Stat-Hypothesis Testing-9

Two types of cells are tested for their length of life and the following data are obtained:

Is there a significant difference in the two means? Value of t for 15 degrees of freedom at 5% level is 2.131.

Solution: Assumptions- (1) the two populations are normal distributions with mean µ1 , µ2 and a common s.d. σ

(2) the two samples are randomly drawn and independent.

H0 : µ1 = µ2 against H1 : µ1 ≠ µ2

Given :

The appropriate test statistic is fisher’s t which, under H0 follows t distribution with (n₁ + n₂ – 2) d.f.

An estimate of the common but unknown s.d. (σ) is obtained from:

s = √ (n₁S₁² + n₂S₂²) / (n₁ + n₂ – 2)

where S₁ and S₂ are the sample s.d.

= √ (9 * 121 + 8 * 144) / (9 + 8 – 2)

= - 40 / (12.2 * 0.486)

= - 6.7

d.f. = 9 + 8 – 2 = 15

Since the alternative hypothesis is both sided, the test is two-tailed.

The critical region is given by both the tails of t distribution.

At 5% level

Critical region : | t | ≥ 2.131

Since the observed value falls in the critical region

(because | t | = | - 6.7 | = 6.7 is larger than 2.131),

we reject the null hypothesis at 5% level of significance and

conclude that there is ‘significant difference in the two means’.

Confidence interval for the difference (µ1 - µ2 ) is given by (x_m1 – x_m2) ± t_0.05 * s√(1/n₁ + 1/n₂)

95% confidence interval = (600 – 640) ± 2.131 * 12.2 √(1/9 + 1/8)

= - 40 ± 12.63

= - 27.37 and – 52.63

Thus, 95% confidence interval for (µ2 - µ1 )is 27.37 to 52.63