This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ISDS Homework 5. Solutions  Confidence Intervals Panayiotis Skordi 1. The temperature readings for 20 winter days in Grand Rapids, Michigan are normally distributed with mean 5.5 degrees and a standard deviation of 1.5. Determine the 90% confidence interval estimate for the winter mean temperature. Answer 5 . 5 = x 5 . 1 = = s SD 20 = n 1 . = Use the t distribution as the standard deviation of the population is unknown and the sample size is small, less than 30. The margin of error n s t E 2 = 729 . 1 % 5 2 = = t t since we have 19 degrees of freedom (201). 90% 5% 5% 1.729 1.729 90% Confidence Interval 90% The 5799 . 20 5 . 1 ) 729 . 1 ( 2 = = = n s t E The lower limit of the interval is 9200 . 4 5799 . 5 . 5 = = E x The upper limit of the interval is 0799 . 6 5799 . 5 . 5 = + = + E x The 90% confidence interval estimate for the winter mean temperature is (4.9200,6.0799) . 1 2. A sample of 49 measurements of tensile strength (roof hanger) are calculated to have a mean of 2.45 and a standard deviation of 0.25. Determine the 95% confidence interval for the measurements of all hangers. Answer 45 . 2 = x 25 . = = s SD 49 = n 05 . = The margin of error n s z E 2 = 96 . 1 % 5 . 2 2 = = z z since 95% 2.5% 2.5% 1.96 1.96 95% Confidence Interval 95% The 07 . 49 25 . ) 96 . 1 ( 2 = = = n s z E The lower limit of the interval is 38 . 2 07 . 45 . 2 = = E x The upper limit of the interval is 52 . 2 07 . 45 . 2 = + = + E x The 95% confidence interval for the measurements of all hangers is (2.38, 2.52) . 2 3. If we repeatedly draw samples of size 100 from the population of teenagers, 95% of the values of sample means will be such that the population mean amount of time teenagers spend on the internet would be somewhere between 6.206 hours and 6.794, and 5% of the values of the sample mean will produce intervals that would not include the population mean. a. Determine the 99% confidence interval estimate of the population mean. b. Determine the 90% confidence interval estimate of the population mean. c. Determine the 95% confidence interval estimate of the population mean if the sample size is changed to 300. 95% 2.5% 2.5% 6.206 6.794 95% Confidence Interval 95% Now 96 . 1 % 5 . 2 2 = = z z since We know it is z ( and not t) as the sample size is large (100). 95% 2.5% 2.5% 1.96 1.96 95% Confidence Interval 95% 100 96 . 1 794 . 6 + = x 100 96 . 1 206 . 6  = x Adding these equations together gives: x 2 00 . 13 = So therefore 5 . 6 = x Since 5 . 6 = x then substituting this value into say the very first equation above gives: 100 96 . 1 5 . 6 794 . 6 + = Solving gives: = 96 . 1 100 ) 5 . 6 794 . 6 ( So then 5 . 1 = 3 a. Determine the 99% confidence interval estimate of the population mean....
View Full
Document
 Fall '07
 Skordi

Click to edit the document details