there is enough evidence that small forms are significantly more eco-friendly.

Two-tailed Test: Small Sample, Unknown, = .05 (Part 1)


Weights in Pounds of a Sample of 20 Plates

22.622.2 23.2 27.4 24.527.026.6 28.1 26.9 24.926.225.3 23.1 24.2 26.125.830.4 28.6 23.5 23.6

X 2551. , S = 2.1933, and n = 20



Critical Values

Rejection Regions

ct 2 093. ct 2 093.

2

025.2

025.

H

H

o

a

:

:

25

25

df n 1 19



tXS

n

2551 25 0

2 1933

20

104. .

. .

Since t do not reject H .o 104 2 093. . ,Critical Values

Non Rejection Region

Rejection Regions

ct 2 093. ct 2 093.

2

025.2

025.

If t reject H .

If t do not reject H .

o

o

2 093

2 093

. ,

. ,

Interpreting Opinion Polls Example1: (Confidence Interval for the population proportion): Suppose...

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