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(1) Example 1 : The lifetime of a certain brand of electric bulb may be
considered a random variable with mean 1200 hours and standard
deviation 250 hours. Using the central limit theorem find the probability
that the average lifetime of 60 bulbs exceeds 1250 hours.
(M.U. 2004, 11)
Sol. : If denotes the average (mean) lifetime of 60 bulbs then by the
central Limit Theorem,
Is a S.N.V. with
If
area between
(2) Example 2 :If
are Poisson variates with the parameter
, use the central limit theorem to estimate
where
and
(M.U. 2010)
Sol. : We know that for a Poisson variate
and Var
If
then by central limit theorem
variate with mean
is anormal
and variance
is a S.N.V.
When
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When
(3) Example 3 : A distribution has unknown mean and variance
Using central limit theorem find the size of the sample such that the
probability that difference between sample mean and the population mean
will be less than
.
(M.U. 2004)
Sol. : We have
and Var
.
If is the sample mean then
is a S.N.V.
We know from then table that
Hence, n must be at least 24.
(4) Example 4 : A random sample of 50 items gives the mean
and
standard deviation
. Can it be regarded as drawn from a normal
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population with mean
at 5% level of significance ?
(M.U. 1996)
Sol. :
(i)
Null hypothesis
Alternative hypothesis
(ii) Test Statistic : Since the population S.D. is unknown but
sample S.D. s is known and since sample is large
(iii)
(iv)
(v)
Level of Significance :
Critical value : The value of
at 5% level of significance
from the table
.
Decision : Since the computed value of
is less than
the critical value
, the null hypothesis is accepted.
The sample is drawn from the population with mean
(5) Example 5 : Can it be concluded that the average life-span of an Indian is
more that 70 years, if a random sample of 100 Indians has an average lifespan of
years with standard deviation of
years ?
(M.U. 2004)
Sol. :
(i)
Null Hypothesis :
years.
Alternative Hypothesis
years
(ii)
Test Statistic :
Since we are given standard deviation of the sample, we put
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(iii)
Level of Significance :
(iv)
Critical value : The value of
(v)
Decision : Since the computed value of
is greater
than the critical value
, the null hypothesis is
rejected.
The hypothesis is rejected.
at 5% level of significance is
(6) Example 6 : A tyre company claims that the lives of tyres have mean
42,000 kms with S.D. of 4000 kms. A change in the production process is
believed to result in better product. A test sample of 81 new tyres has a
mean life of 42,000 kms. Test at 5% level of significance that the new
product is significantly better than the oild one.
(M.U. 2006, 09)
Sol. :
(i)
Null Hypothesis :
Alternative Hypothesis
(ii) Test Statistic : Since the population S.D. is unknown but
sample S.D. s is known and since the sample is lrage
(iii)
Level of Significance :
(iv)
Critical value : The value of
(v)
Decision : Since the computed value of
is less
than the critical value
, the null hypothesis is
accepted.
There is no improvement.
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at 5% level of significance is
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(7) Example 7 : The average of marks scored by 32 boys is 72 with standard
deviation 8 while that of 36 girls is 70 with standard deviation 6. Test at
1% level of significance whether the boys perform better than the girls.
(M.U. 2004, 09, 10)
Sol. :
(i)
Null Hypothesis :
Alternative Hypothesis
(ii) Calculation of statistic :
(We assume that the standard deviation
of the two
populations are not equal.)
(iii)
Level of Significance :
(iv)
Critical value : The value of
at 1% level of significance
from the table is 2
Decision : Since the computed value of
is less
than the critical value
, the hypothesis is accepted.
Boys do not perform better than the girls .
(v)
(8) Example 8 : Two population have the same mean but the standard
deviation of one is twice that of the other. Show that in samples, each of
size 500, drawn under simple random conditions the difference of the
means, in all probability, will not exceed
, where is the smaller
standard deviation.
(M.U. 2007)
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Sol. : we have
We want
Dividing both sides by S.E.,
For S.N.V.
almost impossible. Hence, for
be
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area lies under the curve. The event is
in all probability will
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