Respuesta :
Using the t-distribution, the 95% confidence interval within which the mean of the total population of such cases is expected lie is:
(112.5, 127.5).
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.
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 parameters for this problem are:
[tex]\overline{x} = 120, s = 15, n = 18[/tex].
Hence the bounds of the interval are:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 120 - 2.1098\frac{15}{\sqrt{18}} = 112.5[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 120 + 2.1098\frac{15}{\sqrt{18}} = 127.5[/tex]
More can be learned about the t-distribution at https://brainly.com/question/25890103
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