Question

T4.025 0 Female undergraduates in randomized groups of 20 took part in a self-esteem s and physical attractiveness. Let x1, x, and x3 be random variables Higher index values mean a more positive infuence on self-esteem study. The study measured an index of self-esteem from the point of view of representing the measure of selif-esteem through x, (competence), "2 (seocial acoeptance), and xy (attractiveness) 20 20 20 18.92 18.64 3.27 3.81 (a) Find a 95% conndence interval for μ1-Pg (Round your answer, to two deomai places.) lower limit upper limit (b) Find a 95% confidence interval for»,-1), (Round your answer, to two donal pace.) lower imit upper limt (E) ,ind a 95% confidence interval for μ2-μ y (Round your answers to two decenal places.) lower Emit upper linma (d) Carernent on the meaning of each org·co hdene intervals found in parts (a). eb), and (e). 서"he 95% condence ean you say ab.a 'N-ve age mnr e competence and attractiveness? between social acceptance and attractiveness nfluence on seif esteem between competence and social acceptance? between

Answer 1

$C.I\:of\:\mu_1-\mu_2=\left ( \bar{X_1}-\bar{X_2} \right )\pm Z_{(1-\alpha /2)}\sqrt{\frac{\sigma _1^2}{n_1}+\frac{\sigma _2^2}{n_2}}\\ =\left ( \bar{X_1}-\bar{X_2} \right )\pm Z_{(1-\alpha /2)}\sqrt{\frac{\sigma _1^2}{n_1}+\frac{\sigma _2^2}{n_2}}\\ a)\\ \mu_1-\mu_2=(18.92-18.64)\pm 1.96\sqrt{\frac{3.27^2}{20}+\frac{3.49^2}{20}}\\ =0.28\pm1.96\sqrt{\frac{22.873}{20}}\\ =0.28\pm 2.096\\ \boldsymbol{(-1.82,2.38)}\\$

$b)\\\mu_1-\mu_3=(18.92-17.98)\pm 1.96\sqrt{\frac{3.27^2}{20}+\frac{3.81^2}{20}}\\ =0.94\pm1.96\sqrt{\frac{25.209}{20}}\\ =0.94\pm 2.20\\ \boldsymbol{(-1.26,3.14)}\\ c)\\ \mu_2-\mu_3=(18.64-17.98)\pm 1.96\sqrt{\frac{3.49^2}{20}+\frac{3.81^2}{20}}\\ =0.66\pm1.96\sqrt{\frac{26.696}{20}}\\ =0.66\pm 2.158\\ \boldsymbol{(-1.50,2.82)}\\$

Conclusion: here from all three confidence interval we conclude that 2nd and 3rd will be good shows for their corrosponding means diffrrences

at 95 % CI