Respuesta :
Here we basically need to fin a confidence interval using the formula:
[tex]\mu\pm z\cdot\frac{S}{\sqrt[]{n}}[/tex]Where μ is the mean if the sample, S is the standard deviation, n is the size and z is known as the z-value and is related to the fact that we are assuming the data is from a normally distributed population. The questions gives us the following values: μ=73.1 S=16.5 and n=29. In order to find z we need the value of n and something known as the significance level that is given by:
[tex]\begin{gathered} \alpha=1-\frac{\text{ confidence level percentage}}{100} \\ \alpha=1-\frac{99}{100}=1-0.99=0.01 \end{gathered}[/tex]Knowing that the sizes of the sample is 29 and the significance level is 0.01 we can find z in a table or using a calculator: z=2.576. Then we can find the confidence interval:
[tex]\begin{gathered} \mu\pm z\cdot\frac{S}{\sqrt[]{n}} \\ 73.1\pm2.576\cdot\frac{16.5}{\sqrt[]{29}} \\ 73.1\pm7.893 \end{gathered}[/tex]If we write this as a tri-linear inequality we have:
[tex]65.207<\mu<80.993[/tex]