The local energy company claims the average annual electricity bill for its subscribers is just $600. A consumer watchdog group wants to dispute this claim. All agree that the standard deviation sigma of annual electricity bills is $150. Some time later, a wealthy activist provides funding for a simple random sample of 250 households. The average annual electricity bill for this sample is $622. Find a 95% confidence interval for the true mean annual electric bill, based on this sample.

Question
Confidence intervals
asked 2021-03-11
The local energy company claims the average annual electricity bill for its subscribers is just $600. A consumer watchdog group wants to dispute this claim. All agree that the standard deviation sigma of annual electricity bills is $150.
Some time later, a wealthy activist provides funding for a simple random sample of 250 households. The average annual electricity bill for this sample is $622. Find a \(95\%\) confidence interval for the true mean annual electric bill, based on this sample.

Answers (1)

2021-03-12
Step 1
It is assumed that the sample mean is 622 and the population standard deviation is 150.
Step 2
From the given information, the confidence level is 0.95 and the level of of significance \(\displaystyle{\left(\alpha\right)}{i}{s}{0.05}{\left(={1}-{0.95}\right)}.\)
The \(95\%\) confidence interval for the true mean is obtained as follows:
\(\displaystyle{95}\%{C}{I}=\overline{{x}}\pm{z}_{{\frac{\alpha}{{2}}}}{\left(\frac{\sigma}{\sqrt{{n}}}\right)}\)
\(\displaystyle={622}\pm{z}_{{\frac{0.05}{{2}}}}{\left(\frac{150}{\sqrt{{250}}}\right)}\)
\(\displaystyle={622}\pm{1.96}{\left(\frac{150}{\sqrt{{250}}}\right)}<\)</span>
\(\displaystyle={\left({603.4058},{640.5942}\right)}\)
0

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