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According to the University of Nevada Center for Logistics Management, 11% of all merchandise sold in the United States gets
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Answer #1

a) Point estimate of the proportions of items returned :   \hat{p} = \frac{11}{100} = 0.11

b) Standard error

SE= \sqrt{\frac{\hat{p} (1-\hat{p} )}{n}} = \sqrt{ \frac{0.11(1-0.11)}{100}}

Margin of error : ME= z_{1-\alpha /2} \times \sqrt{\frac{\hat{p} (1-\hat{p} )}{n}} = 1.96\times \sqrt{ \frac{0.11(1-0.11)}{100}} =0.0613

So, 95%confidence interval is :   

\hat{p} \pm ME \\ =( 0.11 \pm 0.0613) \\ =(0.0487 , 0.1713)

c)   H_0: p_0 = 0.11 \ \ H_1 : p_0 \ne 0.11

  

z value / Test statistics :

z = \frac{\hat{p}- p }{\sqrt{\frac{\hat{p} (1-\hat{p} )}{n}}} = 0 as \hat{p} = p =0.11

z value / critical value = 1.96

p value = 1

Fail to reject the null , yes we don't have enough evidence to prove that Huston is different from the national average.

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