Chapter 9.CR, Problem 12CR

To determine

Construct and interpret a confidence interval for the true difference between the two population means.

Answer to Problem 12CR

Solution:

The confidence interval is $\left(17.9171,31.5029\right)$ and the length timing from Atlanta to Dallas is more than the length timing from Dallas to Atlanta so the flight from Dallas to Atlanta is faster.

Explanation of Solution

Given Information:

A business person wants to determine how much faster the return trip from Dallas to Atlanta is than the initial trip from Atlanta to Dallas. He records the times for his next seven flights to and from Dallas and obtains the following data. Assuming that both population distributions arc approximately normal and the variances of the two populations are not the same.

Atlanta to Dallas	Dallas to Atlanta
1 h 49 min	1 h 32 min
1 h 51 min	1 h 23 min
2 h 03 min	1 h 38 min
1 h 47 min	1 h 26 min
1 h 53 min	1 h 30 min
2 h 01 min	1 h 19 min
1 h 45 min	1 h 28 min

Formula used:

When the standard deviations of both the populations are unknown and assumed to be unequal, the samples taken are independent of each other and the population distribution are approximately normal, then the margin of error is calculates as,

$E=t_{\alpha /2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Where $s_1\text{ and }{s}_2$ are the two sample standard deviations,

And $n_1$ and $n_2$ are two sample sizes.

The number of degrees of freedom for the t-distribution is given by the smaller of the values ${\text{n}}_{\text{1}}-1{\text{ and n}}_{\text{2}}-1$.

Then, the confidence interval for the difference between two population means for independent data sets is given by

$\left({\overline{x}}_1-{\overline{x}}_2\right){\text{- E < μ}}_{\text{1}}{\text{ - μ}}_{\text{2}}\text{ < }\left({\overline{x}}_1-{\overline{x}}_2\right)\text{+ E}$

or

$\left(\left({\overline{x}}_1-{\overline{x}}_2\right)\text{- E,}\left({\overline{x}}_1-{\overline{x}}_2\right)\text{+ E}\right)$

Where ${\overline{x}}_1\text{ and }{\overline{x}}_2$ are the two sample means.

Calculation:

Atlanta to Dallas	Dallas to Atlanta
109 min	92 min
111 min	83 min
123 min	98 min
107 min	86 min
113 min	90 min
121 min	79 min
105 min	88 min

Let the length of time the flight takes from Atlanta to Dallas be sample 1 and the length of time the flight takes from Dallas to Atlanta be sample 2.

For the population whose population variances are unequal, the t –distribution is applied for the minimum degree of freedom .i.e. the minimum of $n_1-1$ and $n_2-1$.

$\begin{array}{rll}{n}_1-1& =& 7-1\\ & =& 6\\ n_2-1& =& 7-1\\ & =& 6\end{array}$

Hence the degree of freedom is 6.

It is given that the level of confidence is 0.90 then the level of significance is,

$\begin{array}{rll}\alpha & =& 1-0.90\\ & =& 0.10\end{array}$

Then,

$\begin{array}{rll}{t}_{\alpha /2}& =& t_{0.10/2}\\ & =& t_{0.05}\\ & =& 1.943\end{array}$

The mean is calculated as,

$\overline{x}=\frac{{\displaystyle \sum x_i}}{n}$

Where $x_i$ is the data at the i^th position,

And n is the total number of observations.

Th mean of the sample 1 is calculated as,

$\begin{array}{rll}{\overline{x}}_1& =& \frac{109+111+123+107+113+121+105}{7}\\ & =& \frac{789}{7}\\ & =& 112.71\end{array}$

The standard deviation is calculated as,

$s_d^2=\frac{{\displaystyle \sum \left(x_i-\overline{x}\right)}}{n-1}$

The standard deviation of sample 1 be $s_1^2$ calculated as,

$s_1^2=\frac{{\displaystyle \sum \left(x_i-{\overline{x}}_1\right)}}{n_1-1}$

Substitute 112.71 for ${\overline{x}}_1$, 7 for $n_1$ and the respective $x_i$ values.

$\begin{array}{rll}{s}_1^2& =& \frac{{\left(109-112.71\right)}^2+{\left(111-112.71\right)}^2+\mathrm{....}+{\left(105-112.71\right)}^2}{7-1}\\ & =& \frac{283.42}{6}\\ & =& 47.23\end{array}$

The mean of the sample 2 is calculated as,

$\begin{array}{rll}{\overline{x}}_2& =& \frac{92+83+98+86+90+79+88}{7}\\ & =& 88\end{array}$

The standard deviation of sample 2 be $s_2^2$ calculated as,

$s_2^2=\frac{{\displaystyle \sum \left(x_i-{\overline{x}}_2\right)}}{n_2-1}$

Substitute 88 for ${\overline{x}}_2$, 7 for $n_2$ and the respective $x_i$ values.

$\begin{array}{rll}{s}_2^2& =& \frac{{\left(92-88\right)}^2+{\left(83-88\right)}^2+{\left(98-88\right)}^2+{\left(86-88\right)}^2+\mathrm{..}+{\left(88-88\right)}^2}{7-1}\\ & =& \frac{230}{6}\\ & =& 38.33\end{array}$

The margin of error is calculated as,

$E=t_{\alpha /2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Substitute $t_{0.05}=1.943,s_1^2=47.23,n_1=7,s_2^2=38.33,n_2=7$ in the above formula.

$\begin{array}{rll}E& =& 1.943\times \sqrt{\frac{47.23}{7}+\frac{38.33}{7}}\\ & =& 6.7929\end{array}$

The confidence interval is,

$\left(\left({\overline{x}}_1-{\overline{x}}_2\right)-E,\left({\overline{x}}_1-{\overline{x}}_2\right)+E\right)$

Substitute 112.71 for ${\overline{x}}_1$, 88 for ${\overline{x}}_2$ and 6.7929 for E in the above formula $\begin{array}{rll}\left(\left({\overline{x}}_1-{\overline{x}}_2\right)-E,\left({\overline{x}}_1-{\overline{x}}_2\right)+E\right)& =& \left(\left(\left(112.71-88\right)-6.7929\right),\left(\left(112.71-88\right)+6.7929\right)\right)\\ & =& \left(17.9171,31.5029\right)\end{array}$

Conclusion:

Both the end points are coming out to be positive that is the difference of the mean is also positive that is the length timing from Atlanta to Dallas is more than the length timing from Dallas to Atlanta so the flight from Dallas to Atlanta is faster.