Question

A random sample of 36 households was selected as part of a study on electricity usage, and the number of kilowatt-hours (kWh) was recorded for each household in the sample for the March quarter of 2019. The average usage was found to be 375kWh. In a very large study (a population) in the March quarter of the previous year it was found that the standard deviation of the usage was 72kWh. Assuming the standard deviation is unchanged and that the usage is normally distributed, provide an expression for calculating a 97% confidence interval for the mean usage in the March quarter of 2019.

Answer 1

Solution :

Given that,

= 375 kWh

s = 72 kWh

n = 36

Degrees of freedom = df = n - 1 = 36 - 1 = 35

At 97% confidence level the t is ,

= 1 - 97% = 1 - 0.97 = 0.03

/ 2 = 0.03 / 2 = 0.015

t _/2,df = t_0.015,35 = 2.262

Margin of error = E = t_/2,df * (s /n)

= 2.262 * (72 / 36)

= 27.144

The 97% confidence interval estimate of the population mean is,

- E < < + E

375 - 27.144 < < 375 + 27.144

347.856 < < 402.144

(347.856,402.144)

The 97% confidence interval is 347.856 kWh to 402.144 kWh