lecture notes test of hypothesis

LECTURE NOTES ON TESTING OF HYPOTHESIS- BY Dr. V.GNANARAJ 1

TESTING OF HYPOTHESIS

DEFINITIONS:

Population: It is the set or collection of objects, actual or hypothetical under study. Mainly

population consists of sets of numbers, measurements or observations which are of interest.

Size: The size of the population ‘n’ is the number of objects or observations in the population.

Sampling: This is the process of drawing samples from a given population.

Statistic: Any function of the random variables constituting a random sample is called a statistic.

Large Sampling: If n > 30, the sampling is said to be large sampling, otherwise it is small

sampling.

Statistical Inference: This deals with the methods of drawing valid or logical generalizations

and predictions about the population using the information contained in the sample alone.

Testing of Hypothesis:

A statistical hypothesis, or just hypothesis, is a claim or assertion either about the value of a

single parameter (population characteristic or characteristic of a probability distribution), about

the values of several parameters, or about the form of an entire probability distribution. One

example of a hypothesis is the claim, where m is the true average inside diameter of a certain

type of PVC pipe. Another example is the statement, where p is the proportion of defective

circuit boards among all circuit boards produced by a certain manufacturer. If m1 and m2 denote

the true average breaking strengths of two different types of twine, one hypothesis is the

assertion that, and another is the statement. Yet another example of a hypothesis is the assertion

that the stopping distance under particular conditions has a normal distribution. The process of

deciding whether to accept or reject the the hypothesis is called the testing of hypothesis.

Null Hypothesis: The null hypothesis formulated for the sake of rejecting it under the

assumptions is true, is called null hypothesis. and is denoted by H0.

Alternative Hypothesis: The opposite of null hypothesis is called alternative hypothesis, and is

denoted by H1.

Level of significance: The probability level below which we reject the hypothesis is called level

of significance.


TESTING OF HYPOTHESES - LARGE SAMPLE - SINGLE MEAN

Critical Region

Steps in Testing of Hypothesis

1. Identify the parameter of interest and describe it in the context of the problem situation.

2. Determine the null value and state the null hypothesis.

3. State the appropriate alternative hypothesis.

4. Give the formula for the computed value of the test statistic (substituting the null value and the

known values of any other parameters, but not those of any samplebased quantities).

5. State the rejection region for the selected significance level a.

6. Compute any necessary sample quantities, substitute into the formula for the test statistic

value, and compute that value.

7. Decide whether H0 should be rejected, and state this conclusion in the problem context.


Two Types of Errors

Decision Table

Null Hypothesis

Decision True False.

Reject H0 Type I Error Correct Decision

Accept H0 Correct Decision Type II Error

Definition: A Type I error for a statistical test occurs if you reject the null hypothesis when it is

true. The probability of making Type I error is denoted by α.

A Type II error for a statistical test occurs if you accept the null hypothesis when it is false. The

probability of making Type II error is denoted by β.

TEST OF HYPOTHESIS FOR LARGE SAMPLES

Critical value Table for Z Test

Test Level of significance

1% 2% 5% 10%

Two tail 2.575 2.33 1.96 1.645

Right tail 2.575 2.33 1.96 1.645

Left tail -2.575 -2.33 -1.96 -1.645

PROBLEM

1) The diameter of a mechanical component is normally distributed with a mean of

approximately 28cm.A standard deviation is found from the samples to be 0.25cm. A sample of

30 components gave mean 27.02. Test the hypotheses for mean = 27.02 against mean ≠ 27.02.

SOLUTION

H0: µ = 27.02 [There is no difference between sample and population mean]

H1: µ ≠ 27.02 [TTT] Two Tail Test

Assume Level Of Significance α = 5% α = 1%

Test of statistic: = √

Given µ = 28, n = 30, x = 27.02 and σ = 0.25


= . .√

ZC = -21.47

Conclusion

|ZC| = 21.47

α = 5%

Zα= Z0.05= 1.96 [TTT]

|ZC| > 1.96

Reject H0

α = 1%

Zα= Z0.01 = 2.58 [TTT]

|ZC| > 2.58

Reject H0

H0 is rejected in both cases.

The difference between x and µ is significant.

2) A trucking firm is suspicious of the claim that the average lifetime of certain tires is atleast

28,000 miles. To check this claim, the firm puts 40 of these tires on its truck and gets mean

lifetime of 27,463 mils with a standard deviation of 1,348 miles. What can it conclude if the

probability a Type I error is to be at most 0.01%?

