Assignments

1. Bookstore
A university bookstore claims that 50% of its customers are satisfied with the service and prices.
a. If this claim is true, what is the probability that in a random sample of 600 customers less than 45% are satisfied?
b. If you would reject the bookstore’s claim if the probability computed in a. will be less than 0.05%, how many customers at least should have expressed satisfaction?

Solution.
a. We denote the (population) probability that is claimed by the bookstore as $p(=0.5)$ and the probability computed by the sample as $\hat{p} (=0.45)$ . Then:
$\displaystyle{P(\hat{p}<0.45)}=$
$\displaystyle{=P(\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}<\frac{0.45-0.50}{\sqrt{0.50(1-0.50)/600}})=}$
$\displaystyle{=P(Z<-2.45)=0.0071}}$
b. In this case we have to compute $x$ from $P(Z<x)=0.05$ . From table 3 in appendix B-8 we find $x=-1.65$ . So, we have to solve $\hat{p}$ from the equation: $\displaystyle{\frac{\hat{p}-0.50}{\sqrt{0.50(1-0.50)/600}}=-1.65}$ and we get: $\hat{p}=0.466=x/600$ and thus the number of customers expressing satisfaction should be at least 280.

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