Respuesta :
Using the t-distribution, the 95% confidence intervals for the given standard errors are given as follows:
- s = 20: (851.922, 868.078).
- s = 40: (843.844, 876.156).
- s = 80: (827.688, 892.312).
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.
The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 26 - 1 = 25 df, is t = 2.0595.
The parameters are:
[tex]\overline{x} = 860, n = 26[/tex].
Hence, with s = 20, the bounds of the interval are:
- [tex]\overline{x} - t\frac{s}{\sqrt{n}} = 860 - 2.0595\frac{20}{\sqrt{26}} = 851.922[/tex]
- [tex]\overline{x} + t\frac{s}{\sqrt{n}} = 860 + 2.0595\frac{20}{\sqrt{26}} = 868.078[/tex]
With s = 40, the bounds of the interval are:
- [tex]\overline{x} - t\frac{s}{\sqrt{n}} = 860 - 2.0595\frac{40}{\sqrt{26}} = 843.844[/tex]
- [tex]\overline{x} + t\frac{s}{\sqrt{n}} = 860 + 2.0595\frac{40}{\sqrt{26}} = 876.156[/tex]
With s = 80, the bounds of the interval are:
- [tex]\overline{x} - t\frac{s}{\sqrt{n}} = 860 - 2.0595\frac{80}{\sqrt{26}} = 827.688[/tex]
- [tex]\overline{x} + t\frac{s}{\sqrt{n}} = 860 + 2.0595\frac{80}{\sqrt{26}} = 892.312[/tex]
More can be learned about the t-distribution at https://brainly.com/question/16162795
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