Lecture 13: Statistical Hypotheses and TestsLecture 13: Statistical Hypotheses and Tests MSU-STT...

Lecture 13: Statistical Hypotheses and Tests

MSU-STT 351-Sum19A

(P. Vellaisamy: MSU-STT 351-Sum19A) Probability & Statistics for Engineers 1 / 29

Statistical Hypotheses and Tests

Basic concepts of TSH

Point estimation or confidence interval estimation respectively deal withobtaining a possible estimate of θ or an interval that contains θ with highprobability. Testing a statistical hypothesis deals with how to make adecision between two competing claims (hypotheses).

Definition: A Statistical hypothesis or simply a hypothesis is an assertionor a claim about one or more population characteristics or about theform of population distribution.



Examples: Some hypotheses are:

(i) Let p denote the proportion of students studying engineering in auniversity.

Hypothesis: p ≥ 0.6.

(ii) Let X denote the height of the students in a class.Hypothesis: X ∼ N(170, 5).

Kinds of Hypotheses:

Null hypothesis H0. It is a hypothesis which is initially assumed (orpresumed) to be true.

Alternative hypothesis H1. It is the hypothesis to be compared againstH0.



Example 1 (Ex 1): For each of the following assertions, state whether it isa legitimate statistical hypothesis and why:

(a) H : σ > 100(b) H : x = 45(c) H : s ≤ .20(d) H : σ1/σ2 < 1(e) H : X − Y = 5

Solution: (a) Yes. It is a statement about the standard deviation.(b) No. The sample median X̃ is not a parameter/distribution.(c) No. The sample standard deviation s is not a parameter.(d) Yes. The assertion is that the standard deviation of Population 2

exceeds that of Population 1.(e) No. X and Y are statistics. Any statement about them cannot be a

hypothesis.



Tests.

A test of hypothesis is a method or procedure of deciding between H0 andH1, based on the sample information. If the sample contains strongevidence against H0, then H0 will be rejected in favor of H1. Otherwise, wewill continue to believe in H0.

Suppose X ∼ N(µ, 10) and consider testing:

(i) H0 : µ = 50 against H1 : µ < 50 (lower tailed).(ii) H0 : µ = 50 against H1 : µ > 50 (upper tailed).(ii) H0 : µ = 50 against H1 : µ , 50 (two-tailed).

A test leads to one of the following decisions:(i) Accept H0;(ii) Reject H0 (Accept H1).

Obviously, there are two-kinds of errors which are inherent in any test.



First we need the following definitons.

Definitions: We define

Type I error =Rejecting H0, when H0 is true (wrongly rejecting H0).

Type II error=Accepting H0, when H1 is true (wrongly accepting H0).

We use the symbols α and β to denote

P(Type I error)= α =P(Reject H0 when H0 is true)=P(Reject H0|H0).

P(Type II error)= β =P(Accepting H0 when H1 is true)=P(Accept H0|H1).

It can be shown that both the errors α and β can not be simultaneouslyminimized. Usually, we fix one of the errors, say α, and choose a test thatminimizes β.

In practice, α = 0.05 or α = 0.01 is usually chosen.



Definitions:(i) The specified value of α is also called “level of significance” or

“significance level” of the test.(ii) A test statistic T is a standardized function of X = (X1, . . . ,Xn), which

helps us to take a decision.(iii) A critical region or rejection region is a set of test statistic values

leading to the rejection of H0.

Remarks: (i) A test with significance level α = .05 means that when thetest repeated several times, only 5% of times the type I error may occur.(ii) The quantity (1 − β) = P(Reject H0|H1) is called the power of the test.(iii) If α decreases, then β increases. Usually, the largest tolerable value ofα is decided based on the problem under consideration.

For a specified α, a test is designed so that P(Type I error) ≤ α and β isminimum (i.e., the power (1 − β) is maximum).



Example 2 (Ex 9): Two companies have applied to provide cable televisionservice in a certain region. Let p denote the proportion of all potentialsubscribers who favor the first company 1 over company 2. Considertesting H0 : p = .5 versus Ha : p , .5 based on a random sample of 25individuals. Let X denote the number in the sample who favor company 1.

(a) Which of the following rejection regions is most appropriate and why?R1 = {X : X ≤ 7

⋃X ≥ 18}, R2 = {X : X ≤ 8}, R3 = {X : X ≥ 17}.

