Use the given confidence level and sample data below to find (a) the margin of error and (b) the confidence interval for the population mean. Assume that the sample is a simple random sample and the population has a normal distribution. 99% confidence; n = 7, = 0.12 , s = 0.04

Use Margin of Error formula

**E =±t**since σ unknown_{α/2 }(s/√n)99% CI ==> α=.01 ==> α/2 =.005

degrees of freedom=n-1=7-1=6

Looking at table for t-distribution at 99%C-Level, critical value = 3.707

Substitute all these values into formula:

a.) E = ±3.707(0.04/√7)

= ±3.707(.0151)

=

**± 0.056**b.) x-bar-E < μ < x-bar+E

0.12-0.056 < μ < 0.12+0.056

**.064 < μ < .176**

**Therefore, you can say with 99% confidence, that the true mean(μ) lies between .064 and .176.**

**NOTE: If you're using Graphing calculator TI-83/TI-83 Plus:**

Enter

**STATS, TESTS, 8: T Interval, Stats,**x-bar:**0.12,**S_{x}:**0.04,**n:**7**, C-Level:**.99, Calculate**