# Hypothesis testing with no stated standard deviation

#### not a math genius

##### New Member
My head is going to explode, can someone help before it is too late?

Te problem:

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?

Let me see, I have the (sorry don't know how to write formula on here)
H0: mu = .23 which is the stated average by the government.
H1: mu not = .23
alpha = 0.05
n= 68
x bar= .309 (percentage of sample 21/68 = .309)

I am at a loss here. Maybe I am stupid, but I don't see enough info to get a standard deviation and complete the problem. Am I missing something? Someobody help me please.

By the way... there are a few more problems I will post seperately because I am dying here.

#### JohnM

##### TS Contributor
For this one you would test:

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