Question

2. A random sample of 30 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 2017. The average usage was found to be 375kWh. From past years in the March quarter the population standard deviation of the usage was 81k Assuming the standard deviation is unchanged and that the usage is normally distributed ; A. Determine the interval of 95% confidence for the average kilowatt-hours for the population. B. Determine the 99% confidence interval. C. With a confidence level of 90%, what would the minimum sample size need to be in order for the true mean of the heights to be less than 20kWh from the sample mean? (This would mean an error amount of 20kWh.)

Answer 1

Answer:

Step-by-step explanation:

Given that n =30, x bar = 375 and sigma = 81

Normal distribution is assumed and population std dev is known

Hence z critical values can be used.

For 95% Z critical=1.96

Margin of error = $1.96(\frac{81}{\sqrt{30} } )=29$

Confidence interval = 375±29

=(346,404)

B) 99% confidence

Margin of error = 2.59*Std error =38

Confidence interval = 375±38

=(337, 413)

C) For 90%

Margin of error = 20

Std error = 20/1.645 = 12.158

Sample size

$n=(\frac{81}{12.158} )^2\\=44.38$

Atleast 44 people should be sample size.