Stat-Hypothesis Testing-10
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A group of 5 people are treated with drug A weigh 42, 39, 48, 60, 41 kg.;
a second group of 7 people from the same community treated with drug B weigh
38, 42, 56, 64, 68, 69, 62 kg. Do you agree with the claim that drug B increases
the weight significantly? (The value of ‘t’ at 5% level of
significance for 10 degrees of freedom is 2.2281).
Solution: This is a problem for testing equality of two population means
based on independent samples.
We assume that the samples are drawn from normal distribution of population
having the same standard deviation σ (unknown) and mean µ1 and µ2 respectively.
The null hypothesis is that people treated with drug A and B has the same
average weight (in the population) and the alternative hypothesis is that
drugs B increase the average weight.
H0 : µ1 = µ2 against H1 : µ1 < µ2
The appropriate statistic is fisher’s t.
Since the alternative hypothesis is one –sided H1: µ1 < µ2,
the null hypothesis H0 will be rejected if t ≤ - t0.05
From the given data, we obtain the following results:
The estimate of σ is ;
s = √ (5 * 58 + 7*926/7) / (5 + 7 – 2)
= √ 121.6
11.03
So, t = (46 – 57) / { 11.03 √(1/5 + 1/7)}
= - 11 / 6.46
= - 1.70
Since the observed value of t , viz. – 1.70 is not less than - 2.2281,
we cannot reject H0 at 5% level fo significance.
Thus the claim is not justified.
