Using the t-distribution, it is found that the 95% confidence interval for the population mean is (1413.18, 1564.82).
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 95% confidence interval, with 65 - 1 = 64 df, is t = 1.9977.
The parameters of the interval are given as follows:
[tex]\overline{x} = 1489, s = 306, n = 65[/tex].
Hence, the bounds of the interval are given by:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 1489 - 1.9977\frac{306}{\sqrt{65}} = 1413.18[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 1489 + 1.9977\frac{306}{\sqrt{65}} = 1564.82[/tex]
The 95% confidence interval for the population mean is (1413.18, 1564.82).
More can be learned about the t-distribution at https://brainly.com/question/16162795
