Answer to Question #260923 in Statistics and Probability for MULO

Question

During a water shortage, restaurant owners were asked not to serve water before meals, unless the customer requested water. Over a three-month period, a restaurant owner observed that 45% of the customers asked for water. After the three-month period, the owner placed a card at each table, explaining the restaurant’s policy of not serving water to customers, unless specifically asked to do so, because of the water shortage. If 53 of the next 150 customers asked for water with their meal, can we conclude that placing the cards at each table decreased the number of customers requesting water before their meal? At the 0.02 level of significance, use the following hypothesis test

Answer 1

We have to check hypothesis about the probability of event(a customer ask for a water)

$H{\scriptscriptstyle 0}:p=0.45$

$H{\scriptscriptstyle 1}:p<0.45$

The appropriate test value for such a situatuon is calculated the next way:

$U={\frac{({\frac M n}-p{\scriptscriptstyle 0})*\sqrt{n}} {\sqrt{p{\scriptscriptstyle 0}(1-p{\scriptscriptstyle 0)}}}}$ , where n - number of experiments, M - number of succesfull experiments , $p{\scriptscriptstyle 0}$ - supposed probability

$Ф(Cr) = {\frac{1-2a} 2}$ , where Cr - critical value, Ф - Laplace function, a - level of significance

In the given case:

$U={\frac{({\frac {53} {150}}-0.45)*\sqrt{150}} {\sqrt{0.45*0.55}}} = {\frac {-1.184} {0.497}} = -2.38$

$Ф(Cr) = {\frac{1-2*0.02} 2}=0.48$ -> Cr = 2.06

The null hypothesis is rejected if U < -Cr

-2.38 < -2.06, so we reject the null hypothesis

There are statistically significant evidence that placing the cards decreasing the amount of requests at the 0.02 significance level.