Soln:

Ho: µ = x (there is no difference in population mean and sample mean)

H1: µ > x (There is significance difference in population mean and sample mean)

Left Tail Test.

LOS: α=0.01


Test Statistic: = √

= √

Conclusion:

Zα=-1.645

Zc< Zα

Accept Ho 3) Benzene in the air workers breathe can cause cancer. Samples are taken to check the benzene

content of the air. 35 specimens of air from one location in the plant gave a mean content of

0.760 ppm, and the standard deviation of benzene content was estimated on the basis of the

sample to be 0.45 ppm. Benzene contents in this case are found to be normally distributed.

Is there evidence at the 1% level of significance that the true mean of benzene content is equal to

1.00 ppm?

Soln: Ho: benzene content is equal to 1.00 ppm

H1: benzene content is less than 1.00 ppm

Left Tail Test.

Given: µ = 1 ppm, n = 35 x = 0.76 ppm, σ = 0.45 ppm

l.o.s α = 1%

Test Statistic: = √

= ..√

= - 3.15 Z 0.01 = -2.33 Zc < Z 0.01 Accept Ho


TEST OF HYPOTHESIS - LARGE SAMPLE - DIFFERENCE OF MEANS

Null Hypothesis : : !"### = !$###

Test Statistic Value : %& = !'" !'$()"$

*" +)$$*$

Alternative Hypothesis Rejection Region for Level α

": !"### > !$### % ≥ %. (upper-tailed)

": !"### < !$### % ≤ %. (lower – tailed)

": !"### ≠ !$### either % ≤ %.$ . or % ≥ %.$ (two – tailed)

PROBLEMS

1) 41 cars equipped with standard carburettors were for gas usage and yielded an average

of 8.1 km/litre with a standard deviation of 1.2 km/litre.21 of these cars were then chosen

randomly, fitted with special carburettors and tested, yielding an average of 8.8km/litre

with a standard deviation of 0.9 km/litre . At the 5% level of significance , does the new

carburettor decrease gas usage?

DATA GIVEN:

n₁ = 41

n₂ = 21

x₁ = 8.1 km/litre

x₂ =8.8 km/litre

σ₁ =1.2 km/litre

σ₂ = 0.9 km/litre₀

α = 5 %

SOLUTION:

H₀ : x₁ = x₂ ( no change )


H₁ : x₁ > x₂ ( new carburettor decreases gas usage )

Thus , from H₁ , it is a right tailed test.

Since α = 5% , Zα = 1.645

5 = 6 7₁7₂9:6;₁<₁ +;₂<₂ 9

= 6..9

:6. +.= 9

= ..>+.

= - 9.5

Now, Z > Zα at 5 % significance level.

Thus H₀ is rejected and H₁ is accepted .

Thus, the new carburettors decrease gas usage.

2) An investigation of 2 kinds of photocopying machine showed that 80 failures of 1st kind

of machine took average of 75.2 mins to repair with S.D of 20 mins when 80 failures of 2nd

machine took average of 82.8 mins to repair with S.D of 22 mins.Test the null hypothesis μ" = μ$ against μ"@μ$ at 5% L.O.S.

GIVEN:

A =75.2

B=20

C =80

A =82.8

B =22

C =80

1)D: A = A [There is no difference between two means]


2)D:A ≠ A [two tail test]

3)Level of significance α=5%

4)Test statistics (5)=

:EF + G where B = HE+HEH+H

=>. .

: F + G = -2.89

|5| =2.89

5)Table value:|5J|=1.96

CONCLUSION:

since|5| > |5J|

Reject D

Therefore, the difference between A andA is significant at 5%L.O.S

TEST OF HYPOTHESIS - LARGE SAMPLE - TEST FOR SINGLE PROPORTION

Null Hypothesis: : K = L

Test Statistic: = 6MN9:6OP9

where K = !* , P = probability of success of population.


": K > L % ≥ %. (upper-tailed)

": K < L % ≤ %. (lower – tailed)

": K ≠ L either % ≤ %.$ . or % ≥ %.$ (two – tailed)


PROBLEMS

1) A manufacture of light bulbs claims that on the average 2% of the bulbs manufactured

by him are defective. A random sample of 400 bulbs contained 13 defectives.on the basis of

this sample can you support the manufacturer’s claim at 5% LOS.

Solution:

H0: P = 0.02 i.e .2% of the products are defective

H1: p > P.

One tailed test(right tailed) test is to be used.

