No mean and standard deviation for confidence interval.


New Member
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.


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


New Member
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.