Using the t-distribution, the 96% confidence interval is given as follows:
(18.96, 21.04).
The confidence interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
In which:
The critical value, using a t-distribution calculator, for a two-tailed 96% confidence interval, with 100 - 1 = 99 df, is t = 2.0812.
The parameters are given as follows:
[tex]\overline{x} = 20, s = 5, n = 100[/tex]
Hence the bounds of the interval are:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 20 - 2.0812\frac{5}{\sqrt{100}} = 18.96[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 20 + 2.0812\frac{5}{\sqrt{100}} = 21.04[/tex]
More can be learned about the t-distribution at https://brainly.com/question/16162795
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