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Sagot :
Using the t-distribution, as we have the standard deviation for the sample, it is found that the 98% confidence interval for the true mean age of Summer Olympians since the 1980 Olympics is (20.6, 24.8). It means that we are 98% sure that the true age of all Olympians is between these two values.
What is a t-distribution confidence interval?
The confidence interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
In which:
- [tex]\overline{x}[/tex] is the sample mean.
- t is the critical value.
- n is the sample size.
- s is the standard deviation for the sample.
In this problem, with the help of a t-distribution calculator, with a two-tailed significance level of 0.02 and 24 - 1 = 23 degrees of freedom, the parameters are:
[tex]\overline{x} = 22.7, s = 4.15, n = 24, t = 2.5[/tex]
Then, the interval is:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 22.7 - 2.5\frac{4.15}{\sqrt{24}} = 20.6[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 22.7 + 2.5\frac{4.15}{\sqrt{24}} = 24.8[/tex]
The 98% confidence interval for the true mean age of Summer Olympians since the 1980 Olympics is (20.6, 24.8). It means that we are 98% sure that the true age of all Olympians is between these two values.
More can be learned about the t-distribution at https://brainly.com/question/16162795
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