Question

Use the "Lap time" data for Lap$4$(see Appendix C) to test the claim that Terri finishes Lap $4$ , on average, in less than $129$ seconds. Use all twenty races given.

Short answer

The Terri's average lap time is less than $129$seconds.

Step-by-step solution

Introduction 

Null hypotheses is,

${\mathrm{H}}_0:\mathrm{\mu }\ge 129\text{seconds,}$

Alternative hypotheses is,

${\mathrm{H}}_0:\mathrm{\mu }<129\text{seconds,}$

Sample mean is,

$\overline{\mathrm{x}}=\frac{132+129+\cdots +129+128}{20}$

$=129.5$

Standard sample deviation,

$\mathrm{s}=\sum _{\mathrm{j}=1}^{\mathrm{n}}\frac{{\left({\mathrm{x}}_{\mathrm{j}}-\overline{\mathrm{x}}\right)}^2}{\mathrm{n}-1}$

$=\sum _{\mathrm{j}=1}^{20}\frac{{\left({\mathrm{x}}_{\mathrm{j}}-129.5\right)}^2}{19}$

$=1.85$

$\mathrm{n}-1=19$

Test static is,

${\mathrm{t}}_{19}=\frac{\overline{\mathrm{x}}-\mathrm{\mu }}{\mathrm{s}/\sqrt{\mathrm{n}}}$

$=\frac{129.5-129}{1.85/\sqrt{20}}$

$=1.209$

$\mathrm{p}$-value is,

$\mathrm{p}-\mathrm{value}=\mathrm{P}\{\stackrel{-}{\mathrm{x}}<129.5\}$

$=\mathrm{P}\{{\mathrm{t}}_{19}<1.209\}$

$=0.879$

Explanation

We do not reject the null hypothesis when the p-value is bigger than the established alpha value.

$\mathrm{p}-\mathrm{value}=0.879>0.05$

$=\mathrm{\alpha }$

The mean and variance of random variable is,

localid="1649936919611" $\overline{\mathrm{x}}-{\mathrm{t}}_{\frac{\mathrm{a}}{2},\mathrm{n}-1}\frac{\mathrm{s}}{\sqrt{\mathrm{n}}}\le \mathrm{\mu }\le \overline{\mathrm{x}}+{\mathrm{t}}_{\frac{\mathrm{\alpha }}{2},\mathrm{n}-1}\frac{\mathrm{s}}{\sqrt{\mathrm{n}}}.....1$

where

${\mathrm{t}}_{\frac{\mathrm{\alpha }}{2},\mathrm{n}-1}$is the upper.

$100\frac{\mathrm{\alpha }}{2}$percentage point of the$\mathrm{t}$distribution.

With $\mathrm{n}-1$degrees of freedom.

$\frac{\mathrm{\alpha }}{2}=0.025$

$\Rightarrow {\mathrm{t}}_{\frac{\mathrm{a}}{2},\mathrm{n}-1}$

localid="1649936900332" $={\mathrm{t}}_{0.025,19}=2.09........2$

From equation $1$and $2$

$129.5-2.09\frac{1.85}{\sqrt{20}}\le \mu \le 129.5+2.09\frac{1.85}{\sqrt{20}}$

$129.5-2.09\times 0.41\le \mathrm{\mu }\le 129.5+2.09\times 0.41$

Population mean is,

$128.68\le \mathrm{\mu }\le 130.37$

Conclusion