This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**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 . = α 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 α = 729 . 1 % 5 2 = = t t α since we have 19 degrees of freedom (20-1). 90% 5% 5% –1.729 1.729 90% Confidence Interval 90% The 5799 . 20 5 . 1 ) 729 . 1 ( 2 = = = n s t E α 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 . = α The margin of error n s z E 2 α = 96 . 1 % 5 . 2 2 = = z z α 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 α 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 μ 95% Confidence Interval 95% Now 96 . 1 % 5 . 2 2 = = z z α 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 σ + = x 100 96 . 1 206 . 6 σ- = 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 σ + = Solving gives: σ =- 96 . 1 100 ) 5 . 6 794 . 6 ( So then 5 . 1 = σ 3 a. Determine the 99% confidence interval estimate of the population mean....

View
Full Document