SOLUTION:
From given data,
Here are summary statistics for randomly selected weights of newborn girls: n=163, =28.6 hg, s = 6.9 hg. Construct a confidence interval estimate of the mean.
Use a 95% confidence level. Are these results very different from the confidence interval 27.6 hg < < 30.0 hg with only 20 sample values, = 28.8 hg, and s = 2.5 hg?
n = Sample Size = 163
=Sample Mean = 28.6 hg
s = Sample standard deviation = 6.9 hg
What is the confidence interval for the population mean ?
SE = s / sqrt(n)
SE = 6.9 / sqrt(163) =0.540449
Take
95% confidence interval
95/100 = 0.95
Significance level = = 1-0.95 = 0.05
/2 = 0.05/2 = 0.025
Degree of freedom = df = n - 1 = 163 - 1 = 162
From Table,
critical values of t
t/2,df = t0.025,162 = 1.9747
Confidence Interval:
t/2,df * SE
28.6 (1.9747 * 0.540449)
28.6 1.0672246403
( 28.6 - 1.0672246403 , 28.6+1.0672246403)
(27.6 , 29.7)
27.6 hg < < 29.7 hg (Rounded to one decimal place.)
Are the results between the two confidence intervals very different?
We have the confidence interval 27.6 hg < < 30.0 hg.
Answer : option (A)
No, because the confidence interval limits are similar.
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