Question

1. you are testing whether students at your school are overweight. you sample 25 students and measure how much more they weigh than the average weight for their height. the mean of your sample is 5 lbs, and we know that σ = 5. find a 90% confidence interval and interpret.

Explanation:

Step1: Find the z - score

For a 90% confidence interval, the significance level \(\alpha=1 - 0.90=0.10\), and \(\alpha/2=0.05\). The z - score \(z_{\alpha/2}\) corresponding to an area of \(1-\alpha/2 = 0.95\) in the right - tail of the standard normal distribution. Using a standard normal table or calculator, \(z_{0.05}\approx1.645\).

Step2: Calculate the margin of error \(E\)

The formula for the margin of error for a population mean when the population standard deviation \(\sigma\) is known is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\). Given \(n = 25\), \(\sigma=5\), and \(z_{\alpha/2}=1.645\).

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Step3: Calculate the confidence interval

The formula for the confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu <\bar{x}+E\). Given \(\bar{x} = 5\)

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Interpretation:

We are 90% confident that the True population mean of the amount students at the school weigh more than the average weight for their height lies between \(3.355\) lbs and \(6.645\) lbs.

Answer:

The 90% confidence interval is \((3.355,6.645)\)