Respuesta :
Using the t-distribution, the 98% confidence interval for the mean of the population is (99.21, 110.79).
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.
Researching the problem on the internet, the data-set is given as follows:
104, 102, 103, 105, 125, 106, 107, 110, 110, 95, 108, 86, 112, 101, 101.
The parameters are given as follows:
[tex]\overline{x} = 105, s = 8.54, n = 15[/tex].
The critical value, using a t-distribution calculator, for a two-tailed 98% confidence interval, with 15- 1 = 14 df, is t = 2.6245.
Hence the bounds of the interval are:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 105 - 2.6245\frac{8.54}{\sqrt{15}} = 99.21[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 105 + 2.6245\frac{8.54}{\sqrt{15}} = 110.79[/tex]
More can be learned about the t-distribution at https://brainly.com/question/16162795
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