Let LOS be 5%.therefore ,Zα=1.645

= 6MN9:6OP9

, where R = H = = 0.0325

P=0.02, Q=1 - P=0.98

= 60.0325 − 0.029:60.02 ∗ 0.989400

Z=1.785

Z=1.79(approx)

Z > Zα

Therefore H0 is rejected

Therefore the claim cannot be supported.

2) A foundry produces steel forgings used in automobile manufacturing. We wish to test

the hypothesis that the fraction conforming or fallout from this process is 10%. In a

random sample of 250 forgings, 41 were found to be nonconforming. What are your

conclusions using α=0.05?

Solution:

H0:P=0.1 i.e.10%of the products are conforming.

H1:p≠P.

Two tailed test is to be used.


Let LOS be 5% .Therefore, Zα=1.96

= 6MN9:6OP9

, where R = H = > = 0.164

Z=(p-P)/√(PQ)/n

Where p=x/n=41/250=0.164

P=0.1, Q=1-P=1-0.1=0.9

= 60.164 − 0.19:60.1 ∗ 0.99250

Z=3.37

|Z|=3.37, Z=1.79(approx)

|Z| > |Zα|

H0 is rejected and H1 is accepted.

Thus the foundry produces steel forgings were found to be nonconforming.

3) A new rocket launching is considered for deployment of small and short range rockets.

The existing system has 80% successful launches. A sample of 40 experimental launches is

made with new system and 34 are successful. Would you claim that the new system is

better?

Soln:

Given: n=40, x=34, P=80%

Also n > 30 .So, this is Test for Large Samples and Test for Single proportion.

P==

>=0.8

Q=1-P =1->=

> =0.2

p= H =

= = 0.8


1) Null Hypothesis (Ho): There is no difference between proportions. p = P

2) Alternate Hypothesis (H1: There is difference between proportions. p>P

This is Single Right tail test.

3) Level of Significance(LOS):

α=5% =0.05 (Assuming)

4) Test Statistic (Zc):

Zc = MN:OP

Zc= .>.

:.×.

Zc=.>

:._

Zc=.>. >>>

Zc=0.791

4) Conclusion:

Zα=Z0.05=1.645

If α=1%

Zα=2.33


TEST OF HYPOTHESIS - LARGE SAMPLE - TEST FOR DIFFERENCE OF

PROPORTIONS

Null Hypothesis: : K" = K$

Test Statistic: = 6MM9:6OP9

where K" = !"*" , K$ = !$*$ , P = probability of success of population. L = *"!" +*$!$ *"+*$


": K" > K$ % ≥ %. (upper-tailed)

": K" < K" % ≤ %. (lower – tailed)

": K" ≠ K$ either % ≤ %.$ . or % ≥ %.$ (two – tailed)

1) A study shows that 16 out of 200 submersible pumps produced on one assembly line

required extensive adjustments before they could be shipped,while the same was true for 14

of 400 pumps produced on another assembly line.At 0.01 LOS,does this support the claim

that the second production line does superior work?

Solution:

16 fails out of 200 14 fails out of 400

R = 184200 = 0.92

R = = 0.965

H0: p1 = p2

H1: p1 < p2 One tailed (left tailed) test is to be used.

Let LOS be 0.01.i.e Zα = -2.33,

` = HM+HMH+H = 0.95


P=0.95

Q=0.05

Formula:

= R1 − R :`a F 1b1 + 1b2G

= 0.92 − 0.965:60.95 ∗ 0.059 ∗ F 1200 + 1400G

= −0.045:0.0475 ∗ F 60080000G

= .>∗ >.

= −2.3893.

Now |Z| > |Zα|

The difference between p1 and p2 is significant.

i.e.H0 is rejected and H1 is accepted.

i.e., This does not support the claim that the second production line.

2) A study shows that 16 out of 200 submersible pumps produced on one assembly line

required extensive adjustments before they could be shipped, while the same was true for

14 of 400 pumps produced on another assembly line. At 0.01 LOS,does this support the

claim that the second production line does superior work?

Solution:

In sample ( 1) 16 fails out of 200. In sample (2) 14 fails out of 400

So, R = p1=0.92 R = p2 = 0.965

H0: p1 = p2

H1: p1 < p2 One tailed (left tailed) test is to be used.

Let LOS be 0.01.i.e Zα = -2.33,


` = A + A C + C

` = 184 + 386600

` = 570600

P=0.95

Q=0.05

Formula:

= R − R :`a F 1C + 1C G

= 0.92 − 0.965:60.95 ∗ 0.059 ∗ F 1200 + 1400G

= −0.045:0.0475 ∗ F 60080000G

= .>∗ >.

