Hypothesis of Small Tests

7/28/2019 Hypothesis of Small Tests

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Introduction

• When the sample size is small we only

deal with normal populations.

• For non-normal (e.g. binomial) populationsdifferent techniques are necessary

Piyoosh Bajoria 3



Summary of Test Statistics to be Used in a

Hypothesis Test about a Population Mean

n > 30 ?

σ known ?Popul.

approx.

normal?

σ known ?

Use s toestimate σ

Use s to

Estimate σ

Increase nto > 30 /

x z

n

/

x z

s n

/

x z

n

/

xt

s n

Yes

Yes

Yes

Yes

No

No

No

No

4Piyoosh Bajoria



Student’s t Dist r ibu t ion

• For small samples (n < 30) from normal

populations, we have• z =x¯ - µ/ σ/√n

• If σ is unknown, we use s instead; but we

no more have a Z distribution• Assumptions.

• 1. Sampled population is normal

• 2. Small random sample (n < 30)• 3. σ is unknown

• t =x¯ - µ/s/√n Piyoosh Bajoria 5



Properties of the t Distr ibu t ion : • (i) It has n - 1 degrees of freedom (df)

• (ii) Like the normal distribution it has a symmetric

mound-shaped probability distribution

• (iii) More variable (flat) than the normal distribution

• (iv) The distribution depends on the degrees of freedom.

Moreover, as n becomes larger, t converges to Z.

• (v) Critical values (tail probabilities) are obtained from

the t table

• Examples.

• (i) Find t 0.05,5 = 2.015

• (ii) Find t 0.005,8 = 3.355

• (iii) Find t 0.025,26 = 2.056

Piyoosh Bajoria 6

S ll S l I f Ab t P l ti M



Small-Sample Inferences About a Population Mean

• Parameter of interest: µ

• Sample data: n, x¯, s

• Other information: µ0 = target value ;α • Point estimator: x¯

• Estimator mean: µ x¯ = µ

• Estimated standard error σ x¯ = s/√n • Confidence Interval for µ:

• x ± t α / 2 ,n-1 (s/√n )

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Student’s t test • Test: H0 : µ = µ0

• Ha : 1) µ > µ0; 2) µ < µ0; 3) µ = µ0.

• Critical value: either t α ,n-1 or t α / 2 ,n-1

• T.S. : t = x¯-µ0 /s/√n

• RR:1) Reject H0 if t > t α ,n-1

• 2) Reject H0 if t < -t α ,n-1

• 3) Reject H0 if t > t α / 2 ,n-1 or t < -t α / 2,n-1• Decision: 1) if observed value is in RR: “Reject H0”

• 2) if observed value is not in RR: “Do not reject H0”

• Conclusion: At 100α% significance level there is (in)sufficient statistical evidence to “favor Ha” .

• Assumptions.1. Small sample (n < 30)• 2. Sample is randomly selected

• 3. Normal population

• 4. Unknown variance

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Example • Example: It is claimed that weight loss in a new diet

program is at least 20 pounds during the first month.

Formulate &Test the appropriate hypothesis• Sample data: n = 25, x¯ =19.3, s2 = 25, µ0 = 20, α =0.05

• Critical value: t 0.05,24 = -1.711

• H0 : µ ≥ 20 (µ is greater than or equal to 20 )

• Ha : µ < 20,• T.S.:t =(x¯ - µ0 )/s/√n=(19.3 – 20)/ 5 /√25 =-0.7

• RR: Reject H 0 if t <-1.711

• Decision: Accept H 0

• Conclusion: At 5% significance level there is sufficient

statistical evidence to conclude that weight loss in a new

diet program exceeds 20 pounds per first month.

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Two Means: Independent Samples

• Parameter of interest: µ1 - µ2

• Sample data:Sample 1: n1, x 1, s1 ;Sample 2: n2 , x 2 , s2

• Point estimator: X¯ 1 – X¯ 2

• Estimator mean: µ X¯1 – X¯2 = µ1 - µ2

• Assumptions.

• 1. Normal populations 2. Small samples ( n1 < 30; n2 < 30)

• 3. Samples are randomly selected4. Samples are independent

• 5. Variances are equal with common variance• σ 2 = σ 2

1 = σ2 2

• Pooled estimator for σ.