(b) Describe type I and type II errors in this problem.(c) What is the probability distribution of the test statistic X, when H0 istrue? Use it to compute the probability of a type I error.(d) Compute the probability of a type II error for the selected critical regionwhen p = 0.4.(e) Using the selected region, what would you conclude if 6 of the 25queried favored company 1?



Solution: (a) R1 is the most appropriate, because X either too large or toosmall whcih contradicts p = .5 and supports p , .5.(b) A type I error occurs when deciding one of the two companies favoredover the other when in fact there is a 50-50 split in the population. A type IIerror involves judging the split to be 50-50 when it is not.(c) Note X has a binomial distribution with n = 25 and p = 0.5, whern H0

is true. That is, X ∼ Bin(25, .5). Hence,α = P(Type-I error) = P(X ≤ 7 or X ≥ 18)

= B(7|25, .5) + 1 − B(17|25, .5) = 0.044.

(d) Note β(p) = P(Type-II error) = P(Accept H0|p). Hence,β(.4) = P(8 ≤ X ≤ 17 | p = .4) (use Table A.1)

= B(17|25, 0.4) − B(7|25, 0.4) = 0.999 − 0.154 = 0.0845.

(e) X = 6 is in the rejection region R1, so H0 is rejected in favour of H1.



Example 3 (Ex 12): A new design for the braking system on a certain typeof car has been proposed. For the current system, the true averagebraking distance µ0 at 40 mph is known to be 120 ft. The new design beimplemented only if sample data strongly indicates a reduction in trueaverage braking distance.

(a) Define the parameter of interest and state the relevant hypotheses.(b) Suppose braking distance for the new system is normally distributedwith σ = 10. Let X denote the sample mean of 36 observations. Which ofthe following three rejection is appropriate: R1 = {x : x ≥ 124.80}, R2 ={x : x ≤ 115.20}, R3 = {x; either x ≥ 125.13 or x ≤ 114.87}?(c) What is the significance level for the appropriate region of part (b)?How would you change the region to obtain a test with α = .001?(d) What is the probability that the new design is not implemented when µ(say) is actually 115 ft and the appropriate region from part (b) is used?



Solution: (a) Let µ = true average braking distance for the new design at40 mph. The hypotheses are H0 : µ = 120 vs H1 : µ < 120.(b) R2 should be used, since support for H1 is provided only by an X valuesubstantially smaller than 120.(c) Note σx = (σ/

√n) = (10/6) = 1.6667. So,

α = P(X ≤ 115.20| µ = 120) = P(Z ≤

115.20 − 1201.6667

)= P(Z ≤ −2.88) = 0.002.

Let {X ≤ c} be the CR so that P(X ≤ c) = .001. Since z.001 = 3.08, wehave P(Z ≤ −3.08) = 0.001. Hence,

P(Z ≤

c − 1201.6667

)= .001 =⇒ c = 120 − 3.08(1.6667) = 114.87.

(d) The probability of type-II error isβ(115) = P(X > 115.2|µ = 115) = P(Z > .12) = .4522.



Example 4: Let µ=population mean. Suppose we want to test H0 : µ = 50against H1 : µ < 50.Note, sample mean X gives information about µ and X ∼ N(µ, σ√

n).

When σ is not known and n is large, (X−µ)√

nS � N(0, 1) distribution.

When H0 is true (and n is large), the statistic Z =(X−50)

√n

S � N(0, 1).

Suppose Z = 3. Then (X−µ)√

nS = 3⇒ X = 50 + 3 S√

n= 50 + 3SD(x),

which is not in favor of H0.

Suppose Z = .5 Then X = 50 + (.5) s√n. In this case, x is only 1

2 SD largerthan µ0 which does not contradict H0.

The statistic Z is called the test statistic. We may reject H0 for smallvalues of Z. That is, reject H0 if Z ≤ zα, for some suitable zα.

Remark. For one-sided H1 : µ > 20, reject H0 if X > 20 + σ0zα√

n.



The General Procedure:Let X ∼ N(µ, σ). Consider testing

H0 : µ = µ0 vs H1 : µ , µ0.

Case (i). σ = σ0 is known (see Example 2).In this case, use the test statistic

Z =(X − µ0)

√n

σ0∼ N(0, 1)

and reject H0 if the observed value of Z satisfies

z =∣∣∣∣(x − µ0)

√n

σ0

∣∣∣∣ ≥ zα/2.

Case (ii). σ is unknown; n is large.In this case, use

Z =(X − µ0)

√n

S∼ N(0, 1).



Hence, reject H0 if ∣∣∣∣(X − µ0)√

nS

∣∣∣∣ ≥ zα/2.