= −2.3893.

Now |Z|>|Zα|

The difference between : p1 & p2 is significant.

i.e.H0 is rejected and H1 is accepted.

i.e., this does not support the claim that the second production line.


TEST OF HYPOTHESIS -SMALL SAMPLE - TEST FOR SINGLE MEAN

Null Hypothesis: : !' = e

Test Statistic: f = 69: gh

or f = 69:i

where !' = ∑ !* , k = "* ∑6! − !'9$, k = "*" ∑6! − !'9$ , P = probability of success of

population.

Alteative Hypothesis Rejection Region for Level α

": !' > e % ≥ %. (upper-tailed)

": !' < e % ≤ %. (lower – tailed)

": !' ≠ e either % ≤ %.$ . or % ≥ %.$ (two – tailed)

1) High sulphur content in steel is very undesirable, giving corrosion problems among

other disadvantages. If the sulphur content becomes too high, steps have to be taken. Five

successive independent specimens in a steel-making process give values of % sulphur of

0.0307, 0.0324, 0.0314, 0.0311, and 0.0307. Do the data give evidence at 5% los that the true

mean % sulphur is above 0.0300?

Soln.

Given data µ=0.0300, n = 5

Null Hypothesis H0: µ = 0.0300

Alternative Hypothesis H1: µ ≠ 0.0300

Calculation of sample mean and standard deviation:

x x-ẋ (x-ẋ)2

0.0307 -0.00056 3.136*10-7

0.0324 +0.00114 1.2996*10-6

0.0314 -0.00014 1.96*10-8

0.0311 -0.00016 2.56*10-8


0.0307 -0.00056 3.136*10-7

∑x = 0.1563

A = ∑ H = 0.03126

∑6A − A9= -2.8*10-4

∑6A − A9 =19.72*10-7

A = 0.03126, B = ∑ 6 9###H

f = g√

f = . .l.∗h√

tα = 3.25

t= 4.02

t > t0.05=2.31

Conclusion: Hence, the Null Hypothesis is rejected. The mean of data differ from true mean %

sulphur of 0.0300.

2) The average daily amount of scrap from a particular manufacturing process is 25.5 kg

with a standard deviation of 1.6 kg .A modification of the process is tried in an attempt to

reduce this amount .During a 10 day trial period ,the kg of scrap produced each day were

25,21.9,23.5,25.2,22,23,24.5,25,26.1,22.8. From the nature of the modification no change in

day to day variability of the amount of scrap will result. The normal distribution will

apply. A first glance at the figures suggest that the modification is effective in reducing he

scrap level .Does a significant test confirm this at the 1% level.

SOLUTION:

µ=25.5 kg

s=1.6 kg

n=10

The mean of the sample is given by,


X=∑

H=

m

=23.9

H0: X=µ

H1: X≠µ

Two tail test is to be used .Let LOS be 1%.

t=nµE/√H=

.m >.>./√

t=..> = -3.00

t0.01=3.16

ᵞ=n-1=9; |f| < f.

3 < 3.16

H0 is accepted and H1 is rejected at 1% level.

3) The standard deviation of a particular dimension on a machine part is known to be

0.0053 inches. Four parts coming off the production line are measured giving readings of

2.747, 2.740, 2.750, and 2.749 inches. The population mean is supposed to be 2.740 inches.

The normal distribution applies. Is the sample mean significantly larger than 2.740 inches

at the 1% level of significance?

GIVEN:

Standard deviation s= 0.0053 inches

n = 4

population mean µ = 2.740 inches

SOLUTION:

x =

H∑ Ap

=( 2.747+2.740+2.750+2.749)

= 2.7465 inches

D: A = q


There is no difference between sample & population mean

D: : A > q

One tail (Right tail) test is to be used. Let LOS be 1%.

t = rg√h

= 6 .> .9.s69

= .>.

=2.32

f = 2.32

γ= n-1

= 4-1

γ =3

Let L.O.S be 1%

For γ =3 f.= 5.841 ( from t-table)

We see that |f| < f. (γ =3)

Therefore ,D is accepted.

CONCLUSION:

There is no difference between sample and population mean .

4) A manufacturer of fluorescent lamps claims that his lamps have an average luminous flux of

3,600 lm at rated voltage and frequency and that 90% of all lamps produced by an automatic

process have a luminous flux higher than 3,300 lm. The luminous flux of the lamps follows a

normal distribution. What standard deviation is implied by the manufacturer’s claim? Assume

that this standard deviation does not change. A random sample of l0 lamps is tested and gives a

sample mean of 3,470 lm. At the 5% level of significance can we conclude that the mean

luminous flux is significantly less than what the manufacturer claims? State your null hypothesis

and alternative hypothesis.


