Question

1. + -/14 points My Notes The ideal (daytime) noise-level for hospitals is 45 decibels with a standard deviation of 11 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 11 db. All answers to two places after the decimal. (a) A 99% confidence interval for the actual mean noise level in hospitals is (49) db, 49) db). (b) We can be 90% confident that the actual mean noise level in hospitals is db with a margin of error of 49) (c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between 49 db and 49 db. (d) A 99.9% confidence interval for the actual mean noise level in hospitals is (4.90 db, 49 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 49 db and 49) db. (f) We are 95% confident that the actual mean noise level in hospitals is db, with a margin of error of 4.9 db. (9) How many hospitals must we examine to have 95% confidence that we have the margin of error to within 0.5 db? (h) How many hospitals must we examine to have 99.9% confidence that we have the margin of error to within 0.5 db?

Answer 1

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Dear student, I don't understand the questions number c and e, may due to communi cation problem, a part from that I provide all answers. Details to calculate confidance interval Conly for you). n I o x Zaza

DATE ----....... TAGE where nr sample mean, o standard deviation , n is sample side and Z is table value for various Significance level obtained from Normal table.. 990% con fidence level. 4 7 - 11 x 2.576. 175 = ( 4 3.7 to 50-3.) (6) 90% confidence level. 4 x 1.645 7 1 1 75 = ( 44.9, 4901) Margin of error = 14X1645 = 2.09

- - DATE PAGE: (C.) I don't understand the question But I assuming the following, Here n = 81 and significance level 2 % 98% confidence level is. 47 & 1 181 x 2.33 ( 4415, 49.85). (d.) 99.9% confidence interval is. 47 t x 3.291. 1 175 ( 42.8, 51.2):

Halfhof sample This question is also I don't get it. But I assuming the following: Half of sample = 75 = 37.5=38 95% confidence level is, 4 7 + 1 x 1.96. za (43.5 top 5005) (f) 95% confidence level is. 47 I 775 - (44•5, 49.8) Margins of exxoy - 21196 = 2:49 1 *1*96 YOY l) x1.96

DATO PAGE . ---- (9) Margin of error = 7/2 o 005 = 196 x h = 196X1) 0.5 : 43.12 1 n = 1859.33 44 an n = 1860 (sample size). en is 1860 hospitals to be examined- (bOn5 = 3.291 X !! → n = 5243 :5243 hospital to be examine