Question

Use the data of Exercise 19 to calculate a \(95\% CI\) for the difference between true average stopping distance for cars equipped with system \(1\) and cars equipped with system \(2\). Does the interval suggest that precise information about the value of this difference is available?

Short answer

the solution is

\(( - 20.402, - 6.798).\)

Step-by-step solution

calculate degree of freedom

Assume that the populations are normally distributed and that the samples are chosen separately. The

Two-sample confidence interval for \({\mu _1} - {\mu _2}\)

With confidence level \(100(1 - \alpha )\% is \)

\(\bar x - \bar y \pm {t_{\alpha /2,\nu }} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \)

\(\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}}\)

The number must be rounded to the nearest integer. An upper/lower bound is obtained by substituting \({z_{\alpha /2}}\) with \({z_\alpha } \) and \( \pm \)with only \( + \) and \( - \).

The degrees of freedom can be calculated using the formula below.

\(\begin{array}{l}\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}} = \frac{{{{\left( {\frac{{5.0{3^2}}}{6} + \frac{{5.3{8^2}}}{6}} \right)}^2}}}{{\frac{{{{\left( {5.0{3^2}/6} \right)}^2}}}{{6 - 1}} + \frac{{{{\left( {5.3{8^2}/6} \right)}^2}}}{{6 - 1}}}}\\ = \frac{{{{(4.2168 + 4.8241)}^2}}}{{\frac{{{{(4.2168)}^2}}}{5} + \frac{{4.824{1^2}}}{5}}} = 9.96\end{array}\)

probability

The number must be rounded down to 9 in order to acquire the matching degrees of freedom, implying that.

\(\nu = 9\)

The only value that is lacking is $t alpha / 2, nu$. The value of \({t_{\alpha /2,\nu }}. For 95\% \) confidence interval is \(\alpha \), \(0.05\) hence \(\alpha /2 = 0.025\). The value is based on the table in the appendix.

\({t_{\alpha /2,v}} = {t_{0.025,9}} = 2.262.\)

Finally, the \(95\) confidence interval can be computed

\(\begin{array}{l}\left( {\bar x - \bar y - {t_{\alpha /2,\nu }} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} ,\bar x + \bar y - {t_{\alpha /2,\nu }} \times \sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} } \right)\\ = \left( {115.7 - 129.3 - 2.262 \times \sqrt {\frac{{5.0{3^2}}}{6} + \frac{{5.3{8^2}}}{6}} ,115.7 - 129.3 + 2.262 \times \sqrt {\frac{{5.0{3^2}}}{6} + \frac{{5.3{8^2}}}{6}} } \right)\\ = ( - 13.6 - 2.262 \times 3.007, - 13.6 + 2.262 \times 3.007)\\ = ( - 20.402, - 6.798).\\\\\end{array}\)

conclusion

\(( - 20.402, - 6.798).\)