Answer to Question #311384 in Statistics and Probability for Steve

Question

A clothing company produces men’s jeans. The jeans are made and sold with either a regular cut or a boot cut. In an effort to estimate the proportion of their men’s jeans market in Chilanga City that prefers boot-cut jeans, the analyst takes a random sample of 212 jeans sales from the company’s two Chilanga City retail outlets. Only 34 of the sales were for boot-cut jeans. Construct a 90% confidence interval to estimate the proportion of the population in Chilanga City who prefer boot-cut jeans.

Answer 1

Confidence interval for proportion can be estimated the following way

$(p-Cr*\sqrt{{\frac {p(1-p)} n}}, p+Cr*\sqrt{{\frac {p(1-p)} n}})$ , where p - sample proportion value, Cr - -critical value, n - sample size. Since the sample size is big(>30), then it is appropriate to use z-value as Cr, so

$P(Z>Cr)={\frac {1-\alpha} 2}=0.05\implies Cr=1.64$ ($\alpha$ - confidence level)

$({\frac {34} {212}}-1.64*\sqrt {{\frac {{\frac {34} {212}}(1-{\frac {34} {212}})} {212}}},{\frac {34} {212}}+1.64*\sqrt {{\frac {{\frac {34} {212}}(1-{\frac {34} {212}})} {212}}})=(0.16-0.04, 0.16+0.04)=(0.12, 0.2)$