<PRE><B>Question 441142</B>
The standard deviation of test scores on a certain achievement test is 11.4. A random sample of 80 scores on this test had a mean of 73.6 . Based on this sample, find a 90% confidence interval for the true mean of all scores. Then complete the table below. 
Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place.
----
x-bar = 73.6
ME = zs/sqrt(n) = 1.645*[11.4/sqrt(80)] = 2.0966 

What is the lower limit of the 90% confidence interval
:: 73.6-2.0966  
What is the upper limit of the 90% confidence interval
:: 73.6+2.0966
===================
Cheers,
Stan H.
</PRE>