Case (iii). σ is unknown and n is small:In this case, use the statistic

T =(X − µ0)

√n

S∼ tn−1

and reject H0 if

|T | =∣∣∣∣(X − µ0)

√n

S

∣∣∣∣ > tα/2,n−1,

where tα/2,n is t-critical value such that

P(T > tα/2,n) =α

2.



Example 5: Let X ∼ N(µ, σ0), where σ0 is known. Let X1, . . . ,Xn be arandom sample on X . Suppose we want to test

H0 : µ = 20 versus H1 : µ , 20

Then reject H0 if |X | is large or equivalently if∣∣∣∣ X − µ

σ0/√

n

∣∣∣∣ > k .

What should be the value k so as to have significance level α.



We wantP(|Z | > k |H0) = α (Prob. of Type-I error)

⇒ P(∣∣∣∣(X − µ0)

σ0/√

n

∣∣∣∣ > k |H0

)= α

⇒ P(∣∣∣∣(X − 20)

σ0/√

n

∣∣∣∣ > k |H0

)= α (Under H0, µ = 20)

⇒ k = zα/2.

Therefore, reject H0 if√

n(X − 20)

σ0< −zα/2 or

√n(X − 20)

σ0> zα/2.

That is, reject H0 if

X < 20 − zα/2σ0√

nor X > 20 + zα/2

σ0√

n.



Tests Concerning a Population Mean

Case 1: Normal Population with known σ = σ0.

Null Hypothesis: H0 : µ = µ0.

Test statistic: Z =(X−µ0)

σ0/√

n.

Alternative Hypothesis Rejection RegionH1 : µ > µ0 Z ≥ zα (upper tailed test)H1 : µ < µ0 Z ≤ −zα (lower tailed test)H1 : µ , µ0 either Z ≥ zα/2 or Z ≤ −zα/2 (two tailed test)



Probability of Type-II Error:

Consider testing H1 : µ = µ0 against H1 : µ > µ0, with CR={Z ≥ zα}. Thenthe probability of type-II error when µ = µ1 > µ0 is

β(µ1) = P(Accept H0| H1)

= P((X − µ0)

√n

σ0≤ zα| µ1

)= P

(X ≤

(µ0 +

√nzασ0

)| µ1

)= P

(Z ≤ (µ0 − µ1)

√n

σ0+ zα

)= Φ(zα + (µ0 − µ1)

√n

σ0).



Similarly, the β(µ1) can be computed for other forms of H1. The results aregiven below. Note µ1 belongs to H1.

Let Φ(z) denote the cdf of standard normal. The following table presentsthe probability of Type-II errors.

Alternative Hypothesis P(Type-II error)H1 : µ > µ0 Φ(zα + µ0−µ1

σ0/√

n)

H1 : µ < µ0 1 − Φ(−zα + µ0−µ1

σ0/√

n)

H1 : µ , µ0 Φ(zα + µ0−µ1

σ0/√

n) − Φ(−zα + µ0−µ1

σ0/√

n)



The sample size n for which a level-α test has β(µ1) = β1, at thealternative value µ1, is given by (for upper-tailed H1 : µ > µ0)

β1 = Φ(zα +µ0 − µ1

σ0/√

n)

which implies (note the definiton of zα and the CDF Φ)(zα +

µ0 − µ1

σ0/√

n

)= −zβ1 .

Solving for n, we get

n =

[σ0(zα + zβ1)

µ0 − µ1

]2, for a one-tailed test;

[σ0(zα/2 + zβ1)

µ0 − µ1

]2, for a two-tailed test.



Case II Unknown σ or Non-normal population

When n is large, use the fact (X−µ)√

nS � N(0, 1) and the normal test.

Case III: A Normal Population: Small sample and σ unknownStudent’s t-Tests for µ = µ0

Consider testing µ = µ0 (given, say µ0 = 100). Let X1, . . . ,Xn be a randomsample from N(µ, σ2). Then as discussed earlier,

T =(X − µ)

√n

S

follows t-distribution with (n − 1) df . This can be used to test

H0 : µ = µ0

and it is called the one-sample t-test.



The One-Sample t test

Null hypothesis: H0 : µ = µ0.

Test statistic: T =(X − µ0)

√n

S.

