Question

The ideal (daytime) noise-level for hospitals is 45 decibels with a standard deviation of 10 db; which is to say, this may not be true. A simple random sample of 75 hospitals at a moment during the day gives a mean noise level of 47 db. Assume that the standard deviation of noise level for all hospitals is really 10 db. All answers to two places after the decimal. g

Answer 1

Answer:

(a) A 99% confidence interval for the actual mean noise level in hospitals is (44.02 db, 49.98 db).

(b) We can be 90% confident that the actual mean noise level in hospitals is 47 db with a margin of error of 1.89 db.

(c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between 44.41 db and 49.59 db.

Step-by-step explanation:

The problem is incomplete. The questions are:

(a) A 99% confidence interval for the actual mean noise level in hospitals is (44.02 db, 49.98 db).

For a 99% CI, the value of z is z=2.58

Then, the confidence interval for the mean is:

$M-z\sigma/\sqrt{n}\leq\mu\leq M-z\sigma/\sqrt{n}\\\\47-2.58*10/\sqrt{75} \leq\mu\leq47+2.58*10/\sqrt{75}\\\\47-2.98\leq\mu\leq47+2.98\\\\44.02\leq\mu\leq 49.98$

(b) We can be 90% confident that the actual mean noise level in hospitals is 47 db with a margin of error of 1.89 db.

For a 90% CI, the value of z is z=1.64.

Then, we can calculate the margin of error as:

$E=z*\sigma/\sqrt{n}=1.64*10/\sqrt{75}=1.89$

(c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between 44.41 db and 49.59 db.

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:

$X_1=M+z_1*\sigma/\sqrt{n}=47+(-2.33)*10/\sqrt{81}=47-2.59=44.41\\\\\\ X_2=M+z_2*\sigma/\sqrt{n}=47+(2.33)*10/\sqrt{81}=47+2.59=49.59$