2 problems

[davidzhou176]

The owner of a restaurant that serves continental food wants to study characteristics of customers of his restaurant. In particular, he decides to focus on two variables: the amount of money spent by customers and whether customers order dessert. The results from a sample of 60 customers are as follows:
Amount of money spent (in sample): Average = $38.54, Std. Dev = $7.26
18 customers purchase dessert
i. Set up a 95% confidence interval estimate of the population mean amount spent per customer in the restaurant.

___s_______
E= tu/2 sq root n

=1.96*7.26/7.745
=1.83
confidence interval is

36.703<U<40.377




ii. Set up a 90% confidence interval estimate of the population proportion of customers who purchase dessert.
^ ^
P-E<P<P+E

E=1.645*square root(0.3)(0.7)/60
=0.097

0.203<P<0.397



If the owner of a competing restaurant wants to conduct a similar survey in her restaurant (and does not have access to the information obtained by the owner of the first restaurant1), answer these questions:
iii. If she wants to have 95% confidence of estimating the population mean amount spent in her restaurant to within +-$1.50 and the (population) standard deviation is assumed to be $8, what sample size is needed?

n >= ((Za)^2) * (s/d)^2
>＝1.96^2 ×（8/1.5)^2
>＝3.8416×28.44
>==109
iv. If she wants to have 90% confidence of estimating the true proportion of customers who purchase dessert to within +-4%, what sample size is needed?

n >= ((p*q)/d^2) * Z^2
>=(0.3*0.7)/(0.04^2)*1.645^2
>=35.5
>=36

v. Based on your answers to (c) and (d), what sample size should be taken?

i will take 109. because the confidence interval 95% is much higher than 90%.



am i currect??????

i am not sure...........

[vinux]

More than two problems. Nice problem and you have done it well.
Sample size you can take 110 ( I guess the actual value is 109.2559) for the safer side.

[davidzhou176]

thanks a lot.

now i can face my professor...........