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 81 hospitals at a moment during the day gives a mean noise level of 47 db. Assume that the standard deviation of noise level is really 10 db. All answers to two places after the decimal.
(a) A 99% confidence interval for the actual mean noise level in hospitals is (44.13 db, 49.87 db).
(b) We can be 90% confident that the actual mean noise level in hospitals is 47 db with a margin of error of 1.83 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.
(d) A 99.9% confidence interval for the actual mean noise level in hospitals is ( 43.35 db, 50.65 db).
(e) Assuming our sample of hospitals is among the most typical half of such samples, the actual mean noise level in hospitals is between   db and  db. HELP WITH E!! ALL OTHERS I FIGURED OUT
(f) We are 95% confident that the actual mean noise level in hospitals is 47 db, with a margin of error of 2.18 db.
(g) How many hospitals must we examine to have 95% confidence that we have the margin of error to within 1 db? 384.16  
(h) How many hospitals must we examine to have 99.9% confidence that we have the margin of error to within 1 db? 1082.41

Answer 1

(a)

Since we know that population standard deviation, we will use z value for confidence interval .

z value for 99% confidence interval is  2.58

Standard error of mean = = = 1.11

Margin of error = 1.11 * 2.58 = 2.86

99% confidence interval for the actual mean noise level is,

(47 - 2.86, 47 + 2.86)

(44.14 db, 49.86 db)

(b)

z value for 90% confidence interval is  1.645

Margin of error = 1.11 * 1.645 = 1.83 db

(c)

If our sample (of 81 hospitals) is among the most unusual 2% of samples, then = 0.02

Confidence interval = 1 - = 1 - 0.02 = 0.98

z value for 98% confidence interval is  2.33

Margin of error = 1.11 * 2.33 = 2.59 db

98% confidence interval for the actual mean noise level is,

(47 - 2.59, 47 + 2.59)

(44.41 db, 49.59 db)

(d)

z value for 99.9% confidence interval is 3.29

Margin of error = 1.11 * 3.29 = 3.65 db

98% confidence interval for the actual mean noise level is,

(47 - 3.65, 47 + 3.65)

(43.35 db, 50.65 db)

(e)

If our sample of hospitals is among the most typical half of such samples, then = 0.5

Confidence interval = 1 - = 1 - 0.5 = 0.5

z value for 50% confidence interval is 0.67

Margin of error = 1.11 * 0.67 = 0.74 db

50% confidence interval for the actual mean noise level is,

(47 - 0.74, 47 + 0.74)

(46.26 db, 47.74 db)

Assuming our sample of hospitals is among the most typical half of such samples, the actual mean noise level in hospitals is between 46.26  db and 47.74 db