(a) Testing of A Single Mean. Here sample is small (n < 30) and cr is unknown. 1. Set up H0 : µ = µ0 2. Set up H1 : µ > µ0 or µ < µ or µ ≠ µ0 3. Test statistic

distribution with *(n *- 1) degrees of freedom.

**4. **Set up level of significance *a *and the critical value as !tab (from table of *t *- distribution).

**5. **Compute the statistic say t_{cal}

**6. **Decisions

All t_{tab} based on *(n *- I) degrees of freedom.

*Example 5:** the mean breaking strength of a certain kind of metallic rope is 160pounds. If six pieces of ropes (randomly selected from different rolls) have a mean breaking strength of 154.3 pounds with a standard deviation of 6.4 pounds, test the null hypothesis *1..1. = *160 pounds against the alternative hypothesis *1..1. < *160 pounds at 1% level of significance. Assume that the population follows normal distribution.*

**Solution**. 1. H0 : µ= 160 pounds

2. H1: µ< 160 pounds

3. Since *n *= 6, the test statistic is taken as

**Example 5.** The mean breaking strength of a certain kind of metallic rope is 160 pounds. If six pieces of ropes (randomly selected from different rolls) have a mean breaking strength of 154.3 pounds with a standard deviation of 6.4 pounds, test the null hypothesis 1-L = 160 pounds against the alternative hypothesis 1-L < 160 pounds at 1% level of significance. Assume that the population follows normal distribution.

**Solution**. 1. H_{0} : µ = 160 pounds

2. H_{1} : µ < 160 pounds

3. Since n = 6, the test statistic is taken as

4. α = O.OI, H1 indicates it is left tailed test. Critical value at 6- 1 i.e., 5 degrees of freedom is -3.365

6. Decision

Since t_{cal} > -3.365

- it lies in the acceptance region
- H
_{0}is accepted - Mean breaking strength of the metallic rope can be taken as I60 pounds.

**(b) Testing of Difference of Two Means**. Here n_{l},n_{2} or both are small (< 30) and the population variances are unknown but equal and two populations follow normal distribution.

**1. ** Set up H_{0} : µ_{1}- µ** _{2}** = k

**2.** Set up H_{1 :} µ_{1}- µ** _{2}** > k, or µ

_{1}- µ

**< k or µ**

_{2}_{1}- µ

**≠ k**

_{2}where ,

Here, the statistic follows t - distribution with n1 + n2- 2 degrees of freedom.

4. Set up the level of significance, a. and the critical value say t13b at n1 + n2 - 2 degrees of freedom.

5. Compute the test statistic as t_{cal}·

Example 6. The following are the number of sales with a sample of 6 sales people of gas lighter in a city A and a sample of 8 sales people of gas lighter in another city B made over a certain fixed period of time :

City A : 63, 48, 54, 44, 59, 52

City B : 41, 52, 3850, 66, 54, 44, 61

Assuming that the populations sampled can be approximated closely with normal distributions having the same variance, test H_{0} : µ_{1} = µ_{2} against H_{1 : } µ_{1} ≠ µ_{2} at the 5% level of significance.

**Solution.** 1. H_{0} : µ_{1} = µ_{2}

2. H_{1 : } µ_{1} ≠ µ_{2}