## Lesson 14: Estimation with Confidence Intervals

October 22, 2018

Review:

- Sampling Distributions
- Google Survey for Colorado Governor’s Race

Presentation:

**How to Estimate a Population Mean (μ)**- Conduct a Simple Random Sample (SRS) of size
*n* - Calculate the sample mean, x
*-bar = (∑x)/n* - Calculate the sample standard deviation, s =
*√(∑(x – x-bar)²/(n-1)* - If
, we can*n*is sufficiently large (≥30)- Assume
*x-bar*is Normally distributed - Estimate the population mean,
(point estimate)**μ = x-bar** - Calculate margin of error,
*m = z*σ*- where
*z*= 1.96 for 95% confidence - where
*σ = s/(√n)*

- where

- Assume
- Estimation with confidence interval for
*μ =**x-bar ± m*

- Conduct a Simple Random Sample (SRS) of size
- Watch video
- Example: A sample of size n = 400 produced the sample mean, x-bar = 36.0 and standard deviation, s = 9.0. Construct a 95% confidence interval to estimate the population mean.
- n = 400
- x-bar = 36.0
- s = 9.0
- m = 1.96*(9/(√400)) = 0.882
- Estimate for μ = 36.0 ± 0.882
- 95% confidence interval: [35.118, 36.882]

- Demonstrate (in Sheets) calculation of Pueblo voter turnout using 2014 turnout percentages by County.

Activity:

**Problem 1**. A sample of size n = 100 produced the sample mean, x-bar = 16.0 and standard deviation, s = 3.0. Construct a 95% confidence interval to estimate the population mean.**Problem 2**. An operation manager at a large plant observed 120 workers assembling an electronic component. The average time needed for assembly was 16.2 minutes with a standard deviation of 3.6 minutes. Construct a 95% confidence interval to estimate the mean assembly time.**Problem 3**. A computer technician installs new hard drives on 64 different computers. The average installation time is 42 minutes with a standard deviation of 5 minutes. Construct a 95% confidence interval for the mean installation time.**Problem 4**. A research firm conducted a survey of regular smokers to estimate the average amount spent per week on cigarettes. A sample of 49 regular smokers revealed average spending on cigarettes to be $21.55 with a standard deviation of $5.21. Construct a 95% confidence interval to estimate mean weekly cigarette spending.

Assignment:

- In Sheets, use 2014 voter turnout percentages and current voter registration data to estimate voter turnout in your
. Provide a point estimate with a 95% confidence interval. [Yes, you already estimated voter turnout using linear regression. This is an alternative approach.]**assigned county** - Read pp. 353-362, Introduction to Inference and Estimating with Confidence