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

Step-by-step 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 :

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

Hence, the minimum sample size required =263