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A randomly selected sample of n =51 men in Brazil had an average lifespan of 59 years. The standard deviation was 10 years. Calculate a 98% confidence interval for the average lifespan for all men in Brazil.

User Mitali
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4 votes

Answer:


59-2.40(10)/(√(51))=55.639


59+ 2.40(10)/(√(51))=62.361

The 98% confidence interval would be given by (55.639;62.361)

Explanation:

Information given


\bar X= 59 represent the sample mean


s= 10 represent the sample deviation


n= 51 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:


\bar X \pm t_(\alpha/2)(s)/(√(n)) (1)

The degrees of freedom are given by:


df=n-1=51-1=50

The Confidence is 0.98 or 98%, the significance would be
\alpha=0.02 and
\alpha/2 =0.01,and the critical value would be
t_(\alpha/2)=2.40

And replacing we got:


59-2.40(10)/(√(51))=55.639


59+ 2.40(10)/(√(51))=62.361

The 98% confidence interval would be given by (55.639;62.361)

User Sam Luther
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