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Sagot :
ANSWER:
Lower limit: 4.73
Upper limit: 4.78
STEP-BY-STEP EXPLANATION:
Given the following data set:
[tex]4.72,4.74,4.83,4.75,4.73,4.87,4.69,4.7,4.76,4.7,4.79,4.76[/tex]We calculate the mean and standard deviation:
[tex]\begin{gathered} \mu=\frac{4.72+4.74+4.83+4.75+4.73+4.87+4.69+4.7+4.76+4.7+4.79+4.76}{12}=\frac{57.04}{12}=4.753 \\ \\ \sigma=\sqrt{\frac{\lparen4.72-4.753^2+\left(4.74-4.753\right)^2+\lparen4.83-4.753)^2+\left(4.75-4.753\right)^2+\left(4.73-4.753\right)^2+\lparen4.87-4.753)^2+\left(4.69-4.753\right)^2+\lparen4.7-4.753)^2+\left(4.76-4.753\right)^2+\left(4.7-4.753\right)^2+\left(4.79-4.753\right)^2+\left(4.76-4.753\right)^2}{12-1}} \\ \\ \sigma=0.054 \end{gathered}[/tex]The critical limit of 90% confidence interval is 1.645
We can determine the limits as follows:
[tex]\begin{gathered} \text{ Lower limit: }\mu-Z\cdot\frac{\sigma}{\sqrt{n}}=4.753-1.645\cdot\frac{0.054}{\sqrt{12}}=4.727\cong4.73 \\ \\ \text{ Upper limit:: }\mu+Z\cdot\frac{\sigma}{\sqrt{n}}=4.753+1.645\cdot\frac{0.054}{\sqrt{12}}=4.778\cong4.78 \end{gathered}[/tex]
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