Confidence Interval

PROBLEM:

A study conducted by Runzheimer International showed that Paris is the
most expensive place to live of the 12 European Community cities.
 Paris ranks second in housing expense, with a rental unit of six to
nice rooms costing an average of $4,292 a month. Suppose a company's CEO
believes this figure is too high and decides to conduct her own survey.
Her assistant contacts the owners of 55 randomly selected rental units
of six to nice rooms and finds that the sample average cost is $4008,
with a standard deviation of $386.

Using sample results and level of confidence of 0.01, test to determine
whether the figure published by Runzheimer International is too high.

Was it a one- or two-tailed test?

MY RESPONSE SO FAR

Level of confidence of 0.01

a=0.01

a/2=0.005

n=55

df=55-1=54

t 0.005,54 (I think this is my problem)

( =$4150.01

ợ = $386.

a=$4008

b= $4,292 a month

s/(54=52.04



MY problem is that I cannot figure where t 0.005, 54 is on the chart.