population proportions...

#1
Hello,

Suppose you have a problem to solve in which 71% of Internet users connect using standard telephone lines. The population proportion is .71.

Now, we want to find the probability that a sample proportion from a random sample of 350 Internet users will be at least .75.

Can someone help me determine how to solve this? I'm not asking for the answer, just the process. I'm wondering if this is a two-part problem. First, we may need to determine the standard deviation of "p" by taking the square root of "p(p-1)/n" where "n" is the sample size. Then I don't know. :confused:

Please help. :(
 

JohnM

TS Contributor
#2
You can use the normal distribution where:

z = (p - P) / SEp

p = sample proportion
P = population proportion
SEp = standard error of the proportion = sqrt(P(1-P)/n)
 
#3
Hello,

I don't think it's working. The answer to this problem is supposed to be .0495, but I need to figure out how that was done. To play around, I did a couple of equations:

z = (.75-.71)/[sqrt (.71)*(1-.71)/350 ] = 1.65

z = (.75-.71)/[sqrt (.75)*(1-.75)/350 ] = 1.74

Even after looking up these two figures on the z table, I didn't find anything close to .0495. Sorry, but I'm lost. :(
 

JohnM

TS Contributor
#4
z = 1.65 is correct.

The probability of z >= 1.65 is .0495 (4.95% of the area under the normal curve is above z = 1.65).