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

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
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1 Answer

Use Margin of Error formula E =±tα/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, Sx : 0.04, n: 7, C-Level: .99, Calculate