Braking Reaction Times: Hypothesis Test Use the sample data from Exercise 6 with a 0.01 significance level to test the claim that males and females have the same mean.

Refer to Exercise 6 for the brake reaction times in males and females.

The significance level is 0.01.

The sampled number of males and females are given:

\(\begin{array}{l}{n_M} = 36\\{n_F} = 36\end{array}\)

The hypotheses for testing the claim that males and females have the same mean are statedbelow:

\(\begin{array}{l}{H_o}:{\mu _M} - {\mu _F} = 0\\{H_a}:{\mu _M} - {\mu _F} \ne 0\end{array}\)

Here,\({\mu _M},{\mu _F}\)are the true mean values of braking reaction times for males and females, respectively.

Thus, the test is two-tailed.

The sample mean for males and females is computed below:

\(\begin{aligned} {{\bar x}_M} &= \frac{{\sum {{x_i}} }}{{{n_m}}}\\ &= \frac{{28 + 30 + 31 + ... + 63}}{{36}}\\ &= 44.3611\end{aligned}\)

\(\begin{aligned} {{\bar x}_F} &= \frac{{\sum {{x_j}} }}{{{n_F}}}\\ &= \frac{{22 + 24 + 34 + ... + 80}}{{36}}\\ &= 54.2778\end{aligned}\)

The sample standard deviation for males and females is computed below:

\(\begin{aligned}{c}{s_M} &= \sqrt {\frac{{\sum {{{\left( {{x_i} - {{\bar x}_M}} \right)}^2}} }}{{{n_M} - 1}}} \\ &= \sqrt {\frac{{{{\left( {28 - 44.36} \right)}^2} + {{\left( {30 - 44.36} \right)}^2} + ... + {{\left( {63 - 44.36} \right)}^2}}}{{36 - 1}}} \\ &= 9.4722\end{aligned}\)

\(\begin{aligned}{c}{s_F} &= \sqrt {\frac{{\sum {{{\left( {{x_j} - {{\bar x}_F}} \right)}^2}} }}{{{n_F} - 1}}} \\ &= \sqrt {\frac{{{{\left( {22 - 54.28} \right)}^2} + {{\left( {24 - 54.28} \right)}^2} + ... + {{\left( {80 - 54.28} \right)}^2}}}{{36 - 1}}} \\ &= 15.6106\end{aligned}\)

The test statistic is computedbelow:

\(\begin{aligned} t &= \frac{{\left( {{{\bar x}_M} - {{\bar x}_F}} \right) - \left( {{\mu _M} - {\mu _F}} \right)}}{{\sqrt {\frac{{s_M^2}}{{{n_M}}} + \frac{{s_F^2}}{{{n_F}}}} }}\\ &= \frac{{\left( {44.3611 - 54.2778} \right) - 0}}{{\sqrt {\frac{{{{9.4722}^2}}}{{36}} + \frac{{{{15.6106}^2}}}{{36}}} }}\\ &= - 3.2585\end{aligned}\)

The test statistic is –3.2585.

The degree of freedom is computedbelow:

\(\begin{aligned} df &= \frac{{{{\left( {\frac{{s_M^2}}{{{n_M}}} + \frac{{s_F^2}}{{{n_F}}}} \right)}^2}}}{{\frac{{{{\left( {\frac{{s_M^2}}{{{n_M}}}} \right)}^2}}}{{{n_M} - 1}} + \frac{{{{\left( {\frac{{s_F^2}}{{{n_F}}}} \right)}^2}}}{{{n_F} - 1}}}}\\ &= \frac{{{{\left( {\frac{{{{9.4722}^2}}}{{36}} + \frac{{{{15.6106}^2}}}{{36}}} \right)}^2}}}{{\frac{{{{\left( {\frac{{{{9.4722}^2}}}{{36}}} \right)}^2}}}{{36 - 1}} + \frac{{{{\left( {\frac{{{{15.6106}^2}}}{{36}}} \right)}^2}}}{{36 - 1}}}}\\ &= 57.696\\ &\approx 58\end{aligned}\)

Thus, the degree of freedom is 58.

The p-value is obtained from the t-distribution table at 58 degrees of freedom.

\(\begin{aligned} p - value &= 2P\left( {T < - 3.2585} \right)\\ &= 0.0019\end{aligned}\)

Since the p-value is lesser than 0.01, the null hypothesis is rejected.

Thus, there is not enough evidence to support the claim that the braking reaction time has the same mean for males and females.