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The electric cooperative needs to know the mean household usage of electricity by its non-commercial customers in kWh per day. They would like the estimate to have a maximum error of 0.15 kWh. A previous study found that for an average family the standard deviation is 1.9 kWh and the mean is 16.7 kWh per day. If they are using a 98% level of confidence, how large of a sample is required to estimate the mean usage of electricity? Round your answer up to the next integer

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JeanaShupp

Answer: 263

Step-by-step explanation:

As per given , we have

Population standard deviation : [tex]\sigma=1.9\text{ kWh}[/tex]

maximum error : E=0.15 kWh

Significance level : [tex]\alpha=1-0.98=0.02[/tex]

Critical value for 98% confidence interval : [tex]z_{\alpha/2}=1.28[/tex]

Formula to find the sample size :

[tex]n=(\dfrac{z_{\alpha/2}\cdot \sigma}{E})^2\\\\=(\dfrac{1.28\times1.9}{0.15})^2\\\\=262.872177778\approx263[/tex]

Hence, the minimum sample size required =263

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