Confidence Interval Use the departure delay times for Flight 3 and construct a 95% confidence interval estimate of the population mean. Write a brief statement that interprets the confidence interval.

Refer to exercise 1 to obtain the departure delay times for flight 3.

The formula for 95% confidence interval for the population mean is,

\(\left( {\bar x - {t_{\frac{\alpha }{2}}} \times \left( {\frac{s}{{\sqrt n }}} \right),\bar x + {t_{\frac{\alpha }{2}}} \times \left( {\frac{s}{{\sqrt n }}} \right)} \right)\)

Where,

\[\bar x\]is the sample mean;

s is the standard deviation of the sample;

n is the sample size;

\({t_{\frac{\alpha }{2}}}\)is the critical value obtained at degrees of freedom \(\left( {n - 1} \right)\) and \(\alpha \) level of confidence.

The sample mean is computed as follows:

\(\begin{array}{c}\bar x = \frac{{\sum {{x_i}} }}{n}\\ = \frac{{22 + \left( { - 11} \right) + 7 + ... + 8}}{8}\\ = \frac{{16}}{8}\\ = 2\;\min \end{array}\)

The sample standard deviation is computed as follows:

\(\begin{array}{c}s = \sqrt {\frac{{\sum {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {22 - 2} \right)}^2} + {{\left( { - 11 - 2} \right)}^2} + ... + {{\left( {8 - 2} \right)}^2}}}{{8 - 1}}} \\ = 10.6\end{array}\)

For 95% confidence level, the value of\(\alpha = 0.05\).

Using the t-table, the two-tailed critical value is obtained as 2.36.

Substitute the values in the formula,

\[\begin{array}{c}\left( {2.0 - 2.36\left( {\frac{{10.6}}{{\sqrt 8 }}} \right),2.0 + 2.36\left( {\frac{{10.6}}{{\sqrt 8 }}} \right)} \right) = \left( {2.0 - 8.84,2.0 + 7.8.84} \right)\\ = \left( { - 6.8,10.8} \right)\end{array}\]

Thus, the 95% confidence interval for the population mean is between -6.8 min and 10.8 min.

It can be inferred with a 95% confidence level that the actual population mean departure delay for Fight 3 would lie between -6.8 to 10.8 minutes.