Solution

Let X be a random Variable, corresponding to the lumens of the bulb

Population Mean q = 3600

`6t > 33009 = 0.9

`63300 < t < ∞9 = 0.9

This follows Normal distribution

Convert this to Standard Normal Distribution

= t − qv

Thus

` w3300 − 3600v < < ∞x = 0.9

` w0 < < 300v x = 0.4

From table

300v = 1.28

v = 234.375

The small sample test follows t distribution

Assume H0: There is no difference in the population and sample mean

Assume H1: The Population mean is greater than the sample mean (Right tail Test)

Assume LOS: LOS=0.05 at DOF=10-1=9

Test statistic:

f = A − qv√C

= .l√ = 0.945


This value is greater than the table value t0.05,9=2.626

Thus Accept H0.

We accept alternate hypothesis. And the claim of the company is false.

5) A new process has been developed for applying photoresist to 125-mm silicon wafers used in

manufacturing integrated circuits. The wafers were tested, and the following photoresist

thickness measurements ( angstroms X 1000) were observed: 13.3987, 13.3957, 13.3902,

13.4015, 14.4001, 13.3918, 13.3965, and 13.3925. Test the hypothesis that mean thickness is

13.4 X 1000 angstroms. use l.o.s of significance as 5%.

Ans:

6)The output of a power supply is assumed to be normally distributed. Sixteen observations

taken on voltages are as follows: 10.35, 9.3, 10.0, 9.96, 11.65, 12.00, 11.25, 9.58, 11.54, 9.95,

10.28, 8.37, 10.44, 9.25, 9.38, and 10.85. Test the hypothesis that mean voltage is equal to 12 V

against a two sided alternative using 5% l.o.s.

Solution:


7) A process of making certain ball bearing is under control if the diameter of the bearing has

mean of 0.5 cm. I f a random sample of 10 of these bearings has a mean diameter of 0.5060 cm.

and s.d of 0.0040 cm, is the calim under control?

µ=xH :0 , µ>xH :1 α = 0.005.

f = g√ = 4.7434

tα = 3.25

t > tα Reject H0

f. >,> = 2.57

|f| > f. >,>

H0 is rejected.

8) The electrical resistances of components are measured as they are produced. A sample of six

items gives a sample mean of 2.62 ohms and a sample standard deviation of 0.121 ohms. If the

population mean is 2.80 ohms. Test the hypothesis that whether there is any significant

difference in mean resistance at 5% l.o.s.


SMALL SAMPLE - TEST FOR DIFFERENCE OF MEANS

Null Hypothesis: : !"### = !$###

Test Statistic: f = 6########9:g +g

or

+

−+

+

−=

2121

2

22

2

11

21

11

2 nnnn

snsn

xxt

where !' = ∑ !* , k = "* ∑6! − !'9$, k = "*" ∑6! − !'9$ , P = probability of success of

population.

Alteative Hypothesis Rejection Region for Level α

": !' > e % ≥ %. (upper-tailed)

": !' < e % ≤ %. (lower – tailed)

": !' ≠ e either % ≤ %.$ . or % ≥ %.$ (two – tailed)

PROBLEMS

1) Consider the following problem. The Engineering Department at Sims Software, Inc.,

recently developed two chemical solutions designed to increase the usable life of computer

disks. A sample of disks treated with first solution lasted 86, 78, 66, 83, 84, 81, 84, 109, 65,

and 102 hours. Those treated with the second solution lasted 91, 71, 75, 76, 87, 79, 73, 76,

79, 78, 87, 90, 76, and 72 hours. At the 0.10 significance level, can we conclude that there is

a difference in the length of time the two types of treatment lasted?

SOLUTION: In this problem, we do not know the mean and standard deviation for the samples.

Thus, we calculate the mean and standard deviation for each of the samples before beginning our

5-step procedure. Using the TI-84 calculator or Statdisk, we find the means and standard

deviations (rounded to one-decimal place accuracy) as listed in the table below.

Treatment Sample Size Mean Standard Deviation

1 10 83.8 13.7

2 14 79.3 6.7


Step 1

Null hypothesis : D: A### = A ###

From “…there is a difference…,” we write the alternate hypothesis as

Step 2

D: A### ≠ A ###

Step 3.

Select a level of significance.

