Answer to Question #171505 in Statistics and Probability for CHOI ELISA

Question

An experiment is performed to determine whether the average nicotine content of one kind of cigarette exceeds that of another kind by 0.20 milligram. If =50 cigarettes of the first kind had an average nicotine content of =2.61 milligrams with a standard deviation of =0.12 milligram, whereas =40 cigarettes of the other kind had an average nicotine content of = 2.38 milligrams with a standard deviation of =0.14 milligram, test the null hypotheses - = 0.20 against the alternative hypotheses - ¹ 0.20 at the 0.05 level of significance. 

Answer 1

$H_0 : μ_1 -μ_2 = 0.20 \\ H_1 : μ_1 -μ_2 > 0.20 \\ α = 0.05 \\ \bar{x_1}= 2.61 \\ s_1 = 0.12 \\ n_1 = 50 \\ \bar{x_2}= 2.38 \\ s_2 = 0.14 \\ n_2 = 40 \\ δ = 0.20$

We have independent random samples from two normal populations having mean and known variances. We want to teat the null hypothesis $μ_1 -μ_2 = δ$ , where δ is a given constant, against of the alternative $μ_1 -μ_2 > δ$ . The appropriate test statistic for this test is

Test concerning differences between means

$z = \frac{\bar{x_1}-\bar{x_2}-δ}{\sqrt{\frac{σ_1^2}{n_1} + \frac{σ_1^2}{n_2}}}$

When we deal with independent random sample from populations with unknown variances which may not even be normal, the test can still be used with $s_1$ substituted for $σ_1$ and $s_2$ substituted for $σ_2$ so long as both samples are large enough for the central limit theorem to be invoked.

$z = \frac{2.61-2.38-0.2}{\sqrt{\frac{(0.12)^2}{50} + \frac{(0.14)^2}{40}}} = 1.08$

Reject $H_0$ if z≥1.96

Since z=1.08 < 1.96, we cannot reject the null hypothesis.