Question

1-

You are the operations manager for an airline and you are considering a higher fare level for passengers in aisle seats. How many randomly selected air passengers must you​ survey? Assume that you want to be

9090​%

confident that the sample percentage is within

1.51.5

percentage points of the true population percentage. Complete parts​ (a) and​ (b) below.

a. Assume that nothing is known about the percentage of passengers who prefer aisle seats.

nequals=nothing

​(Round up to the nearest​ integer.)

b. Assume that a prior survey suggests that about

3333​%

of air passengers prefer an aisle seat.

nequals=nothing

​(Round up to the nearest​ integer.)

2-

A food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm). Listed below are the amounts of mercury​ (ppm) found in tuna sushi sampled at different stores in a major city. Construct a

9999​%

confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna​ sushi?

0.500.50  

0.740.74  

0.110.11  

0.980.98  

1.321.32  

0.520.52  

0.890.89

  

What is the confidence interval estimate of the population mean

muμ​?

nothing

ppmless than<muμless than<nothing

ppm

​(Round to three decimal places as​ needed.)

Does it appear that there is too much mercury in tuna​ sushi?

A.

​No, because it is not possible that the mean is greater than 1 ppm.​ Also, at least one of the sample values is less than 1​ ppm, so at least some of the fish are safe.

B.

​Yes, because it is possible that the mean is greater than 1 ppm.​ Also, at least one of the sample values exceeds 1​ ppm, so at least some of the fish have too much mercury.

C.

​No, because it is possible that the mean is not greater than 1 ppm.​ Also, at least one of the sample values is less than 1​ ppm, so at least some of the fish are safe.

D.

​Yes, because it is possible that the mean is not greater than 1 ppm.​ Also, at least one of the sample values exceeds 1​ ppm, so at least some of the fish have too much mercury.

Answer 1

Gunition Given Alisema Corlidence interval frr): 904 09 confidence that the sample perontage is eithiolt1-50% Iof papoution of postopens whes porten an aide from pant (e) prior suiwer supports se havelp) 5. Powd of significance (d):1-01 1-09 we know that (0) = P14Zone horn (ov) n= pruzz here P = 0.33 -15% s 9:1-P 0.015 The :1-0.33 -0.67 2005 - 11.645) from standing romal z table Substituting all Valses in inte Cam fcanner - cslca

D: (0-31)(0.67) * ( be bus ..^=(0.33) (0,67)*10006.777 In = 2659.120 ô n 12659 CS Scanned with CamScanner

2nd question

answer

Standard Deviation, s: 0.39101211770286 Count, N: 7 Sum, Ex 5.06 Mean, X 0.72285714285714 Variance, s2 0.15289047619048 Steps 1 N --- Šar- Vi - T), VN - 1 is 52 - E(X; - x)2 - N-1 - (0.50 - 0.72285714285714)2 + ... + (0.89 -0.72285714285714)2 0.91734285714286 7 .1 = 0.15289047619048 S = 10.15289047619048 = 0.39101211770286

(river dola 6.50 ,0.74, 0.1, 0.78,0·23,0:52,0.3 S 0.391 Construct 92. comdistance intemul estimate of the Pop Jetion mpon M. d: 1-ci 6005,6 to + valore (night taily 3.7074). 1.0.99 & from standard nomad + or normal 01 -(0.42284 3.7074 * 0.3911 V hele (0.1126 3.7074 ito 10:1724 + o. sual : 10.7227-0.5479,07728 +0.3.78] CS CamScanner Scanned with .

= (0.145 , 1-271) The CI estimate of the population meam M is, 0.175ppm 2M € 19.271) ppm. CS Scannertion 'B' is correct, CamScanner

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