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Sagot :
The reasonable 95% confidence intervals for the population mean difference is (7.83, 12.13) , the correct option is (a) .
In the question ,
it is given that ,
90% confidence interval is given as (8.21 , 11.75)
using this interval , we have to find the 95% confidence interval .
Now, [tex]x_{d}[/tex] - E = 8.21
[tex]x_{d}[/tex] + E = 11.75 , where E is the margin of Error .
Solving the equations ,
we get ,
E = 1.77 and [tex]x_{d}[/tex] = 9.98
So , the margin of error E is given as
E = [tex]t_{\frac{0.10}{2},15 } \times \frac{S.D.}{\sqrt{n} }[/tex]
we have to find standard deviation from this ;
given n = 16 and [tex]t_{\frac{0.10}{2},15 }[/tex] = 1.7531
Substituting in the margin of error equation ,
we get ,
S.D. = (1.77 × 4)/1.7531
= 4.0386
So , the 95% confidence interval is given as [tex]x_{d}[/tex] ± [tex]t_{\frac{0.10}{2},14 } \times \frac{S.D.}{\sqrt{n} }[/tex] .
from t table , [tex]t_{\frac{0.10}{2},14 }[/tex] = 2.1314
Substituting ,
we get ,
CI = (9.98 ± 2.1314×4.0386/√16)
= (9.98 ± 2.1519)
= ( 9.98 - 2.1519 , 9.98 + 2.1519)
= (7.83 , 12.13)
Therefore , the correct option is (a) (7.83 , 12.13) .
The given question is incomplete , the complete question is
Based on the 90% confidence interval of (8.21, 11.75) points, which of the following are reasonable 95% confidence intervals for the population mean difference in exam scores before and after taking an exam prep course? Select any that may apply:
(a) (7.83, 12.13)
(b) (7.53, 11.83)
(c) (8.45, 11.51)
(d) (8.63, 11.33)
Learn more about Confidence Interval here
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