Question

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.

Answer 1

Answer:

[895.05; 940.81]kWh

Step-by-step explanation:

Hello!

Be X: monthly electricity usage by a residential customer

X~N

σ²: 12100KWh²

To estimate the population mean using a 98% CI you have to use the following formula

[X[bar]±$Z_{1-\alpha /2}$ * $\frac{Sigma}{\sqrt{n} }$]

1-α: 0.98

α0.02

α/2: 0.01

1-α/2:0.99

$Z_{1-\alpha /2}= Z_{0.99}= 2.326$

n= 125

X[bar]= ∑X/n= 228565/125= 917.93

[917.93±2.326 * $\frac{110}{\sqrt{125} }$]

[895.05; 940.81]

Using a 98% confidence level you'd expect that the interval [895.05; 940.81]kWh will include the average monthly electricity usage of residential customers.

I hope this helps!

Raw data:

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