Using the t-distribution, as we have the standard deviation for the sample, we could claim that the mean weight in pounds is in the interval (79.1, 85.7).
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 99% confidence interval, with 100 - 1 = 99 df, is t = 2.6259.
The other parameters have values given by:
[tex]\overline{x} = 82.4, s = 12.6, n = 100[/tex].
We could claim that the mean is any value in the 99% confidence interval, which has bounds given by:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 82.4 - 2.6259\frac{12.6}{\sqrt{100}} = 79.1[/tex]
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 82.4 + 2.6259\frac{12.6}{\sqrt{100}} = 85.7[/tex]
More can be learned about the t-distribution at https://brainly.com/question/16162795
