contestada

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 0.12 gallons. A previous study found that for an average family the variance is 5.29 gallons and the mean is 17 gallons per day. If they are using a 85% level of confidence, how large of a sample is required to estimate the mean usage of water

Respuesta :

Answer:

[tex]n=(\frac{1.440(2.3)}{0.12})^2 =761.76 \approx 762[/tex]

So the answer for this case would be n=762 rounded up to the nearest integer

Step-by-step explanation:

Information given

[tex]\bar X = 17[/tex] represent the mean

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma= \sqrt{5.29}= 2.3[/tex] represent the standard deviation

Solution to the problem

The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]    (a)

And on this case we have that ME =0.12 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex]   (b)

The confidence level is 85%, the significance level would be [tex] \alpha=1-0.85 = 0.15[/tex] and [tex]\alpha/2 =0.075[/tex] the critical value for this case would be [tex]z_{\alpha/2}=1.440[/tex], replacing into formula (b) we got:

[tex]n=(\frac{1.440(2.3)}{0.12})^2 =761.76 \approx 762[/tex]

So the answer for this case would be n=762 rounded up to the nearest integer

Otras preguntas