can you check question 5

14 X : 52.0 5: 17,6 90% 1.645 NV= 5S 5:20 - 1.645 (135- 1 - 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 90 The lower limits of our Interval is 1:2 ) and upper um + ( 5. 20 + 3 . 93 ) Is 9.13 The average age population is greater than or equal because the alpha was @ set at 0.05 level this estimate has a sob chance of being wrong 4. In your new job in the Albany County Planning Office, your supervisor wants to know if the average age of adults (persons age 18 and over) in Albany County is over age 50. You don't have a very big budget, so to determine your answer to this question, you collect a random 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? 210845 2 72,516 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.? 90 95 (1 ) 8 : 5.20 2 * = 3 1 38 (3) X : 2.30 5= 2170 6-0.30 N= 10 152 1 44 N = 37 2.58 N = 58 102 CI= X = 2 (UNT CI = X + 2 (UN-T. 2 = 1. 645 Ci - 5.20 - 2.51 /2.70 3 , 38 - 1:96 /:30 6 . 3.38 H.96 ( .1 )