Question

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

Answer 1

Answer: 263

Explanation:

As per given , we have

Population standard deviation :
`\sigma=1.9\text{ kWh}`

maximum error : E=0.15 kWh

Significance level :
`\alpha=1-0.98=0.02`

Critical value for 98% confidence interval :
`z_(\alpha/2)=1.28`

Formula to find the sample size :

`n=((z_(\alpha/2)\cdot \sigma)/(E))^2\\\\=((1.28*1.9)/(0.15))^2\\\\=262.872177778\approx263`

Hence, the minimum sample size required =263