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A department store manager to estimate at a %90 confidence level the mean about by all customers at this store

Sagot :

Answer:

The answer is "287.19"

Step-by-step explanation:

Please find the complete question in the attached file.

[tex]\alpha=1-0.90=0.10\\\\\frac{\alpha}{2}=0.05\\\\Z=1.64\\\\Marginal \ error=Z \times \frac{\sigma_x}{\sqrt{n}}\\\\n=(\frac{Z \times \sigma_x}{error})^2\\\\n=(\frac{1.64 \times 31}{3})^2\\\\n=287.1895\approx 287.19[/tex]

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