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

Question #171505

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. 


1
Expert's answer
2021-03-22T04:59:14-0400

H0:μ1μ2=0.20H1:μ1μ2>0.20α=0.05x1ˉ=2.61s1=0.12n1=50x2ˉ=2.38s2=0.14n2=40δ=0.20H_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 μ1μ2=δμ_1 -μ_2 = δ , where δ is a given constant, against of the alternative μ1μ2>δμ_1 -μ_2 > δ . The appropriate test statistic for this test is

Test concerning differences between means

z=x1ˉx2ˉδσ12n1+σ12n2z = \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 s1s_1 substituted for σ1σ_1 and s2s_2 substituted for σ2σ_2 so long as both samples are large enough for the central limit theorem to be invoked.

z=2.612.380.2(0.12)250+(0.14)240=1.08z = \frac{2.61-2.38-0.2}{\sqrt{\frac{(0.12)^2}{50} + \frac{(0.14)^2}{40}}} = 1.08

Reject H0H_0 if z≥1.96

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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment