A sample of 56 research cotton samples resulted in a sample average percentage e
ID: 3356839 • Letter: A
Question
A sample of 56 research cotton samples resulted in a sample average percentage elongation of 8.13 and a sample standard deviation of 1.42. Calculate a 95% large-sample CI for the true average percentage elongation . (Round your answers to three decimal places.)
(___ , ___)
What assumptions are you making about the distribution of percentage elongation?
o We assume the distribution of percentage elongation is uniform.
o We assume the distribution of percentage elongation is normal with the value of known.
o We make no assumptions about the distribution of percentage elongation.
o We assume the distribution of percentage elongation is normal with the value of unknown.
Explanation / Answer
Mean is 8.13 and s is 1.42, the standard error, SE for 56 sample size is s/sqrt(N)=1.42/sqrt(56)=0.1898
For 95% confidence, the z value is 1.96
thus lower limit is mean-SE*z =8.13-1.96*0.1898=7.758
Upper limit is mean+SE*z=8.13+1.96*0.1898 =8.502
We are assuming that the distribution is normal with s known. thus answer is B
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