Hypothesis testing

The problem is:
During 2008, college work study students earned a mean of $1478. Assume that a sample consisting of 45 of the work study students at a large university was found to have earned a mean of $1503 during that year, with a standard deviation of $210. Would a one-tail test at the 0.05 level suggest the average earnings of this university's work study students were significantly higher than the national mean?

So far I've gotten:
u =< 5.00
u > 5.00
z = -1.8855

If anyone could help me along, that'd be great.

jakej78 said:

The problem is:
During 2008, college work study students earned a mean of $1478. Assume that a sample consisting of 45 of the work study students at a large university was found to have earned a mean of $1503 during that year, with a standard deviation of $210. Would a one-tail test at the 0.05 level suggest the average earnings of this university's work study students were significantly higher than the national mean?

So far I've gotten: I don't understand what you have done
u =< 5.00
u > 5.00
z = -1.8855

If anyone could help me along, that'd be great.

Click to expand...

The population mean for all students everywhere was $1478, but they did not tell you the population standard deviation. So you have to use the std.dev. of the observed sample as an estimator of the population std.dev. The distribution of sample means of size N=45 has mean equal population mean = $1478, and standard deviation equal to population divided by sqrt(45) = $31.30.

The null hypothesis is that the observation is from a distribution with mean $1478, and standard deviation $31.30
The alternate hypothesis is mean > $1478.
The critical z-value for a tail of area 0.05 is 1.645.

OK?