Stated in the problem as 10% or α = 0.10

Step 4: Test Statistic

Since the two sets of data are not related or paired or matched, we consider the two samples to be

independent.

Thus, we calculate the test statistic with the following formula.

( )

2

2

2

1

2

1

21

n

s

n

s

xxt

+

−=

( )

9599.0

14

7.6

10

7.13

3.798.83

22=

+

−=t

For the degrees of freedom, we use the smaller of n1 – 1 or n2 – 1. Since n1 – 1 = 10 – 1 = 9 and

n2 – 1 = 14 – 1 = 13, we use df = 9 and α = 0.10 for a two-tailed test to find critical values

t0.10,9,13 = 1.833, from Table

Since the computed t = 0.960 is between the critical t-values of –1.833 and 1.833, we do not

reject the null hypothesis that the means are statistically the same or equal at the 10% level of

significance. Thus, we conclude that the lengths of time the two types of treatment lasted are

equal.


2. Two companies produce resistors. Resistors from company A give a sample of size 9 with

sample mean 4025 ohms and estimated standard deviation 42.6 ohms. A shipment from

company B gives a sample of size l3 with sample mean 3980 ohms and estimated standard

deviation 30.6 ohms. Resistances are approximately normally distributed.

At 5% level of significance, is there a difference in the mean values of the resistors produced by

the two companies?

Solution:

Given:

µ = 4000Ω A### = 4025Ω

n1 = 9 ; A ### = 3980Ω

n2 = 13 ; s1 = 42.6Ω

; s2 = 30.6Ω

n1 < 30 ; n2 < 30

So this is test for small sample and normally distributed. So, t-test for difference mean.

H0 : A### = A ###( There is no difference in mean )

H1 : A### ≠ A ### ( There is difference in mean )

( Two tail test)

f = ####:EF + G where B = 6H9E+6H9E

H+H

B = 6m9 .+69.m+ = 1287.72

f = 4025 − 3980:1287.72 F19 + 112G


t = 2.89

Table value:

γ = n1 + n2 -2 = 20

One tail, so t(α,γ) = t0.05,20 = 2.09

Since t > tα,γ

So H0 is rejected.

3) Two chemical processes for manufacturing the same product are being compared under

the same conditions. Yield from Process A gives an average value of 96.2 from six runs, and

the estimated standard deviation of yield is 2.75. Yield from Process B gives an average

value of 93.3 from seven runs, and the estimated standard deviation is 3.35. Yields follow a

normal distribution. Is the difference between the mean yields statistically significant? Use

the 5% level of significance, and show rejection regions for the difference of mean yields on

a sketch. (12E11)

Solution:

.35.3,3.93,7,6,75.2,2.96 222111 ====== sxnnsxHere

211210 :: xxHandxxH ≠=

Two tail test to be used. LOS = 5%.

+

−+

+

−=

2121

2

22

2

11

21

11

2 nnnn

snsn

xxt

+

−+

×+×

−=

7

1

6

1

276

))35.3(7())75.2(6(

3.932.96

22

( )309.011

558.78375.45

9.2

+=


865.1

9.2=

t = 1.55

Also 11276221 =−+=−+= nnν

11=ν , 05.0=α

From the t-table,

20.211,05.0, == tt να

)11,05.0(tt <

So H0 is accepted and H1 is rejected.

That is two sample means are do not differ significantly at 5% LOS.

4) Two companies produce resistors with a nominal resistance of 4000 ohms. Resistors

from company A give a sample of size 9 with sample mean 4025 ohms and estimated

standard deviation 42.6 ohms. A shipment from company B gives a sample of size l3 with

sample mean 3980 ohms and estimated standard deviation 30.6 ohms. Resistances are

approximately normally distributed.

At 5% level of significance, is there a difference in the mean values of the resistors produced by

the two companies?

Soln:

Given:

µ = 4000Ω A### = 4025Ω

n1 = 9 ; A ### = 3980Ω

n2 = 13 ; s1 = 42.6Ω

; s2 = 30.6Ω

n1 < 30 ; n2 < 30

So this is test for small sample and normally distributed. So, t-test for difference mean.

H0 : A### = A ###( There is no difference in mean )

H1 : A### ≠ A ### ( There is difference in mean )


( Two tail test)

LOS: α = 5% = 0.05

f = ####:EF + G where B = 6H9E+6H9E

H+H

B = 6m9 .+69.m+ = 1287.72

f = 4025 − 3980:1287.72 F19 + 112G

t = 2.89

Table value:

γ = n1 + n2 -2 = 20

One tail, so t(α,γ) = t0.05,20 = 2.09

Since t > tα,γ

So H0 is rejected.

