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Sagot :
Given that:
- The sample size is 28 city residents:
[tex]n=28[/tex]- The mean of the time (in years) they had lived at their present address was:
[tex]\mu=9.3[/tex]- The standard deviation (in years) of the population was:
[tex]\sigma=2[/tex]Then, you need to use the following formula for calculating the Confidence Interval given the Mean:
[tex]C.I.=\mu\pm z\frac{\sigma}{\sqrt{n}}[/tex]Where μ is the sample mean, σ is the standard deviation, "z" is the z-score, and "n" is the sample size.
By definition, the z-score for a 90% confidence interval is:
[tex]z=1.645[/tex]Therefore, you can substitute values into the formula and evaluate:
[tex]C.I.=9.3\pm(1.645)(\frac{2}{\sqrt{28}})[/tex]You get that the lowest value is:
[tex]9.3-(1.645)(\frac{2}{\sqrt{28}})\approx8.678[/tex]And the highest value is:
[tex]9.3+(1.645)(\frac{2}{\sqrt{28}})\approx9.922[/tex]Hence, the answer is:
[tex]From\text{ }8.678\text{ }to\text{ }9.922[/tex]
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