Question

Comparing Two Means Treating the data as samples from larger populations, test the claim that there is a difference between the mean departure delay time for Flight 3 and Flight 21.

Short answer

Enough statistical evidence is not available to conclude that there is a difference between the mean departure delay times of flights 3 and 21 at 0.05 level of significance.

Step-by-step solution

Given information

Refer to the data in Exercise 1 for flights 3 and 21.

Flight 3	Flight 21
22	18
–11	60
7	142
0	–1
–5	–11
3	–1
–8	47
8	13

Step 2:Identify the hypothesis

The hypothesis formulated totest the difference between the mean values of the delay times between two flights is shown below:

\(\begin{array}{l}{H_o}:{\mu _3} = {\mu _{21}}\\{H_a}:{\mu _3} \ne {\mu _{21}}\end{array}\)

\[\]

Here, are true mean measures for population delay times of flights 3 and 21, respectively.

Compute the sample mean and sample standard deviation

The sample means for flights 3 and 21 are calculated below:

\(\begin{array}{c}{{\bar x}_3} = \frac{{22 + \left( { - 11} \right) + 7 + ... + 8}}{8}\\ = 2\\{{\bar x}_{21}} = \frac{{18 + 60 + 142 + ... + 13}}{8}\\ = 33.375\end{array}\)

The sample standard deviations for flights 3 and 21 are calculated below:

\(\begin{array}{c}{s_3} = \sqrt {\frac{{\sum {\left( {{x_i} - {{\bar x}_3}} \right)} }}{{{n_3} - 1}}} \\ = \sqrt {\frac{{{{\left( {22 - 2} \right)}^2} + {{\left( { - 11 - 2} \right)}^2} + ... + {{\left( {8 - 2} \right)}^2}}}{{8 - 1}}} \\ = 10.583\end{array}\)

\(\begin{array}{c}{s_{21}} = \sqrt {\frac{{\sum {\left( {{x_i} - {{\bar x}_{21}}} \right)} }}{{{n_{21}} - 1}}} \\ = \sqrt {\frac{{{{\left( {18 - 33.375} \right)}^2} + {{\left( {60 - 33.375} \right)}^2} + ... + {{\left( {13 - 33.375} \right)}^2}}}{{8 - 1}}} \\ = 50.253\end{array}\)

Thus, the sample standard deviation for flight 3 is 10.583 and flight 21 is 50.253.

Compute the test statistic

The test statistic is computed below:

\(\begin{array}{c}t = \frac{{\left( {{{\bar x}_3} - {{\bar x}_{21}}} \right) - \left( {{\mu _3} - {\mu _{21}}} \right)}}{{\sqrt {\frac{{s_3^2}}{{{n_3}}} + \frac{{s_{21}^2}}{{{n_{12}}}}} }}\\ = \frac{{\left( {2 - 33.375} \right) - 0}}{{\sqrt {\frac{{{{10.583}^2}}}{8} + \frac{{{{50.253}^2}}}{8}} }}\\ = - 1.728\end{array}\)

Thus, the test statistic is –1.728.

Compute the degrees of freedom

The degree of freedom is computed below:

\(\begin{array}{c}df = \frac{{{{\left( {\frac{{s_3^2}}{{{n_3}}} + \frac{{s_{21}^2}}{{{n_{21}}}}} \right)}^2}}}{{\frac{{{{\left( {\frac{{s_3^2}}{{{n_3}}}} \right)}^2}}}{{{n_3} - 1}} + \frac{{{{\left( {\frac{{s_{21}^2}}{{{n_{21}}}}} \right)}^2}}}{{{n_{21}} - 1}}}}\\ = \frac{{{{\left( {\frac{{{{10.583}^2}}}{8} + \frac{{{{50.253}^2}}}{8}} \right)}^2}}}{{\frac{{{{\left( {\frac{{{{10.583}^2}}}{8}} \right)}^2}}}{{8 - 1}} + \frac{{{{\left( {\frac{{{{50.253}^2}}}{8}} \right)}^2}}}{{8 - 1}}}}\\ \approx 7.620\end{array}\)

Thus, the degree of freedom (df) is 7.62.

Compute the p-value

The p-value is computed using the t-table as shown below.

\(\begin{array}{c}P - value = 2P\left( {{t_{df}} < - 1.728} \right)\\ = 2\left( {0.0621} \right)\\ = 0.1241\end{array}\)

As the p-value is greater than 0.05 significance level, the null hypothesis fails to be rejected.

State the conclusion

There is not enough evidence to support the claim that there is a difference between the mean departure delay times of flights 3 and 21 at the 0.05 level of significance.