Question

A random sample of n measurements was selected from a population with unknown mean u and standard deviation o = 15 for each of the situations in parts a through d. Calculate a 99% confidence interval for u for each of these situations. a. n=50, x=31 b. n=250, x= 112 c. n= 120, x = 16 d. n= 120, x=5.21 e. Is the assumption that the underlying population of measurements is normally distributed necessary to ensure the validity of the confidence intervals in parts a through d? Explain.

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Answer 1

m=50, X=31 Since n e x is Nomelly distributed by Central Linit I Theown (C.LT) 990. Confidence itewel is - 77 2.58 5 ha = (x-2. Spray 3125803) = (31 - 2.58 x 15 , 31+ 2.58 x 15 ) => [ 25.526, 36.473) 12.58 is 2x, using Normal Table where a 0.01 as 997 Confidence ütevel is Lanoluetech 1- 2.58x15, 31+ 2.58 x 15 150 5. n=25o, X=112 n>30, 7 Nound by C.LT 9911_esis- & 2.5863 (112 - 2.58x15 , 112 +2.58x15 => 1 109.552, 114,443) 125 1250 n=120, P=16 n73o. Xa Hormal by C.LT 1994. com is '- 52.586 =) ( 16–2:58x15 , 16+2.58 XIS 720 120 =) (12.467, 19,532) in 120 n = 120, X= 5.21 n>30, x ~Normal by (.LT 99-1.6I iszt + 288 521–2258x 15, 5-21 +2.58 x 15 V120 1.677, 8.742) no assumption is not necessary. Because by Central Limit Theaum all averages are assymptotically normally Distributed for n> 3o .. e