The following table presents the critical regions for different alternativehypotheses:

Alternative Hypothesis Rejection Region for a Level α TestH1 : µ > µ0 T ≥ tα,n−1 (upper tailed test)H1 : µ < µ0 T ≤ −tα,n−1 (lower tailed test)H1 : µ , µ0 T ≥ tα/2,n−1 or T ≤ −tα/2,n−1 (two tailed test)



Example 6 (Ex 16): Let the test statistic Tν have a t distribution with ν d.f.,when H0 is true. Find the significance level α for each of the followingsituations:(a) H1 : µ > µ0, df=15, rejection region t ≥ 3.7(b) H1 : µ < µ0, n = 24, rejection region t ≤ −2.6.(c) H1 : µ , µ0, n = 30, rejection region t ≥ 1.7. or t ≤ −1.7

Solution: (a) Using Table A.8 for t-tail areas, we get

α = P(T15 ≥ 3.7) = 0.001.

Similarly, since Tν is symmetric about origin,(b) For df = 23, α = P(T23 ≤ −2.5) = P(T23 ≥ 2.5) = 0.01.(c) When df = 29,

α = P(T29 ≥ 1.7) + P(T29 ≤ −1.7)

= 2P((T29 ≥ 1.7) = 2(0.05) = 0.10.



Example 7 (Ex 18): The drying time for a test specimen is normallydistributed with σ = 9. The hypotheses H0 : µ = 75 versus H0 : µ < 75 areto be tested using a random sample of n = 25 observations.

(a) How many standard deviation (of X ) below the null value (75) isX = 72.3?

(b) If X = 72.3, what is the conclusion using α = 0.01?

(c) What is α for the test procedure that rejects H0 when Z ≤ −2.88?

(d) For the test procedure of part (c), what is β(70)?

(e) If the test procedure of the part (c) is used, what n is necessary toensure that β(70) = 0.01?

(f) If a level 0.01 test is used with n = 100, what is the probability of a typeI error when µ = 76?



Solution: (a) Note SD of (X̄) = (σ/√

n) = 9/5 = 1.8. Hence,Z = 72.3−75

1.8 = −1.5. So, 72.3 is 1.5 SD’s (of X ) below 75.

(b) When α = .01, H0 will be rejected if Z ≤ −2.33. SinceZ = −1.5 ≥ −2.33, H0 is not rejected.

(c) α = area under standard normal curve below -2.88 isΦ(−2.88) = 0.0020.

(d) Φ(− 2.88 + 75−70

9/5

)= Φ(−.1) = 0.4602. So, using the formula on p. 19,

we get β(70) = 1 − Φ(−.1) = 0.5398.

(e) n =[9(2.88 + 2.33)

75 − 70

]2= 87.95, so use n = 88.

(f) Zero. By definition, a type-I error can only occur when H0 is true, butµ = 76 means that H0 is actually false.



Example 8 (Ex 24): The sample observations on stabilized viscosity ofasphalt specimens are 2781, 2900, 3013, 2856 and 2888. Suppose that fora particular application, it is required that true average viscosity be 3000.Does this requirement appear to have been satisfied? State and test theappropriate hypotheses.

Solution: When nothing is specified, take α = 0.05 and go for two-tailed

test. To test H0 : µ = 3000 versus H1 : µ , 3000; T =X − 3000

S/√

n. Reject

H0 if the observed value |T | > t.025,4 = 2.776. The observed value is

T =2887.6 − 3000

84/√

5= −2.99 < −2.776,

Hence, we reject H0 and the requirement is not satisfied.



Example 9 (Ex 29): The amount of shaft wear (.0001 in) after a fixedmileage was determined for each of n = 8 internal combustion engineshaving copper lead as a bearing material, resulting in X = 3.72 andS = 1.25.

(a) Assuming that the distribution of shaft wear is normal with mean µ, usethe t test at level .05 to test H0 : µ = 3.50 versus H0 : µ > 3.50.

(b) When σ = 1.25 is known, what test you will use and find its type IIerror probability β(µ′) of the test for the alternative µ′ = 4.00?



Solution: (a) For n = 8, n − 1 = 7, and t.05,7 = 1.895. So, H0 is rejectedat level 0.05 if the observed value of T ≥ 1.895.

Sinces√

n=

1.25√

8= 0.442, we get

T =3.72 − 3.50

.442= .498.

this does not exceed 1.895, and so H0 is not rejected.

(b) When σ is known, we will use Z-test. Its power, when µ = µ1 = 0.4, is

β(µ′) = Φ(zα + (µ0 − µ1)

√nσ

)

= Φ(1.645 −.5√

81.25

)

= Φ(.5136) = 0.6963.



Home work:

Section 8.1: 2(a, c, e), 6, 11, 13

Section 8.2: 15, 19, 23, 25, 30


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