Two-sided confidence intervals and tests

Question

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?

Answer 1

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

Answer 2

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.