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# Test of Hypotheses for Population Mean (Population Variance Is Known)

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 σ2 be respectively the mean and the variance of the population under study, where σ2 is known. If µ0 is an admissible value of µ, then frame the null hypothesis as H0: µ = µ0 and choose the suitable alternative hypothesis from

(i) H1: µ ≠ µ0 (ii) H1: µ > µ0 (iii) H1: µ < µ0

Step 2 : Let (X 1, X2 , …, Xn) be a random sample of n observations drawn from the population, where n is large (n ≥ 30).

Step 3 : Let the level of significance be α.

Step 4 : Consider the test statistic under H0. Here, represents the sample mean,  which is defined in  Note 2. The approximate sampling distribution of the test statistic under H0 is the N(0,1) distribution.

Step 5 : Calculate the value of Z for the given sample (x1, x2, ..., xn) as Step 6 : Find the critical value, ze, corresponding to α and H1 from the following table Step 7 : Decide on H0 choosing the suitable rejection rule from the following table corresponding to H1. ### Example 1.7

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.

### Solution:

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: H0: μ = 2000

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

Alternative hypothesis: H1 : 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 H0

Under the null hypothesis H0, Z follows the N(0,1) distribution.

Step 5 : Calculation of Test Statistic

The value of Z under H0 is calculated from = –3.33

Thus; |z0| = 3.33

Step 6 : Critical value

Since H1 is a two-sided alternative, the critical value at α = 0.05 is ze = z0.025 = 1.96.

(see Table 1.6).

Step 7 : Decision

Since H1 is a two-sided alternative, elements of the critical region are determined by the rejection rule |z0| ≥ ze . Thus, it is a two-tailed test. For the given sample information, the rejection rule holds i.e., |z0| = 3.33 > ze = 1.96. Hence, H0 is rejected in favour of H 1: μ ≠ 2000. Thus, the mean life span of the LED bulbs is significantly different from 2000 hours.

Example 1.8

The mean breaking strength of cables supplied by a manufacturer is 1900 n/m2 with a standard deviation of 120 n/m2. The manufacturer introduced a new technique in the manufacturing process and claimed that the breaking strength of the cables has increased. In order to test the claim, a sample of 60 cables is tested. It is found that the mean breaking strength of the sampled cables is 1960 n/m2. Can we support the claim at 1% level of significance?

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/m2. The null and alternative hypotheses are

Null hypothesis H0: μ = 1900

i.e., the mean breaking strength of the cables is not significantly different from 1900n/m2.

Alternative hypothesis: H1: μ > 1900

i.e., the mean breaking strength of the cables is significantly more than 1900n/m2.

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 H0

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 H0 is calculated from z0  Thus, z0 = 3.87

Step 6 : Critical value

Since H1 is a one-sided (right) alternative hypothesis, the critical value at α = 0.01 level of significance is ze = z0.01= 2.33 (see Table 1.6)

Step 7 : Decision

Since H1 is a one- sided (right) alternative, elements of the critical region are determined by the rejection rule z0 > ze. Thus, it is a right-tailed test. For the given sample information, the observed value z0 = 3.87 is greater than the critical value ze = 2.33. Hence, the null hypothesis H0 is rejected. Therefore, the mean breaking strength of the cables is significantly more than 1900 n/m2.

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