SMALL SAMPLE - DIFFERENCE OF VARIANCES

1) Two different lighting techniques are compared by measuring the intensity of light at selected

locations in areas lighted by the two methods.If 15 measurement in the first area ad a S.D of 2.7

foot candles and 21 measurement in the second area had a S.D of 4.2 foot candles can it be

concluded that the lighting in the second area is less uniform?Use 5% l.o.s.

Solution:

Given: S=2.7 foot candle, S =4.2 foot candle, n=1, n =21, α=5%

First we set up a null hypothesis

H: σ =σ

H: σ@ σ


σ = h

= >>

= >

= 1.07

σ =h

=

=

=1.05

σ >σ

F=σσ

=..>

=1.01

F=1.01

F.>,6, 9=2.28

F < Fα Accept Null Hypothesis

Yes.It can be concluded that the lighting in the second is less uniform.

3)The following are the Brinell hardness values obtained for samples of two magnesium

alloys before testing.

Alloy 1: 66.3 63.5 64.9 61.8 64.3 64.7 65.1 64.5 68.4 63.2

Alloy 2: 71.3 60.4 62.6 63.9 68.8 70.1 64.8 68.9 65.8 66.2

Test whether the two samples came from same normal population.


Solution:

Mean A### = ∑ H

= .

=64.67

A ### = A C

= .

= 66.28

B = ∑ H − F∑ HG

=15.80

B2 = ∑ HH − F∑ HG

= 10.93

v = HEH

= 6>.9m

= 17.56

v = HEH

= 6.m9m

= 12.14

= vv

= .>.

= 1.44


H0 )" = )$

H1 )" ≠ )$

.F 0.05, (9,9)=1.58

Accept Ho

t= 6!" − !$9 ÷ √(k"$ − k$$)/√* − "

= −". " ÷ √17.56-12.14/√9

=2.20

. t(., $* − $)

. t5%(18)= $. ""

Reject Ho

They are not from same population

CHI SQUARE DISTRIBUTION

1. Three different shops are used to repair electric motors. One hundred motors were sent for sent

for each shop. When a motor is returned, it is put in use and then the repair is classified as

complete, requiring an adjustment or an incomplete repair. The column totals are fixed 100 each

and the grand total is 300. Shop I produced 78 complete repairs, 15 minor adjustments, and 7

incomplete repairs. Shop II produced 56, 30, and 14 respectively; while Shop III produced 54,31,

15 complete, minor adjustment, and incomplete repairs respectively. Is here any significant

difference between repairs and Shops at 5% l.o.s.

Solution:

Given data

shop I Shop II Shop III Total

Repaired 78 56 54 188

Repair with Minor

adjustment

15 30 31 76

Without repair 7 14 15 36


Total 100 100 100 300

H0-There is no significance difference between the shop and repair

H1- There is significance difference between shop and repair

Oi Ei Oi – Ei ( − ) ²

p

78 (188*100)/300=62.6 15.4 3.780

56 (188*100)/300=62.6 6.6 0.695

54 (188*100)/300=62.6 8.6 1.180

51 (76*100)/300=25.3 10.3 4.193

30 (76*100)/300=25.3 4.7 0.873

31 (76*100)/300=25.3 5.7 1.284

7 (36*100)/300=12 5 2.083

14 (36*100)/300=12 2 0.333

15 (36*100)/300=12 3 0.750

χ²= 15.171

At 5% l.o.s. from table value, χ² with d.o.f. (γ=8) = 15.507

Calculated χ² value =15.171

Calculated value< table value

H0 is accepted.

2) The following table gives the two wheeler accidents occurred during one week.

Days: Sun Mon Tue Wed Thu Fri Sat

Accidents: 14 16 8 12 11 9 14


Can you say that the accidents are uniformly distributed over the week? Use 5% LOS.

Solution:

H0: Accidents occur uniformly over the week.

H1 : Accidents does not occur uniformly over the week.

l.o.s α = 5%

Total number of accidents=84

Based on H0 the expected number of accidents on any day=84÷7=12

Oi 14 16 8 12 11 9 14

Ei 12 12 12 12 12 12 12

$ =Ʃ 69$

=1/12(4+16+16+0+1+9+4)

=4.166

Since ƩEi= ƩOi,

d.o.s = γ = 7-1=6.

From the $ table, .$ (γ =6)=12.592.

Since $ < .$ , H0 is accepted .That is accidents may be regarded to occur uniformly over the

week.