• s = √[(n1 - 1)s2 1 + (n2 - 1)s2

2 ] /(n1 + n2 – 2)

• Estimator standard error:

• σ X¯1 – X¯2 = σ √ (1/n1+1/n2 )

• Confidence Interval:

• ( x¯ 1 - x¯ 2 ) ± (t α /2 ,n1+n2-2 )(s √1/ n1+1/n2 )

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Two Means: Independent Samples

• Test:H0 : µ1 = µ2

• Ha : 1)µ1 > µ2 or

2) µ1 < µ2 or

3) µ1 = µ2

• T.S. :t =( x¯ 1 - x¯ 2 ) /s √1/n1+ 1/n2

• RR: 1) Reject H0 if t > t α ,n1+n2-2

2) Reject H0 if t < - t α ,n1+n2-2

3) Reject H0 if t > t α /2,n1+n2-2 or t < - t α /2,n1+n2-2



Example.(Comparison of two weight loss programs)

• Refer to the weight loss example. Test the hypothesis that weight

loss in a new diet program is different from that of an old program.

We are told that that the observed value of Test Statistics is 2.2 and we know that

• Sample 1 : n1 = 7 ; Sample 2 : n2 = 8 ; α= 0.05

• Solution.

• H0 : µ1 = µ2

• Ha : µ1 ≠ µ2

• T.S. :t =( x−1 - x−2 ) /s √1/n1+ 1/n2 = 2.2

• Critical value: t.025,13 = 2.16

• RR: Reject H0 if t > 2.160 or t < -2.160

• Decision: Reject H0

• Conclusion: At 5% significance level there is sufficient

statistical evidence to conclude that weight loss in the

two diet programs are differentPiyoosh Bajoria 12



ma - amp e n erences ou e erenceBetween

Two Means: Paired Samples• Parameter of interest: µ1 = µ2

• Sample of paired differences data:

• Sample : n = number of pairs, d¯ = sample mean, sd

• Point estimator: d¯

• Estimator mean: µd¯ = µd

• Assumptions.

• 1. Normal populations

• 2. Small samples ( n1 < 30; n2 < 30)

• 3. Samples are randomly selected

• 4. Samples are paired (not independent)

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• Sample standard deviation of the sample

of n paired differencesn

• sd = √ Σ (d i - ¯d)2 / n-1i=1

• Estimator standard error: σd

= sd /√n

• Confidence Interval.

• ¯d ± t α /2 ,n-1sd /√n

Piyoosh Bajoria 14

T t



Test.• H0 : µ1 = µ2

• Ha : 1) µ1 > µ2

• 2) µ1 < µ2 • 3) µ1 = µ2

• T.S. :t =¯d /sd /√n

• RR:

• 1) Reject H0 if t > t α , n-1

• 2) Reject H0 if t < -t α ,n-1

• 3) Reject H0 if t > t α /2,n-1 or t < -t α /2,n-1

Piyoosh Bajoria 15



Example. • A manufacturer wishes to compare wearing qualities of

two different types of tires, A and B. For the comparison

a tire of type A and one of type B are randomly assignedand mounted on the rear wheels of each of five

automobiles. The automobiles are then operated for a

specified number of miles, and the amount of wear is

recorded for each tire. These measurements aretabulated below.

• Automobile Tire A Tire B

• 1 10.6 10.2

• 2 9.8 9.4• 3 12.3 11.8

• 4 9.7 9.1

• 5 8.8 8.3

• Piyoosh Bajoria 16



• x¯ 1 = 10.24 x¯ 2 = 9.76

• Using the previous section test we would have t = 0.57 resulting inan insignificant test which is inconsistent with the data.

• Automobile Tire A Tire B d=A-B

• 1 10.6 10.2 0.4• 2 9.8 9.4 0.4

• 3 12.3 11.8 0.5

• 4 9.7 9.1 0.6

• 5 8.8 8.3 0.5

• x¯1 = 10.24 x¯2 = 9.76 d¯ =0 .48

• Q1: Provide a summary of the data in the above table.

• Sample summary: n = 5, d ¯= .48, sd = .0837

Piyoosh Bajoria 17



• Q2: Do the data provide sufficient evidence to indicate a

difference in average wear for the two tire types.

• Test.

• H0 : µ 1 = µ 2

• Ha : µ 1 ≠ µ 2

• T.S. :t =d ¯/sd /√n

• =.48 – 0/.0837/√5 = 12.8

• RR: Reject H0 if t > 2.776 or t < -2.776 ( t .025,4 = 2.776)

• Decision: Reject H0

• Conclusion: At 5% significance level there is sufficientstatistical evidence to to conclude that the average

amount of wear for type A tire is different from that for

type B tire.

