No mean and standard deviation for confidence interval.

nihal

New Member
#1
According to the government, 23% of all American adults have more than one job. A random survey of 68 adults finds that 21 of them hold more than one job. At the 5% level of significance, can we accept the government's claim?

For this problem the solution would be like this
Ho: P = 0.23
Ha: P <> 0.23

The sample p ("small" p) = 21/68 = 0.309

The standard error of proportions is SEp = sqrt(PQ/n) where

P = .23
Q = 1-.23 = .77
n = 68

Z = (p - P)/SEp

What if the 23% is not given in the question if we are asked to find the solution without the percentage?? similar to this

A market research company is trying to estimate the proportion of consumers that would consider buying a new product. A random sample of 300 people were asked if they would consider buying this product, of which 90 said that they would.

Please help on this.
 

Dason

Ambassador to the humans
#2
When you are dealing with proportions all you need is the sample proportion. This is clearly homework - do you have a book to reference?
 

nihal

New Member
#3
Hi Dason,

I dont have any book for reference. I tried to solve it but failed. If I have to use the formula Z = (p - P)/SEp
P would be 0.3 (as 90 of 300 would be 30%)
Q= 1-0.3=0.7

But what about 'p'? will that be 90/300 according to the problem that i posted in the question?
Unable to figure it out.