3. The following table shows the opinions of voters before and after a Prime Minister election.

Before After Total

For Ruling Party 79 91 170

For Opposition Party 84 66 150

Undecided 37 43 80

Total 200 200 400

Test at 5% LOS whether there has been any change of opinion of voters.

Solution:

H0:NULL HYPOTHESIS:OPINION OF VOTERS ARE INDEPENDENT

DEGRESS OF FREEDOM: σ =(2-1)(3-1)=(1)(2)=2

χ20.05 (σ=2)=5.991

Since χ 2 > χ2

0.05, H0 is rejected. That is, there has been change in the opinion of voters.

O

E ROUNDED E 6 − 9

79 170*200/400 85 (-6)2/85=0.42

91 170*200/400 85 62/85=0.42

84 150*200/400 75 92/75=1.08

66 150*200/400 75 (-9)2/75=1.08

37 80*200/400 40 (-3)2/40=0.25

43 80*200/400 40 32/

/40=0.25

χ2=6.5


4) To determine whether there really is a relationship between an employee’s performance in the

company’s training program and their ultimate success in the job, the company takes a sample of

400 cases from its extensive files and obtain the following results.

Performance

Success

in job

(Employer’s

raring)

Below

Average

Average Above

Average

Poor 23 60 29

Average 28 79 60

Very Good 9 49 63

Use 1% l.o.s to test the hypothesis that is there any relation between performance in the training

and the success in the jobs.

Sol:

Setup null hypothesis:

Ho: There is no relation between performance in the training and the success in the jobs.

H1: There is a relation between performance in the training and the success in the jobs.

Test Statistic : = ∑ 69

χ2 Table:

O rounded E O - E (O- E)2

6 − 9

23 17 6 36 2.11765

60 53 7 49 0.92453

29 43 -14 196 4.55814

28 25 3 9 0.36

79 78 1 1 0.01282

60 63 -3 9 0.14286

9 18 -9 81 4.5

49 57 -8 64 1.12281

63 46 17 289 6.28261

χ2 = 20.0214


Calculated value for χ2=

20.0214

Table value for χ2 (0.05,4)=9.488

Calculated value is greater than the table value.

Therefore, there is a relation between performance in the training and the success in the jobs.

5) Absenteeism of Fourth Semester students from Mathematics IV classes is a major concern to

math instructor because missing class appears to increase. The instructor wondered whether the

absentee rate was the same for every day of the school week. He took a sample of absent

students from three of their Mathematics IV classes during one week of the term. The results of

the survey appear in the table.

Day Mon day Tuesday Wednesday Thursday Friday

No. of Students

Absent

28 22 18 20 32

Determine the null and alternate hypotheses needed to run a goodness-of-fit test.

Solution:

H0 : The absentee rate was the same for every day of the school week.

H1 : the absentee rate was not the same for every day of the school week

l.o.s : α = 5%

ARf BCfB = H EHE H ¡¢E £¤ = > = 24

Observed No.

of Students

Absent

28 22 18 20 32

= 4.084

Expected 24 24 24 24 24

(O – E) 4 -2 -6 -4 8 6 − 9$

0.667 0.167 1.5 o.667 2.667

Table value χ2 (0.05, 4) = 9.488

Since χ2

< χ2 (0.05, 4).

Accept Null Hypothesis.

The absentee rate was uniformly distributed over all week days.


PART A QUESTIONS

1. Define critical region and acceptance region.

2. Mention the steps in the process of testing of hypotheses.

3. Discuss the properties of Chi-Square Distribution.

4. What is Type I and Type II error?

5. Mention the properties of normal distribution curve.

6. Retell the properties of t distribution curve.

7. Distinguish between level of significance and degrees of freedom.

8. Mention the properties of F distribution.

9. In a sample of 600 parts manufactured by a factory, the number of defective parts was

found to be 45. The company claims that only 5% of their products is defective. Is the

claim tenable at 5% LOS.?

10. To Check 108 ambulance service’s claim that at least 40% of its calls are life threatening

emergencies, a random sample was taken from its files and it was found that only 49 out

of 150 calls were life threatening emergencies. Can you justify the claim at 5% LOS.

11. State the advantages of (a) t- distribution (b) F distribution (c) Chi square distribution.

12. Mention the two types in Chi square test.

13. What do you mean by goodness of fit.

lecture notes test of hypothesis

Documents

statistical hypothesis

opposite of null hypothesis

null value

h0 type

population n

test statistic value

h0 correct decision

given population