Question

A Trade Union spokesman claimed that 75% of union members would support the strike if their basic demands were not met. The company spokesman believed that the actual percentage was smaller than the figure. He conducted the study, of which of the 125 samples, 87 trade union members which states will strike.

(a) Perform a hypothesis test with a 90% confidence level. Give analysis and conclusions

(b) Calculate the confidence interval for the percentage of members who say they will strike.

Answer 1

$Let\;\hat p=\frac{87}{125}=0.696, p_{0}=0.75,n=125\\ a) H_{0}: p \geq 0.75\\ H_{1}: p < 0.75\\ Z_{α}=Z_{0.10}=1.282,\\ Z=\frac{\hat p -p_{0}}{\sqrt{\frac{ p_{0}(1-p_{0})}{n}}}\\ =\frac{0.696 -0.75}{\sqrt{\frac{0.75(0.25)}{125}}}=-1.39\\ \text{the rejection region} :\; Z<-1.282\\ \text{The decision is to reject}:\; H_0\\ \text {This mean that that the actual }\\ \text{percentage is smaller than the figure. }\\ b) \text{the confidence interval for the proportion }\\ \text{can be calculated as follows:}\\ \hat p±Z_{\frac{α}{2}}(\sqrt{\frac{\hat p(1-\hat p)}{n}}),\\ Z_{α}=Z_{\frac{0.10}{2}}=Z_{0.05}=1.65,\\ \text{ the lower limit }= 0.696-1.65(\sqrt{\frac{0.696(1-0.696)}{125}})\\ =0.628\\ \text{ the upper limit} = 0.696+1.65(\sqrt{\frac{0.696(1-0.696)}{125}})\\ =0.768\\ \therefore 0.628<p<0.768$