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General Procedure for Test of Hypotheses for Population Mean (Population Variance Is Known) : Procedure Steps, Example Solved Problems

**TEST OF HYPOTHESES FOR POPULATION MEAN (Population Variance Is
Known)**

*Procedure:*

**Step 1 : **Let** ***µ*** **and** ***σ*^{2}** **be
respectively the mean and the variance of the population under study,** **where
*σ*^{2} is known. If *µ _{0}* is an admissible value
of

(i)* H _{1}: µ ≠ µ_{0}* (ii)

**Step 2 : **Let (*X*** **_{1}, X_{2}** ***,
…, X _{n}*) be a random sample of

**Step 3 : **Let the level of significance be** ***α*.

Step 4 : Consider the test statistic under H_{0}.
Here, represents the sample mean,
which is defined in Note 2. The approximate sampling distribution of the
test statistic under *H** _{0}* is the

**Step 5 : **Calculate the value of** ***Z*** **for the given sample
(*x*_{1},** ***x*_{2}, ...,** ***x*_{n})
as

**Step 6 : **Find the critical value,** ***z _{e}*,
corresponding to

**Step 7 : **Decide on** ***H _{0}*

A company producing LED bulbs finds that mean life span of the
population of its bulbs is 2000 hours with a standard derivation of 150 hours.
A sample of 100 bulbs randomly chosen is found to have the mean life span of
1950 hours. Test, at 5% level of significance, whether the mean life span of
the bulbs is significantly different from 2000 hours.

**Step 1 : **Let** ***μ*** **and** ***σ*** **represent
respectively the mean and standard deviation of the probability** **distribution
of the life span of the bulbs. It is given that *σ* = 150 hours. The null
and alternative hypotheses are

**Null hypothesis: ***H*_{0}:** ***μ*** **= 2000

*i.e*., the mean life span of the bulbs is not significantly different
from 2000 hours.

**Alternative hypothesis: ***H*_{1}** **:** ***μ ≠*** **2000

*i.e*., the mean life span of the bulbs is significantly different from
2000 hours.

It is a two-sided alternative hypothesis.

**Step 2 : ****Data**

The given sample information are

Sample size (*n*) = 100, Sample mean (* x*) = 1950
hours

**Step 3 : ****Level of significance **= 5%

**Step 4 : ****Test statistic**

The
test statistic is , under H_{0}

Under
the null hypothesis* H _{0}, Z*
follows the

**Step 5 : Calculation of Test Statistic**

The
value of *Z* under *H*_{0} is calculated from

= –3.33

Thus;
|*z*_{0}| = 3.33

**Step 6 : ****Critical value**

Since *H _{1}* is a two-sided alternative, the
critical value at

(*see* Table 1.6).

**Step 7 : ****Decision**

Since *H _{1}* is a two-sided alternative, elements of
the critical region are determined by the rejection rule |

**Example 1.8**

The mean breaking strength of cables supplied by a manufacturer is
1900 *n/m ^{2}* with a standard deviation of 120

*Solution:*

**Step 1 : **Let** ***μ*** **and** ***σ*** **represent
respectively the mean and standard deviation of the probability** **distribution
of the breaking strength of the cables. It is given that *σ* = 120 *n/m ^{2}*.
The null and alternative hypotheses are

**Null hypothesis ***H*_{0}:** ***μ*** **= 1900

*i.e*., the mean breaking strength of the cables is not significantly
different from* *1900*n/m ^{2}*.

**Alternative hypothesis: ***H*_{1}:** ***μ*** **> 1900

*i.e*., the mean breaking strength of the cables is significantly more
than 1900*n/m ^{2}*.

It may be noted that it is a one-sided (right) alternative hypothesis.

**Step 2 : ****Data**

The given sample information are

Sample size (*n*) = 60. Hence, it is a large sample.

Sample mean ()= 1960

**Step 3 : ****Level of significance**

α= 1%

**Step 4 : ****Test statistic**

The
test statistic is , under H_{0}

Since *n* is large, under the null hypothesis, the sampling
distribution of *Z* is the *N*(0,1) distribution.

**Step 5 : ****Calculation of test statistic**

The value of Z under H_{0} is calculated from z_{0}
=

Thus, z_{0} = 3.87

**Step 6 : ****Critical value**

Since *H _{1}* is a one-sided (right) alternative
hypothesis, the critical value at

**Step 7 : ****Decision**

Since *H _{1}* is a one- sided (right) alternative,
elements of the critical region are determined by the rejection rule

Thus, the manufacturer’s claim that the breaking strength of
cables has increased by the new technique is found valid.

Tags : Procedure Steps, Example Solved Problems | Statistics , 12th Statistics : Chapter 1 : Tests of Significance - Basic Concepts and Large Sample Tests

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12th Statistics : Chapter 1 : Tests of Significance - Basic Concepts and Large Sample Tests : Test of Hypotheses for Population Mean (Population Variance Is Known) | Procedure Steps, Example Solved Problems | Statistics

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