hmarieyoung
New member
- Joined
- May 24, 2010
- Messages
- 1
Hello,
The homework question I'm having difficulty with is (it is part b I don't know if I solved correctly, but I've included part a as well):
You are planning a sample survey of small businesses in your area. You will choose a random sample of businesses listed in the telephone books' Yellow Pages. Previous experience shows that only about 30% of the businesses you contact will respond to your questionnaire.
a) If you contact 150 busineses, what is the probability that 30 or fewer will respond?
p=.3 n=150 mean=45 standard deviation=5.6125
p(x<=30) = p(z<=30-45/5.6125) using formula z= (x-mean)/standard deviation
z=-2.6726 and using Normal Curve areas/z table --> 0.4962
and then 0.5-0.4962 = 0.0038 which is the probability that 30 or fewer will respond
b) How many businesses do you need to contact if you want to have at least 30 responses, with a 98% confidence. (<-- THIS IS THE ONE I'M HAVING PROBLEM WITH)
I've calculated out the z numbers for a 98% confidence interval at +/- 2.32 and used it in the formulat x=mean+z*standard deviation -->to get the interval of numbers the responses would fall between [31.979, 58.021], and using this numbers found the mean=44.9955 (of the two intervals) and used the mean to solve for n in the equation n = mean/p (n=44.9955/.3)=149.985. So, one needs to contact 150 businesses to have at least 30 responses with a 98% confidence. Is this correct?
Thank you!
Heather
The homework question I'm having difficulty with is (it is part b I don't know if I solved correctly, but I've included part a as well):
You are planning a sample survey of small businesses in your area. You will choose a random sample of businesses listed in the telephone books' Yellow Pages. Previous experience shows that only about 30% of the businesses you contact will respond to your questionnaire.
a) If you contact 150 busineses, what is the probability that 30 or fewer will respond?
p=.3 n=150 mean=45 standard deviation=5.6125
p(x<=30) = p(z<=30-45/5.6125) using formula z= (x-mean)/standard deviation
z=-2.6726 and using Normal Curve areas/z table --> 0.4962
and then 0.5-0.4962 = 0.0038 which is the probability that 30 or fewer will respond
b) How many businesses do you need to contact if you want to have at least 30 responses, with a 98% confidence. (<-- THIS IS THE ONE I'M HAVING PROBLEM WITH)
I've calculated out the z numbers for a 98% confidence interval at +/- 2.32 and used it in the formulat x=mean+z*standard deviation -->to get the interval of numbers the responses would fall between [31.979, 58.021], and using this numbers found the mean=44.9955 (of the two intervals) and used the mean to solve for n in the equation n = mean/p (n=44.9955/.3)=149.985. So, one needs to contact 150 businesses to have at least 30 responses with a 98% confidence. Is this correct?
Thank you!
Heather