Two-sided confidence intervals and tests From a sample of 1751 army hospitals, estimate the mean expenses for a full time equivalent employee for all US army hospitals using a 90% confidence interval given x = 6563 and s = 2484.
Work:
1.645(2484/41.845)
1.645(59.362)
97.65
6563+- 97.65
6465.35---6660.65   ANswer

I think I did that one right but I'm a bit confused on the wording of part b...
In 2007 mean expenses for the entire public was believed to be 7000, conduct a test at a = .10 to determine if the mean expenses for an army hospital employee are equal to that of the public. Any ideas on this one? 
 A: Since population variance is not known, go for two tailed t-test:
H0: $\mu_0=7000$
H1: $\mu_1 \ne 7000$
$t_\text{statistic} = \dfrac{6563-7000}{\frac{2484}{\sqrt{1751}}}$
$t_\text{statistic} = -7.361$
Rejec $H_0$ if $t_\text{statistic} < t_\text{critical}$
At  $\alpha = 0.1$, $t_\text{critical} = -1.645$
Sincec  $-7.361 \le -1.645$, Reject $H_0$ and conclude that the mean expenses of a hospital employee is not equal  to that of the public.
Thanks
Satish
A: First, I assume you mean that the sample mean of the $n = 1751$ hospitals is $\bar X = 6563.$ Now, for some clarifications.
Confidence interval. Minitab software will compute a 90% CI from summarized data 
such as yours. Here are results from Minitab's 'one-sample t' procedure. 
 MTB > Onet 1751 6563 2484;
 SUBC>   Test 7000;
 SUBC>   Confidence 90.

 One-Sample T 

 Test of mu = 7000 vs not = 7000

    N    Mean   StDev  SE Mean       90% CI           T      P
 1751  6563.0  2484.0     59.4  (6465.3, 6660.7)  -7.36  0.000

The results of the 90% CI agree with yours and with those obtained
in the Answer by @Satish.  
Hypothesis test. The main reason for my answer is to deal more fully with the test of
$H_0: \mu = 7000$ against $H_a: \mu \ne 7000,$ about which you asked in your question.
The P-value from Minitab is $< .0005$ and so is $H_0$ is
rejected at the 10% level of significance (also at the 5% and 1% levels).
Actually, this is a t-interval with 1750 degrees of freedom because
the population SD $\sigma$ is estimated by the sample SD $S,$ based
on $n = 1751$ observations. For smaller $n$ the 'probability
factor' in the CI would be somewhat larger than 1.645. But for
such a large $n$ as yours, it is OK to use 1.645 (presumably obtained from normal tables).
The exact value for $df = 1750$ from software is 1.645725.
The 'better way' you ask about for smaller samples, is to 
obtain the percentile .95 from a table of Student's t distribution
with $n-1$ degrees of freedom (or from statistical software).
Relationship between CI and HT. More specifically, you may view the 10% CI as an interval of reasonable
(one might say 'non-rejectable') hypothetical values $\mu_0$ based on the given $n, \bar X, S$ and 90% confidence. Because $\mu_0 = 7000$ is not
contained in that interval, the null hypothesis is rejected
at the 10% level. 
This illustrates a general 'duality' between
two-sided confidence intervals with confidence $100(1 - \alpha)$% and
two-sided tests at level $\alpha$ of significance.
