Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.
-The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were: $\begin{array} { l l l l l } 7.1 & 10.7 & 9.4 & 8.6 & 11.4 \end{array}$
$\begin{array} { l l l l l } 7.1 & 6.6 & 11.2 & 10.0 & 12.3 \end{array}$
Determine a $95 \%$ confidence interval for the mean time for all players.
A) $10.91 \mathrm {~min}

Question

Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.
-The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were: $\begin{array} { l l l l l } 7.1 & 10.7 & 9.4 & 8.6 & 11.4 \end{array}$
$\begin{array} { l l l l l } 7.1 & 6.6 & 11.2 & 10.0 & 12.3 \end{array}$
Determine a $95 \%$ confidence interval for the mean time for all players.
A) $10.91 \mathrm {~min} < \mu < 7.97 \mathrm {~min}$
B) $10.81 \mathrm {~min} < \mu < 8.07 \mathrm {~min}$
C) $7.97 \mathrm {~min} < \mu < 10.91 \mathrm {~min}$
D) $8.07 \mathrm {~min} < \mu < 10.81 \mathrm {~min}$

Quizplus · Accepted Answer

The correct answer is D.The sample mean is:$$\bar{x} = \frac{1}{10} \sum_{i=1}^{10} x_i = \frac{1}{10} (7.1 + 10.7 + 9.4 + 8.6 + 11.4 + 7.1 + 6.6 + 11.2 + 10.0 + 12.3) = 9.4$$The sample standard deviation is:$$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} = \sqrt{\frac{1}{9} (7.1 - 9.4)^2 + (10.7 - 9.4)^2 + (9.4 - 9.4)^2 + (8.6 - 9.4)^2 + (11.4 - 9.4)^2 + (7.1 - 9.4)^2 + (6.6 - 9.4)^2 + (11.2 - 9.4)^2 + (10.0 - 9.4)^2 + (12.3 - 9.4)^2} = 1.87$$The 95% confidence interval for the population mean is:$$\bar{x} \pm z_{\alpha/2} \frac{s}{\sqrt{n}} = 9.4 \pm 1.96 \frac{1.87}{\sqrt{10}} = (8.07, 10.81)$$

Use the Given Degree of Confidence and Sample Data to Construct