Here are summary statistics for randomly selected weights of newborn girls:
n=235, x overbar=29.42hg, s=7.37hg. Construct a confidence
interval estimate of the mean. Use a 98% confidence level. Are
these results very different from the confidence interval
27.2hg<μ<31.0hg with only 16sample values, x
overbar=29.1hg, and s equals=2.9hg?
What is the confidence level What is the confidence interval for
the population mean μ?
Are the results between the two confidence intervals very
different?
Solution:
Given that,
= 29.42 ....... Sample mean
s = 7.37 ........Sample standard deviation
n = 235 ....... Sample size
Note that, Population standard deviation() is unknown..So we use t distribution.
Our aim is to construct 98% confidence interval.
c = 0.98 ..Confidence level.
= 1- c = 1- 0.98 = 0.02
/2 = 0.01
Also, d.f = n - 1 = 234
= = 0.01,234 = 2.342
( use t table or t calculator to find this value..)
The margin of error is given by
E = /2,d.f. * ( / n)
= 2.342 * (7.37 / 235 )
= 1.1
Now , confidence interval for mean() is given by:
( - E ) < < ( + E)
(29.42 - 1.1) < < (29.42 + 1.1)
28.32 < < 30.52
Now , compare this interval with a given interval 27.2 hg < μ < 31.0 hg
Yes , the results very different.
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