(a)Test of a Single Mean. Here sample is small which (n < 30) and cr is unknown

- Setting up H0 : µ = µ0 2.
- Set up H1 : µ > µ0 or µ < µ or µ ≠ µ0
- Test statistic

Distribution with *(n *– 1) degrees of freedom

- Setting up level of significance denoted by the
*a*and the critical value is denoted by the !tab (from table of*t*– distribution). - Computation of the statistic saytcal
- Decisions

H_{1} | Reject H_{0}, if |

µ< µ_{0} | t_{cal}>t_{tab}i.e., t_{α} |

µ > µ_{0} | t_{cal}<-t_{tab} i.e., -t_{α} |

µ ≠ µ_{0} | t_{cal}<t_{tab} i.e., t_{α/ 2}Or, t |

All t_{tab} based on *(n *– I) which is the 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 mean breaking strength of 154.3 pounds with a standard deviation of 6.4 pounds, test 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***:*

*H0: µ= 160 pounds**H1: µ< 160 pounds**Since n = 6, Here the test statistic is considered as*

t=

- α = O.OI,

H1 indicates that it is left tailed test. Critical value is 6- 1 which is 5 degrees of the freedom is -3.365

- Computation

t_{cal} == –2.18

- Decision

Since t_{cal}> -3.365

- It is located in the acceptance region
- H
_{0}is accepted - The Mean breaking strength of the metallic rope is considered as I60 pounds.

** **

**b) Testing of Difference of Two Means**. In this n1 and n2 are small (< 30) and the population variances are not known but equal and two populations are followed by normal distribution.

- Setting up H0 : µ1- µ
**2**= k - Setting up H1 : µ1- µ
**2**> k, or µ1- µ**2**< k or µ1- µ**2**≠ k - Test statistic

t =

Here, the statistic is followed by t – distribution with n1 + n2- 2 degrees of freedom.

- Setting up of level of significance, a. and the critical value say t13b at n1 + n2 – 2 degrees of freedom.
- Computation of test statistic as t
_{cal}· - Decision

H_{1} | Reject H_{0}, if |

µ_{1 }– µ_{2 }< k | t_{cal}>t_{tab} i.e., t_{α} |

µ_{1} – µ_{2}> k | t_{cal}< -t_{tab} i.e., -t_{α} |

µ_{1} – µ_{2} ≠ k | t_{cal}<-t_{tab} i.e., -t_{α/ 2}Or, t |

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 sample is closely related with normal distributions which is 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**

- H
_{0}: µ_{1}= µ_{2} - H
_{1 :}µ_{1}≠ µ_{2} - Test statistic: