Answer to Question #198219 in Statistics and Probability for Duke Banda

Question

At a large firm, there is a movement to form a union. Approximately 50% of the entire firm favour unionising, while 30% do not favour unionising. A pro-union leader takes a random sample of 100 workers. Let p1 denote the proportion in this sample that favours a union. An antiunion leader takes an independent random sample of 100 workers. Let p2 denote the proportion in this sample that favours the union. Calculate the probability that p1 exceeds p2 by 0.1 or more. (b) The mean annual income of statisticians of firm 1 is K10, 000 higher than the mean annual income of statisticians of a second firm. Random samples of size n1 = 100 and n2 = 200, respectively, are taken from the two firms. (i) What is the probability that in the two samples the mean annual incomes differ by more than K15, 000? For each firm (population), the standard deviation is K30, 000.

Answer 1

(a) $n=100,\hat{p_1}=0.50,\hat{p_2}=0.30$

(i) Probability that $p_1$ exceeds $p_2$ by 0.1 or more-

$P(p_1-p_2\ge 0.1)=1-P(p_1-p_2<0.1) =1-P(z_{0.1})$

 $z-score =\dfrac{0.1-(0.50-0.30)}{\sqrt{\dfrac{0.50(1-0.50)}{100}+\dfrac{0.30(1-0.30)}{100}}}$

     =-1.4744

$P(z_{0.1})=0.0702$

$P(p_1-p_2\ge 0.1)=1-0.0702=0.9298$

(ii) z-score for 90% confidence interval

$\alpha=0.1,\frac{\alpha}{2}=0.05,z_{\frac{\alpha}{2}} =1.6449$

  Confidence Interval $=\hat{p_1}-\hat{p_2}\pm z\sqrt{\dfrac{\hat{p_1}(1-\hat{p_1})}{n}+\dfrac{\hat{p_2}(1-\hat{p_2})}{n}}$

             $=0.50-0.30\pm 1.6449\sqrt{\dfrac{0.50(1-0.50)}{100}+\dfrac{0.30(1-0.30)}{100}} \\[9pt] =0.20\pm 0.1153 \\[9pt] =(0.0847,0.3153)$

(b) $n_1=100,n_2=200,x_1-x_2=15000,s=30000$

(i) probability that in the two samples the mean annual incomes differ by more than K15, 000-

$z=\dfrac{x_1-x_2}{s\sqrt{(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$

$=\dfrac{15000}{30000\sqrt{(\dfrac{1}{100}+\dfrac{1}{200})}}$

$=\dfrac{141.42}{2\times 17.32}=4.082$

$P(z>4.082)=0.0002$