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To assess the accuracy of a kitchen scale a standard weight known to weigh 1 gram is weighed a total of n times and the mean, , of the weighings is computed.

Suppose the scale readings are Normally distributed with unknown mean, µ, and standard deviation = 0.01 g.

How large should n be so that a 90% confidence interval for µ has a margin of error of ± 0.0001?

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JeanaShupp

Answer: 27061

Step-by-step explanation:

Given : The scale readings are Normally distributed with unknown mean, µ, and standard deviation [tex]\sigma= 0.01[/tex] g.

Confidence interval = 90%

We know that the critical value for 90% confidence interval = z*=1.645 (by z-table)

Margin of error E= ± 0.0001

Now, Required minimum size : n= [tex](\dfrac{z^*\cdot\sigma}{E})^2[/tex]

[tex]=(\dfrac{1.645\cdot0.01}{0.0001})^2[/tex]

[tex]=(\dfrac{0.01645}{0.0001})^2[/tex]

[tex]=(164.5)^2=27060.25\approx27061[/tex]

Hence, n= 27061.

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