Monday, Monday, Monday....

AP EXAM FORMS/$ DUE TOMORROW! This will count as a grade for our class! You either turned your form in on time with EXACT CHANGE (100%), or you did not (0%). Get them in!

Tonight's homework: please complete the following in your textbook...CHECK OUT THE HINTS/TIPS BELOW!

Page 446: #3 (even though this is one problem, I expect to see lots of writing if you want full credit tomorrow!)

• For each scenario (a,b,c,d), define p, p-hat, the sample, the population, and check the 3 conditions (random sample, 10% condition, np and nq>10)
• Conditions = how we determine if we can use the methods of this chapter (a one proportion z-interval)
• The parameter of interest, p, is what we're trying to estimate--look at the question!
• So, for a, our p (what we're estimating) is "the percentage of all cars that may be unsafe"
• P-hat represents our sample percentage, or the percentage we actually collect data to calculate
• For these contexts, we do not know the value of "p." So, when we check our conditions, you will have to use p-hat for the success failure condition!
• In other words, your condition is: n(p-hat)>10 and n(1-phat)>10
• For example, in a, we first find p-hat = 14/134 = .1045...
• Condition: (134)(.1045) > 10 and (134)(1 - .1045) = (134)(.9955) > 10
• If we know a sample is not random, we can say:
• "It's not stated if this is a random sample, we will proceed with caution." OR
• For example, in context b, we KNOW we have a voluntary response sample (not random)--so it would not be correct to state that we can assume this is a random sample!
• The back of the book is very helpful for p, p-hat, sample and proportion...
• The information about "whether we can use the methods of this chapter" is also informative, but we would show our conditions differently (see below)

Here are the answers for the conditions (so you can check yours!--check p, p-hat, population, and sample in the back of the book):

a.) We can assume the police stopped a random sample of 134 cars; 134 cars is less than 10% of all cars; (134)(.1045) > 10 and (134)(.9955) > 10

b.) This is a voluntary response sample (not random), so we will proceed with caution; 602 is less than 10% of all viewers; (602)(.811) > 10 and (602)(.189) > 10

c.) This is a voluntary response sample (not random), so we will proceed with caution; 380 is NOT less than 10% of all parents/households (1245). We CAN NOT use a one-proportion z interval!

d.) We will assume our sample (all freshmen) is representative of all students at this school; 1632 may be less than 10% of all students at this school; (1632)(0.85) > 10 and (1632)(0.15) > 10