91影视

Null hypotheses is,

${\mathrm{H}}_0:\mathrm{渭}鈮�129\text{seconds,}$

Alternative hypotheses is,

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

Sample mean is,

$\stackrel{炉}{\mathrm{x}}=\frac{132+129+鈰�+129+128}{20}$

$=129.5$

Standard sample deviation,

$\mathrm{s}=\underset{\mathrm{j}=1}{\overset{\mathrm{n}}{鈭�}}\frac{{\left({\mathrm{x}}_{\mathrm{j}}-\stackrel{炉}{\mathrm{x}}\right)}^2}{\mathrm{n}-1}$

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

$=1.85$

$\mathrm{n}-1=19$

Test static is,

${\mathrm{t}}_{19}=\frac{\stackrel{炉}{\mathrm{x}}-\mathrm{渭}}{\mathrm{s}/\sqrt{\mathrm{n}}}$

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

$=1.209$

$\mathrm{p}$-value is,

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

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

$=0.879$

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

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

$=\mathrm{伪}$

The mean and variance of random variable is,

localid="1649936919611" $\stackrel{炉}{\mathrm{x}}-{\mathrm{t}}_{\frac{\mathrm{a}}{2},\mathrm{n}-1}\frac{\mathrm{s}}{\sqrt{\mathrm{n}}}鈮�\mathrm{渭}鈮�\stackrel{炉}{\mathrm{x}}+{\mathrm{t}}_{\frac{\mathrm{伪}}{2},\mathrm{n}-1}\frac{\mathrm{s}}{\sqrt{\mathrm{n}}}.....1$

where

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

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

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

$\frac{\mathrm{伪}}{2}=0.025$

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

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

From equation $1$and $2$

$129.5-2.09\frac{1.85}{\sqrt{20}}鈮�渭鈮�129.5+2.09\frac{1.85}{\sqrt{20}}$

$129.5-2.09脳0.41鈮�\mathrm{渭}鈮�129.5+2.09脳0.41$

Population mean is,

$128.68鈮�\mathrm{渭}鈮�130.37$