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Answer:
The confidence interval will be "41.04, 43.81".
Step-by-step explanation:
Given that,
Sample size,
n = 10
Sample total,
423.5
Sample mean,
[tex]\bar X =\frac{423.5}{10}[/tex]
[tex]=42.35[/tex]
Sample variance,
[tex]s = \sqrt{4.5894}[/tex]
[tex]=2.1423[/tex]
[tex]df=n-1[/tex]
[tex]=9[/tex]
[tex]t^*=18331[/tex]
Now,
The margin of error will be:
⇒ [tex]E=\frac{s\times t^*}{\sqrt{n} }[/tex]
[tex]=\frac{2.1423\times 1.8331}{\sqrt{9} }[/tex]
[tex]=\frac{3.928}{3 }[/tex]
[tex]=1.309[/tex]
hence,
The 90% confidence level will be:
= [tex](\bar X-E),(\bar X+E)[/tex]
By substituting the values, we get
= [tex](42.35-1.309),(42.35+1.309)[/tex]
= [tex](41.04),(43.81)[/tex]