Question

in a recent study of 35 ninth - grade students, the mean number of hours per week that they played video games was 16.6. the standard deviation of the population was 2.8.
a. find the the best point estimate of the mean 16.6
b. find the 95% confidence interval of the mean of the time playing video games (write as ordered pair with a space after the comma) (15.67, 17.53)
c. find the 99% confidence interval of the mean time playing video games (write as ordered pair with a space after the comma) (15.38, 17.82)

Explanation:

Step1: Recall the formula for confidence interval

For a population with known standard deviation \(\sigma\), the confidence interval for the population mean \(\mu\) is given by \(\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(z_{\alpha/2}\) is the z - score, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.

Step2: Calculate for 95% confidence interval

• For 95% confidence, \(\alpha = 1 - 0.95=0.05\), so \(\alpha/2 = 0.025\). The \(z\) - score \(z_{\alpha/2}=z_{0.025} = 1.96\) (from standard normal distribution table).
• Given \(\bar{x}=16.6\), \(\sigma = 2.8\), \(n = 35\).
• Margin of error \(E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=1.96\times\frac{2.8}{\sqrt{35}}\approx1.96\times0.473\approx0.93\)
• Lower limit \(=16.6 - 0.93=15.67\)
• Upper limit \(=16.6+ 0.93=17.53\)

Step3: Calculate for 99% confidence interval

• For 99% confidence, \(\alpha=1 - 0.99 = 0.01\), so \(\alpha/2=0.005\). The \(z\) - score \(z_{\alpha/2}=z_{0.005}=2.576\) (from standard normal distribution table).
• Margin of error \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.576\times\frac{2.8}{\sqrt{35}}\approx2.576\times0.473\approx1.22\)
• Lower limit \(=16.6-1.22 = 15.38\)
• Upper limit \(=16.6 + 1.22=17.82\)

Answer:

a. \(16.6\)
b. \((15.67,17.53)\)
c. \((15.38,17.82)\)