Question

1. you sample 12 bugs and find the sample mean is 2.40 cm. you are told that σ = 0.2 cm. find a 95% confidence interval and interpret.
2. you sample 16 students in your school, and they average 13 hours of tv a week. assume σ = 3. find a 99% confidence interval and interpret.

Explanation:

Step1: <Find the z - value>

For a 95% confidence interval, the z - value \(z_{\alpha/2}\) is 1.96. For a 99% confidence interval, the z - value \(z_{\alpha/2}\) is 2.576.

Step2: <Calculate the margin of error for problem 5>

The formula for the margin of error \(E\) is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\).
For problem 5, \(n = 12\), \(\sigma=0.2\), \(z_{\alpha/2}=1.96\)
\(E = 1.96\times\frac{0.2}{\sqrt{12}}\)
\(E=1.96\times\frac{0.2}{3.464}\)
\(E = 1.96\times0.0577\)
\(E\approx0.113\)
The confidence interval is \(\bar{x}-E<\mu <\bar{x} + E\), \(\bar{x} = 2.40\)
\(2.40- 0.113<\mu<2.40 + 0.113\)
\(2.287<\mu<2.513\)

Step3: <Calculate the margin of error for problem 6>

For problem 6, \(n = 16\), \(\sigma = 3\), \(z_{\alpha/2}=2.576\)
\(E=2.576\times\frac{3}{\sqrt{16}}\)
\(E=2.576\times\frac{3}{4}\)
\(E = 1.932\)
The confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\), \(\bar{x}=13\)
\(13 - 1.932<\mu<13+1.932\)
\(11.068<\mu<14.932\)

Step4: <Interpret the confidence intervals>

For problem 5: We are 95% confident that the True population mean length of bugs lies between \(2.287\) cm and \(2.513\) cm.
For problem 6: We are 99% confident that the True population mean number of hours of TV watched per week by students in the school lies between \(11.068\) hours and \(14.932\) hours.

Answer:

Problem 5: The 95% confidence interval is \((2.287,2.513)\). Interpretation: We are 95% confident that the True population mean length of bugs lies between \(2.287\) cm and \(2.513\) cm.
Problem 6: The 99% confidence interval is \((11.068,14.932)\). Interpretation: We are 99% confident that the True population mean number of hours of TV watched per week by students in the school lies between \(11.068\) hours and \(14.932\) hours.