Stats question; Can someone please help with this? Thank you ~~

Question

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

Answer 1

Use Margin of Error formula E =±t_α/2 (s/√n) since σ unknown

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