Wks 8
____________, ____________: _______
Surname (e.g. “Straayer”), Given Name (e.g. “Dave”)
Class time
I think a lot of students in my classes were born in counties other than the U.S.A. I’ll go out on a limb
and say that more than 10% of my students were born in countries other than the U.S.A.
Now collect data: How many students in the classroom? n = _________________
How many born in countries other than the U.S.A.? x = _____________________
Our sample proportion? 𝑝̂ = ____________________
The critical number to do the calculation is the Standard Error. We now don’t have an assumed value for
the population proportion, so we’re going to have to use the sample proportion to calculate the
𝑝̂𝑞̂
Standard Error is: √ 𝑛 which calculates to S.E. = __________________
This means we’d expect samples of this size to have a normal distribution, centered on the correct
proportion p, with a standard deviation of the S.E.
Use the Bell program to find a value of Z (call it Z*) such that 95% of a normal curve is between –Z* and
Z*. Z* = __________________ (Gee, is this about 2?)
This means that 95% of all samples of this size are within Z* (times) S.E. of the correct population
proportion. Z* times S.E. = M.E. = ________________________ (Call this Margin of Error or M.E.)
If 95% of all samples of this size are within M.E. of the correct population proportion, then there is a
95% probability that our sample is within M.E. of the correct population proportion. The set of all
possible population proportions p for which our sample is within M.E. of p is 𝑝̂ ± 𝑀. 𝐸.
Express this set in point estimate ± M.E. form: ________________ ± _______________
Express this same set in interval notation: (_________________ , __________________)
Almost done. Now, what if we wanted to be 99% confident? What Z* traps 99% of a normal curve
between –Z* and +Z*? Z* = ________________
The M.E. for this confidence level is Z* X S.E. = ____________________
The 99% confidence estimate is ( _________________ , ______________)