• Exercise. Construct a 99% confidence interval for thedifference in avera e wear for the two tire t es.18

Inferences About a Population Variance



Inferences About a Population Variance

• Chi-square distribution. When a random

sample of size n is drawn from a normal population with mean µ and standard deviationσ , the sampling distribution of S2 depends on n.The standardized distribution of S2 is called the

chi-square distribution and is given by• χ 2 =(n - 1)s2 / σ 2

• Degrees of freedom (df): ν = n - 1

• Graph: Non-symmetrical and depends on df • Critical values: using χ 2 tables

Piyoosh Bajoria 19



Inferences About a Population Variance • Test.H 0 : σ 2 = σ 2

0

• Ha : σ 2 ≠ σ 2 0 (two-tailed test).

• Ha : σ 2 <σ 2 0 (One –tailed test )

• T.S. : X 2 =(n - 1)s2 / σ 2 0

• RR: (two-tailed test).

• Reject H0 if X 2

> X 2

α /2 or X 2

< X 2

1-α /2 where• X 2 is based on (n - 1) degrees of freedom.

• RR: (One-tailed test).

• Reject H0 if X 2 < X 2 1-α where

• X 2 is based on (n - 1) degrees of freedom• Assumptions.

• 1. Normal population

• 2. Random samplePiyoosh Bajoria 20



Example

• Future Technologies ltd. manufactures high

resolution telescopes .The management wantsits products to have a variation of less than 2 sd

in resolution while focusing on objects which are

beyond 500 light years. When they tested their

newly manufactured telescope for 30 times tofocus on an object 500 light years away they

found sample sd to be 1.46. State the

Hypothesis and test it at 1% level of significance.

Can the management accept to sell thisproduct?

Piyoosh Bajoria 21



Solution• Given ,n=30 and S2=(1.46)2, σ 2

0 =4 We set up hypothesis

• H0: σ2 =4• Ha: σ2< 4

• Significance Level α =1%=.01

• This is a one tailed test

• T.S = X 2 =(n - 1)s2 / σ 2 0= (30-1)(1.46)2/4=15.45• Referring to X 2 tables we find at 29 d.f the value of X 2

that leaves an area of 1-.01 =.99 in the upper tail and .01in the lower tail is 14.256.Since calculated value 15.45 is

in the acceptance region we accept null hypothesis and conclude that sd=2. There fore management will not allow sale of its telescope.

Piyoosh Bajoria 22



• F-distribution. When independent samples are drawn from two

normal populations with equal variances then S2 1 /S

2 2 possesses a

sampling distribution that is known as an

• F distribution. That is

• F = S2 1 /S

2 2

• Degrees of freedom (df): ν1 = n1 - 1; ν 2 = n2 - 1

• Graph: Non-symmetrical and depends on df

• Critical values: using F tables

• Test.• H0 : σ 2

1 = σ 2 2

• Ha : σ 2 1≠ σ 2

2(two-tailed test).

• T.S. :F = S2 1 /S

2 2

• Where S2

1is the larger sample variance.

• Note: F = larger sample variance/ smaller sample variance

• RR: Reject H0 if F > F α /2 where F α /2 is based on (n1 - 1) and (n2 - 1)degrees of freedom.

• Assumptions.

• 1. Normal populations 2. Independent random samples 23

E l



Example • Investment risk is generally measured by the volatility of

possible outcomes of the investment. The most common

method for measuring investment volatility is bycomputing the variance ( or standard deviation) of

possible outcomes. Returns over the past 10 years for

first alternative and 8 years for the second alternative

produced the following data:

• Data Summary:

• Investment 1: n1 = 10, x¯ 1 = 17.8%; s2 1 = 3.21

• Investment 2: n2 = 8, x¯2 = 17.8%; s2 2 = 7.14

• Both populations are assumed to be normallydistributed.

Piyoosh Bajoria 24

S l ti



Solution• Q1: Do the data present suffcient evidence to indicate

that the risks for investments1 and 2 are unequal ?

• Solution.• Test:H0 : σ 2

1 = σ 2 2

• Ha : σ 2 1 ≠ σ2

2 (two-tailed test).

• T.S. :F = S2 1 /S

2 2 =7.14/ 3.21= 2.22

• .RR: Reject H0 if F > F α /2 where

• F α /2,n2-1,n1-1 = F .025,7,9 = 4.20

• Decision: Do not reject H0

• Conclusion: At 5% significance level there is insuffcientstatistical evidence to indicate that the risks for

investments 1 and 2 are unequal.

• Exercise. Do the upper tail test. That is

• Ha : σ2

1 > σ2

2 25

Hypothesis of Small Tests

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