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Hint: We first express the concept of Confidence levels. We take an arbitrary example to understand the concept better.
Complete step by step solution:
Confidence levels are expressed as a percentage (for example, a 95% confidence level). It means that should you repeat an experiment or survey over and over again, 95 percent of the time your results will match the results you get from a population (in other words, your statistics would be sound). Confidence intervals are your results and they are usually numbers.
We constructed a 95% confidence interval for an experiment that found the sample mean temperature for a certain city in August was $\overline{x}=100$, with a population standard deviation of $\sigma =17.50$. There were 40 samples in this experiment.
The $\alpha $-level is $\dfrac{1-.95}{2}=.025$.
The formula is $\overline{x}\pm \left( z\times \dfrac{\sigma }{\sqrt{n}} \right)$. Here n is the number of samples. z is from the standard distribution tables (in the reference), and is $1.96$for a CI of 95%.
So, $\overline{x}\pm \left( z\times \dfrac{\sigma }{\sqrt{n}} \right)=100\pm \left( 1.96\times \dfrac{17.5}{\sqrt{40}} \right)=100\pm 5.42=\left[ 94.58,105.42 \right]$.
Note:
For example, if we survey a group of pet owners to see how many cans of dog food, they purchase a year. We test our statistics at the 99 percent confidence level and get a confidence interval of (200, 300). That means you think they buy between 200 and 300 cans a year. We are super confident (99% is a very high level) that our results are sound, statistically.
Complete step by step solution:
Confidence levels are expressed as a percentage (for example, a 95% confidence level). It means that should you repeat an experiment or survey over and over again, 95 percent of the time your results will match the results you get from a population (in other words, your statistics would be sound). Confidence intervals are your results and they are usually numbers.
We constructed a 95% confidence interval for an experiment that found the sample mean temperature for a certain city in August was $\overline{x}=100$, with a population standard deviation of $\sigma =17.50$. There were 40 samples in this experiment.
The $\alpha $-level is $\dfrac{1-.95}{2}=.025$.
The formula is $\overline{x}\pm \left( z\times \dfrac{\sigma }{\sqrt{n}} \right)$. Here n is the number of samples. z is from the standard distribution tables (in the reference), and is $1.96$for a CI of 95%.
So, $\overline{x}\pm \left( z\times \dfrac{\sigma }{\sqrt{n}} \right)=100\pm \left( 1.96\times \dfrac{17.5}{\sqrt{40}} \right)=100\pm 5.42=\left[ 94.58,105.42 \right]$.
Note:
For example, if we survey a group of pet owners to see how many cans of dog food, they purchase a year. We test our statistics at the 99 percent confidence level and get a confidence interval of (200, 300). That means you think they buy between 200 and 300 cans a year. We are super confident (99% is a very high level) that our results are sound, statistically.
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