**(a) Testing of A Single Mean. (Inferences about a single mean)**

1. Set up H_{0} : *µ *= *µ _{0}*

2. Set up H_{1} : *µ *> *µ _{0}* or

*µ*<

*µ*or

*µ ≠ µ*

_{0}which follows standard normal distribution.

4. Set up the level of significance a. and the critical value as Z_{tab }from the normal table.

5. Compute the statistic, say Z_{cal}

**Note.** If the population S.D. (σ) is not known for large sample we can take the sample S.D. (S) in the test statistic.

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**Example **2. *The mean lifetime of I 00 picture tubes produced by a manufacturing company is estimated to be 5795 hours with a standard deviation of 150 hours. If µ be the mean lifetime of all the picture tubes produced by the company, test the hypothesis *J* µ* = *6000 hours against µ:F. 6000 hours at 5 *% *level of significance.*

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**Solution. **

- H
_{0}:*µ*= 6000 hours - H
_{1}:*µ:F.*6000 hours

Test statistic : Here *n *= 100 *i.e., *large sample. Population S.D. is not given but the sample S.D. is given as 150 hours.

4. α = 0.05, α /2 = 0.025

H_{1} shows, this is two tailed test.

Z α /2 = 1.96 so the critical values are 1.96 and -1.96.

6. Decision :

Since Z_{cal} < - 1.96 *i.e., *it lies in the rejection region

=> H_{0} is rejected

=> The claim produced by the company is not true.

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**Example **3. *A tire company claims that the lives of the tires have mean of 42000 kilometers with standard deviation of 4000 kilometers. A change in the production process is believed to result in a better product. A test sample of 81 new types has a mean life of 42500 kilometers. Test at 5% level of significance that the new product is significantly better than the current one ?*

** **

**Solution. **

- H0 : µ= 42000 km.
- H1 : µ > 42000 km.

- Alt. hypothesis shows, this is right tailed test.

α= 0.05, Z_{α }= 1.64 which is the critical value.

6. Decision :

Since Z_{ca1} < Z_{α} => Zeal lies in the acceptance region

=> H_{0} is accepted

=> H_{1} is rejected

::::> New product is not significantly better than the current one.

(b) **Testing of Difference of Two Means.** Consider two populations having the means µ_{1 }and µ_{2} and the variance σ_{1}^{2} and σ_{2}^{2} respectively.

- Set up Ho : µ
_{1 }- µ_{2}=*k* - Set up H
_{1}: µ_{1 }and µ_{2}>*k*or µ_{1 }and µ_{2}<*k*or µ_{1 }and µ_{2}≠*k*where*k*is a specified constant. - Set up the test statisticwhere, x
_{1}= Means of sample size n1 from the first population.X2 = Means of sample size

*n2*from the second population.4. Set up the level of significance a. and the critical value as Z13b from the normal table.

5. Compute the statistic, says Z

_{cal}.**Note.**When both n1 and n2 are greater than or equal to 30, the population S.D. can be estimated by sample S.D. in the statistic.**Example 4.***A random sample of 100 villages was taken from a district A and the average height of the population per village was found to be 170 cm with a standard deviation of 10 cm. Another random sample of 120 villages was taken from another district B and the average height of the population per village was found to be 176 cm with a standard deviation of 12 cm. Is the difference between averages of the two populations statistically significant ?***Solution.**- Ho : µ
_{1 }- µ_{2}= 0,*i.e.,*there is no significant difference between the means of two populations. - H1 : µ
_{1 }- µ_{2 }≠0 - Test statistic4. Let the level of significance
*α*= 0.05,*α/2*= 0.025,critical values are - 1.96 and 1.96.

5. Computation :

Given

*xi*= 170,*s1*= 10,*n1*= 100

6. Conclusion :

zeal < - 1.96

*i.e.,*zeal lies inside the critical region=> Reject H0

=> There is significant difference between the means of two populations.

- Ho : µ