Using the t-distribution, it is found that the confidence interval is (14.2, 14.8).
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.
- s is the standard deviation for the sample.
The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 18 - 1 = 17 df, is t = 2.1098.
The other parameters are given by:
[tex]\overline{x} = 14.5, s = 0.68, n = 18[/tex].
Hence, the bounds of the interval are given by:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 14.5 - 2.1098\frac{0.68}{\sqrt{18}} = 14.2[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 14.5 + 2.1098\frac{0.68}{\sqrt{18}} = 14.8[/tex]
The confidence interval is (14.2, 14.8).
More can be learned about the t-distribution at https://brainly.com/question/16162795
