Question

In: Math

In order to estimate the average electricity usage per month, a sample of 125 residential customers...

In order to estimate the average electricity usage per month, a sample of 125 residential customers were selected, and the monthly electricity usage was determined using the customers' meter readings. Assume a population variance of 12,100kWh2. Use Excel to find the 98% confidence interval for the mean electricity usage in kilowatt hours. Round your answers to two decimal places and use ascending order.

Electric Usage
765
1139
714
687
1027
1109
749
799
911
631
975
717
1232
806
637
894
856
896
1272
1224
621
606
898
723
817
746
933
595
851
1027
770
685
750
1198
975
678
1050
886
826
1176
583
841
1188
692
733
791
584
1163
593
1234
603
1044
1233
1178
598
904
778
693
590
845
893
1028
975
788
1240
1253
854
1185
1164
741
1058
1053
795
1198
1240
1140
959
938
1008
1035
1085
1100
680
1006
977
1042
1252
943
1165
1014
912
791
612
935
864
953
667
1005
1063
1095
1086
810
1032
970
1099
1229
892
1074
579
754
1007
1116
583
763
1231
966
962
1132
738
1033
697
891
840
725
1031

Solutions

Expert Solution

Solution:

We are given that a sample of 125 residential customers electricity usage per month. We have to use Excel to construct 98% confidence interval for the mean electricity usage in kilowatt hours.

Population Variance =

Thus population standard deviation

Formula:

We use excel commands to find sample mean and margin of Error E

For sample mean:

=AVERAGE( select numbers)

for Margin of Error E

=CONFIDENCE.NORM( alpha , std_dev , Size )

where

alpha = 1 -c = 1- 0.98 = 0.02

Std_dev =

Size = n = 125

Thus we get:

=CONFIDENCE.NORM(0.02, 110 , 125 )

=22.89

Electric Usage
765
1139
714
687
1027
1109
749
799
911
631
975
717
1232
806
637
894
856
896
1272
1224
621
606
898
723
817
746
933
595
851
1027
770
685
750
1198
975
678
1050
886
826
1176
583
841
1188
692
733
791
584
1163
593
1234
603
1044
1233
1178
598
904
778
693
590
845
893
1028
975
788
1240
1253
854
1185
1164
741
1058
1053
795
1198
1240
1140
959
938
1008
1035
1085
1100
680
1006
977
1042
1252
943
1165
1014
912
791
612
935
864
953
667
1005
1063
1095
1086
810
1032
970
1099
1229
892
1074
579
754
1007
1116
583
763
1231
966
962
1132
738
1033
697
891
840
725
1031

Thus

Thus 98% confidence interval for  the mean electricity usage is:


Related Solutions

In order to estimate the average electricity usage per month, a sample of 40 residential customers...
In order to estimate the average electricity usage per month, a sample of 40 residential customers were selected, and the monthly electricity usage was determined using the customers' meter readings. Assume a population variance of 12,100kWh2. Use Excel to find the 98% confidence interval for the mean electricity usage in kilowatt hours. Round your answers to two decimal places and use ascending order. Electric Usage 765 1139 714 687 1027 1109 749 799 911 631 975 717 1232 806 637...
In order to estimate the average electric usage per month, a sample of 47 houses were...
In order to estimate the average electric usage per month, a sample of 47 houses were selected and the electric usage determined. The sample mean is 2,000 KWH. Assume a population standard deviation of 142 kilowatt hours. At 90% confidence, compute the margin of error? In order to estimate the average electric usage per month, a sample of 46 houses were selected and the electric usage determined. The sample mean is 2,000 KWH. Assume a population standard deviation of 127...
The average electricity usage per month for a 2000 ft2 house is 550 kWh and the...
The average electricity usage per month for a 2000 ft2 house is 550 kWh and the standard deviation is 180 kWh. What is the probability that a household will use more than 750 kWh per month? It might help to sketch the distribution and put the important information on the graph. z=? p(x>750)=?
A city planner wants to estimate the average monthly residential water usage in the city. He...
A city planner wants to estimate the average monthly residential water usage in the city. He selected a random sample of 40 households from the city, which gave the mean water usage to be 3411.10 gallons over a one-month period. Based on earlier data, the population standard deviation of the monthly residential water usage in this city is 387.70 gallons. Make a 95% confidence interval for the average monthly residential water usage for all households in this city. Round your...
Python problem Residential and business customers are paying different rates for water usage. Residential customers pay...
Python problem Residential and business customers are paying different rates for water usage. Residential customers pay $0.005 per gallon for the first 6000 gallons. If the usage is more than 6000 gallons, the rate will be $0.007 per gallon after the first 6000 gallons. Business customers pay $0.006 per gallon for the first 8000 gallons. If the usage is more than 8000 gallons, the rate will be $0.008 per gallon after the first 8000 gallons. For example, a residential customer...
Write a MATLAB code to calculate the electricity usage in a residential home. Note: Thing that...
Write a MATLAB code to calculate the electricity usage in a residential home. Note: Thing that impact the charge could be number of rooms, square feet, electric appliances. Plot the electricity charge in terms of the variability.
A random sample of 36 households was selected as part of a study on electricity usage,...
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...
In order to ensure efficient usage of a server, it is necessary to estimate the mean...
In order to ensure efficient usage of a server, it is necessary to estimate the mean number of concurrent users. According to records, the average number of concurrent users at 100 randomly selected times is 37.7. The population standard deviation is σ = 9.2. (a) Construct a 90% confidence interval for the expectation of the number of concurrent users. (b) Conduct a hypothesis test to test whether the true mean number of concurrent users is greater than 35. Based on...
In order to ensure efficient usage of a server, it is necessary to estimate the mean...
In order to ensure efficient usage of a server, it is necessary to estimate the mean number of concurrent users. According to 100 randomly selected times of day, the mean number of concurrent users is 37.7 and the standard deviation is 9.2. a. Find the 99% confidence interval for the mean number of concurrent users. b. Write a statement explaining your confidence interval.
The electric cooperative needs to know the mean household usage of electricity by its non-commercial customers...
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 0.14 kWh. A previous study found that for an average family the variance is 5.29 kWh and the mean is 19.9 kWh per day. If they are using a 80% level of confidence, how large of a sample is required to estimate the mean usage of electricity? Round your...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT