can you check 4 and help me with 5

have a very big budget, so to determine your an sample of 55 (M) adult county residents and calculate a sample mean age ( X ) of 52.0 with a standard deviation (s ) of 17.6. You choose to construct the 90% c.i. (confidence interval) for estimating the population mean age of adults in Albany County. Compute the upper and lower limits of the interval. What is your answer to your supervisor and what is your justification for this answer? 5. Using your same sample, construct a c.i. that has a smaller probability of error (also called alpha, a ). Report the new c.i.; do you reach the same conclusion as in your 90% c.i.? (1 ) F - 5:20 2 X = 3.38 m X = 2.30 13/0 5 = 2170 6- 0.30 52 1. 44 N = 37 2.58 N= 10 N= 58 CI = X - Z WN-1. 102 C ! = x + 2 ( VNT ) 2 = 1.645 ci- 5,20 - 2.51 /2.70 3, 38 - 1:96 (( 3 ) ( 1 = X = 2 JIN- 1 3.38 H . 96 ( . 1 ) 1.2 2130+ 1:64 158 - 1 5:20 - 2:58 ( 0. 45) 4, 039 3.387, 19 0 7.55 5,20 + 1.161 3, 576 - 132) 4.039 2 1 2 6.36 30184 2 MZB. 576 2:30 - 2 216 6 7 8 11 12 13 14 15 S M To 17 18 19 20 21 22 JUNE 30 31 23 24 25 26 27 28 29 W T 2 F 6 7 8 9 10 11 12 13 14 15 16 17 20 21 22 27 SM JULYX- 52.0 15: 17.6 90% . 1.645 IN: 55 5,20 4 1.645 (155-1 2 157- 735 5.20 $ 1.645 ( 2.37) 5.20 + 3.93 Low 1,27 High 9.13 Based on this result, we would estimate that the average age of adults of 5,20 + 3. 93 In Albany county are over 50 The lower limits of our Interval is 1:21 and upper limit ( 5.20 + 3. 93 ) is 9.B the average lage population is greater than or equal because the alpha was a setat 0. 05 level this estimate has a sob chance of being wrong