Using the t-distribution, as we have the standard deviation for the sample, it is found that the 99% confidence interval for the true mean number of visitors on summer days is given by:
c. 92.664 to 103.940 visitors
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 30 - 1 = 29 df, is t = 2.756.
The other parameters are given as follows:
[tex]\overline{x} = 98.3, s = 11.2, n = 30[/tex].
Hence, the bounds of the interval are given by:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 98.3 - 2.756\frac{11.2}{\sqrt{30}} = 92.664[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 98.3 + 2.756\frac{11.2}{\sqrt{30}} = 103.94[/tex]
Hence option c is correct.
More can be learned about the t-distribution at https://brainly.com/question/16162795
