[t-interval] A random sample of size 18 is drawn from a population that is normally distributed. The sample mean is 58.5, and the sample standard deviation is found to be 11.5. Determine a 95% confidence interval about population mean.
A. [54.04,62.96]
B. [52.78,64.22]
C. [53.78,63.22]
D. [53.18,63.81]
Sample size = n = 18
Sample mean = = 58.5
Standard deviation = s = 11.5
We have to construct 95% confidence interval.
Formula is
Here E is a margin of error.
Degrees of freedom = n - 1 = 18 - 1 = 17
Level of significance = 0.05
tc = 2.110 ( Using t table)
So confidence interval is ( 58.5 - 5.7188 , 58.5 + 5.7188) = > ( 52.78 , 64.22)
B. [52.78,64.22]
[t-interval] A random sample of size 18 is drawn from a population that is normally distributed....
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