Question

Hospital Noise Levels For a sample of 6 operating rooms taken in a hospital study, the mean nolse level was 38 decibels and the standard deviation was 11.9. Find the 99% confidence interval of the true mean of the noise levels in the operating rooms. Assume the variable is normally distributed. Round your answers to at least two decimal places. << alle

Answer 1

Here we have

I= 38, s = 11.9, n - 6 and a= 0.01

So $Z_{\alpha/2} = 2.58$

So 99% confidence interval for the mean is given by

S ī Za/2 n

$38 \pm 2.58\times \frac{11.9}{\sqrt 6}$

i.e (25.47 , 50.53)

The answer is 25.47 and  50.53

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