HYPOTHESIS TESTING EXAMPLES

TEST OF HYPOTHESIS CONCERNING NORMAL POPULATION,

INFINITE, LARGE SAMPLES, WITH KNOWN POPULATION

VARIANCE.

( Z TEST)

If the size of sample exceeds 30 it should be

regarded as a large sample.

a) Testing Hypothesis about population mean μ.

Z= x - μ

S.E. of x

Z= x– μ

σp/√n x = sample mean

σp = population s.d.n= sample size

Population size is infinite

.

SOME HYPOTHESIS TESTING EXAMPLES

ILLUSTRATION1 (TWO-TAILED TEST)

The mean lifetime of a sample of 100 light tubes produced by a

company is found to be 1,580 hours Test the hypothesis at 5% level

of significance that the mean lifetime of the tubes produced by the

company is 1,600 hours with standard deviation of 90 hours.

Solution:

z= -2.22.

Critical value of Z = ± 1.96

ILLUSTRATION2 ONE TAILED (UPPER TAILED)

An insurance company is reviewing its current policy rates.

When originally setting the rates they believed that the average

claim amount will be maximum Rs180000. They are concerned

that the true mean is actually higher than this, because they

could potentially lose a lot of money. They randomly select 40

claims, and calculate a sample mean of Rs195000. Assuming

that the standard deviation of claims is Rs50000 and set

α= .05, test to see if the insurance company should be

concerned or not.

SOLUTION

Step 1: Set the null and alternative hypotheses

H0 : μ≤ 180000

H1 : μ > 180000

Step 2: Calculate the test statistic

z= = x – μ

σ/√n

= 1.897

Step 3: Set Rejection Region

1.65

Step 4: Conclude

We can see that 1.897 > 1.65, thus our test statistic

is in the rejection region. Therefore we fail to

accept the null hypothesis. The insurance company

should be concerned about their current

policies.

ILLUSTRATION3 ONE TAILED (LOWER TAILED)

Trying to encourage people to stop driving to

campus, the university claims that on average it

takes at least 30 minutes to find a parking space on

campus. I don’t think it takes so long to find a spot.

In fact I have a sample of the last five times I drove

to campus, and I calculated x = 20. Assuming that

the time it takes to find a parking spot is normal,

and that σ = 6 minutes, then perform a hypothesis

test with level α= 0.10 to see if my claim is correct.

SOLUTION

Step 1: Set the null and alternative hypotheses

H0 : μ ≥ 30

H1 : μ < 30

Step 2: Calculate the test statistic

Z= x– μ

σ/√n

= -3.727

STEP 3: SET REJECTION REGION

STEP 4: CONCLUDE

We can see that -3.727 <-1.28 ( or absolute value is

higher than the critical value) , thus our test

statistic is in the rejection region. Therefore we

Reject the null hypothesis in favor of the alternative.

We can conclude that the mean is significantly less

than 30, thus I have proven that the mean time to

find a parking space is less than 30.

EXERCISE

REJECTION REGION AND CONCLUSION

TEST OF HYPOTHESIS CONCERNING NORMAL

POPULATION, FINITE, LARGE SAMPLES, WITH

KNOWN POPULATION VARIANCE.

( Z TEST)

Z= x– μ

( σ/√n) √(N-n)/(N-1)

N= population size

ILLUSTRATION 4

Suppose we are interested in a population of 20

industrial units of the same size, all of which are

experiencing excessive labour turnover problems.

The past record show that the mean annual turnover

is 320 employees, with a standard deviation of 75

employees. A sample of 5 of these industrial units is

taken at random which gives a mean of annual

turnover as 300 employees. Is the sample mean

consistent with the population mean? Test at 5% level.

TESTING HYPOTHESIS ABOUT DIFFERENCE

BETWEEN TWO POPULATION MEANS

We assume that the populations are normally

distributed.

The null hypothesis is H0 : μ1 = μ2

i.e. H0 : μ1 -μ2 = 0

Z = x1 - x2

√σ1 2/ n1 + σ2

2 / n2

In case σ1 2 and σ2

2 are not known then s1 2 and s2

2 can be

used.

ILLUSTRATION 1

A test given to two groups of boys and girls gave

the following information:

Gender Mean score S.D. Sample Size

Girls 75 10 50

Boys 70 12 100

Is the difference in the mean scores of boys and girls statistically significant?

Test at 1% level.

Z=2.695, table value Z =2.58.

ILLUSTRATION 2

Suppose you are working as a purchase manager

for a company. The following information has been

supplied to you by two manufacturers of electric

bulbs:

Company A Company B

Mean life ( in hours) 1,300 1,288

Standard deviation( in hours) 82 93

Sample size 100 100

Which brand of bulbs are you going to purchase if you desire to take

a risk of 5%

SOLUTION HINT

Take the null hypothesis that there is no significant difference

In the quality of two brands of bulbs i. e. H0 : μ1 = μ2

Z= 0.968

POPULATION NORMAL, POPULATION INFINITE,

SAMPLE SIZE SMALL (30 OR LESS) AND VARIANCE

OF POPULATION UNKNOWN

t statistic.

t = x – μ

( σs/√n) with degree of freedom =(n-1)

σs = ∑(X-X)2

(n-1)

ILLUSTRATION

A tea stall near the New Delhi Railway Station is

making a sales of 500 tea cups per day. Because of

the development of bus stand nearby , it expects to

increase its sales. During the first 12 days after the

start of the bus stand, the daily sales were recorded

which are as under:

550, 570,490, 615,505,580,570,460,600,580,530,526.

On the basis of this sample information, can one

conclude that the tea stall’s sales have increased?

Use 5% level of significance.

t= 3.558

TESTING HYPOTHESIS ABOUT DIFFERENCE

BETWEEN TWO POPULATION MEANS

We assume that the populations are normally

distributed.

Variance of population is unknown.

The null hypothesis is H0 : μ1 = μ2

Small Sample t test

t = x 1 – x 2s√ (1/n1 + 1/n2)

s= ∑(x1 - x1 ) 2 + ∑(x2 – x2 )

2

(n1 + n2 -2 )

with degree of freedom = (n1 + n2 -2 )