PLEASE HELP !! IT IS DUE AT 12!!!
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Answer:
<em>(a) A 99% confidence interval for the actual mean noise level in hospitals is </em><em>(44.02 db, 49.98 db)</em><em>. </em>
<em>(b) We can be 90% confident that the actual mean noise level in hospitals is </em><em>47 db</em><em> with a margin of error of </em><em>1.89 db</em><em>. </em>
<em>(c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between </em><em>44.41 db and 49.59 db</em><em>. </em>
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Step-by-step explanation:
<em>The problem is incomplete. The questions are:</em>
<em />
<em>(a) A 99% confidence interval for the actual mean noise level in hospitals is </em><em>(44.02 db, 49.98 db)</em><em>. </em>
For a 99% CI, the value of z is z=2.58
Then, the confidence interval for the mean is:
<em>(b) We can be 90% confident that the actual mean noise level in hospitals is </em><em>47 db</em><em> with a margin of error of </em><em>1.89 db</em><em>. </em>
For a 90% CI, the value of z is z=1.64.
Then, we can calculate the margin of error as:
<em>(c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between </em><em>44.41 db and 49.59 db</em><em>. </em>
The 2% tails data corresponds, in the standard normal distirbution, to the values of z whose absolute value is higher than 2.33.
The values of db for these critical values are: