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The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day. They would like the estimate to have a maximum error of 0.1 gallons. A previous study found that for an average family the variance is 5.76 gallons and the mean is 19.5 gallons per day. If they are using a 90% level of confidence, how large of a sample is required to estimate the mean usage of water? Round your answer up to the next integer.

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Cricetus

Answer:

The right solution is "1559".

Step-by-step explanation:

The given values are:

Variance,

[tex]\sigma=\sqrt{5.76}[/tex]

  [tex]=2.40[/tex]

Maximum error,

[tex]M.E=0.1[/tex]

At 90% confidence level,

[tex]\alpha=0.1[/tex]

According to the z-table, the critical value will be:

[tex]Zc=1.645[/tex]

Now,

The sample size will be:

⇒ [tex]n=(Zc\times \frac{\sigma}{E} )^2[/tex]

On substituting the values, we get

⇒    [tex]=(1.645\times \frac{2.4}{0.1} )^2[/tex]

⇒    [tex]=(1.645\times 24)^2[/tex]

⇒    [tex]=(39.48)^2[/tex]

⇒    [tex]=1558.67[/tex]

or,

⇒    [tex]=1559[/tex]

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