Question

4. Scores on an exam are normally distributed with a population standard deviation of 5.5. A random sample of 50 scores on the exam has a mean of 28. (a) (5 pts.) Construct 80% confidence interval. (b) (5 pts.) Construct 85% confidence interval. (c) (5 pts.) Construct 92% confidence interval. (d) (5 pts.) When confidence level increases what will happen to the confidence interval.

Answer 1

On an Given that Scores exam are normally distri- buted with population Stondard deviation, o =5.5 Sample size, n = 50 Sample mean, z = 28 (a) We have to find the 80% confidence interval. Significance level, C = 1-0.80 = 0.20 Here the population Variance is given, also given that the population is normal. Then the 80% confidence interval is: X - Z X + 2 Here x = 28 n = 50 x = -2 → a= 0.1 1.282 (using Standard The Za Zool normal table) o = 5.5 Thus the confidence interval is. *O 0.1 V x + Zoon * V 5.5 (28-0.9972 , 28- 1.282 * 5.5 28+ 1.282* V50 V50 28+ 0.9972 ) (24.0028, 28.9972)

Thus the 80% confidence interval is ( 27.0028, 28.9972) (b) We have to find the 85% confidence interval. Then Significance level, d = 1-0.85 = 0.15 Then the 85% Confidence interval is. x + * 2 x= 0.15 Blo 0.075. Then using Standard normal table, we can find Zody 20.075 1.44 Given that ã = 28, D= 50, and 0 = 5.5 Then the 85% confidence intewal is : 7 - 20.015* Vož +70015 in) * 28 = 1.44* 5.5 28+ 1.44 V50 5.5 V50 (28- 1.12 28+1.12) (26.88 , 29.12) Thus the 85% confidence interval is 29.12) (26.88

(C) kle have to construct a 92%. Confidence inter- val. Then 1-0.92 0.08 → = 0,08 - 0.04 92% The confidence interval is : * (x - Za to ) a Here a 10.04 28, n = 50 and population SD, T-5.5 = 1.751 Cuising Standard normal table we can find this 20:04) 92% confidence interval is Thus the (x - 70.04 x + Z004 V ( 28 -1.751 * 5.5 + 1- 751* 5.5 150 V50 ( 28- 1.362 28 + 1. 1.362) ( 26.638, 29.362) 3 Thus the 92% confidence interval is. ( 26.638, 29.362) - (d) Increasing the confidence level increases the Margin of eryOY = Z *

That is, for 80% confidence level, Margin of eyyor=0.9972 for 85% confidence level, Margin of exvoy = 1.12 and for 92% confidence level, Margin of exror = 1.362 As the confidence level increases, the margin of erroy also increases, and makes the confid- ence interval wider. Thus, when confidence level increases, the Confidence interval becomes widey.