Question

(1 point) Test the given claim using the a = 0.05 significance level and assuming that the populations are normally distributed. Claim: The treatment population and the placebo population have the same mean. Treatment group: n=9, T = 78, 8 = 5.7. Placebo group: n = 7,1 = 71, s = 5.4. (a) The test statistic is 2.51 (b) What is the conservative estimate for the degrees of freedom? df = 13.36 (C) The negative critical value is -6.35 (d) The positive critical value is 20.35 (e) Is there sufficient evidence to warrant the rejection of the claim that the treatment and placebo populations have the same mean? A. Yes B. No Note. Voiran earn partial credit on this problem

Answer 1

Claim : The treatment population and placebo population have same mean.

The hypothesis are :

H0 : $\mu _{T}=\mu _{P}$ v/s H1: $\mu _{T}\neq \mu _{P}$

a) The test statistic is,

$t=\frac{\overline{x}_{T}-\overline{x}_{P}}{\sqrt{\frac{s_{T}^{2}}{n_{T}}+\frac{s_{P}^{2}}{n_{P}}}}$

$=\frac{78-71}{\sqrt{\frac{5.7^{2}}{9}+\frac{5.4^{2}}{7}}}$

= 2.510

b)

$df=\frac{\left ( \frac{s_{T}^{2}}{n_{T}}+\frac{s_{P}^{2}}{n_{P}} \right )^{2}}{\frac{(s_{T}^{2}/n_{T})^{2}}{n_{T}-1}+\frac{(s_{P}^{2}/n_{P})^{2}}{n_{p}-1}}$

(5.7 1912 (5.47/7)

= 13.37 $\approx$ 13

$\alpha$ = 0.05

c) The negative critical value is,

---------( using excel formula " =t.inv( 0.025, 13) " )

ta/2,df = 0.05/2.13 = -2.160

d) The positive critical value is,

---------( using excel formula " =t.inv( 0.975, 13) " )

t1-a/2,df = t1-0.05/2,13 = 2.160

e)

Here calculated value of t > critical value of t .

Hence we reject null hypothesis.

There is sufficient evidence to warrant the rejection of the claim that the treatment and placebo populations have same mean.