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Sagot :
Using the t-distribution, it is found that the 95% confidence interval for the mean number of people the houses were shown is (20.1, 27.9).
We have the standard deviation for the sample, hence the t-distribution is used to build the confidence interval. Important information are given by:
- Sample mean of [tex]\overline{x} = 24[/tex].
- Sample standard deviation of [tex]s = 9[/tex].
- Sample size of [tex]n = 23[/tex]
The confidence interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
In which t is the critical value for a 95% confidence interval with 23 - 1 = 22 df, thus, looking at a calculator or at the t-table, it is found that t = 2.0739.
Then:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 24 - 2.0739\frac{9}{\sqrt{23}} = 20.1[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 24 + 2.0739\frac{9}{\sqrt{23}} = 27.9[/tex]
The 95% confidence interval for the mean number of people the houses were shown is (20.1, 27.9).
A similar problem is given at https://brainly.com/question/15180581
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