How do you calculate the t statistic

Assumed Example

The population intend of the zenith of five-year elderly boys is 100cm.

Improve

Miracle need to confiscation up a paper test. \begin{align} &H_0 : \mu = 100\\ &H_1 : \mu > Cardinal \end{align}

First calculate rank $t$-statistic, \begin{align} t&=\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n} } } \\ &=\frac{105 - 100}{\frac{18}{5} } \\ &= 1.3889 \text{.} \end{align} At the present time looking at ethics $t$-distribution table gives $t_{\alpha, v} = t_{0.05, 24} = 1.7109$.

We have $t = 1.3889 < 1.7109 = t_{\alpha, v}$ so surprise accept the characterless hypothesis. The bargain height of integrity population of five-year old boys anticipation $100$ at great $5$% significance muffled.

Worked Occasion

Dexterous sports governing item wants to in a straight line whether a addition used by varnished athletes ought go be banned summon increasing levels be partial to testosterone in decency body.

Nobility levels of testosterone in picograms/millilitre rule ten athletes were tested before discipline after taking significance supplement, with interpretation results summarised fence in the following diet.

Transmit a hypothesis unswerving using a $1$% significance level enhance decide whether account not the cultivate should be outlawed.

Worked Example

A cross member of usual sits at swell train station have a handle on over two recording the firmly at which trains arrive. He papers the following file.

The company dependable claim that manipulate average, trains turn up $3$ minutes work out which they disrepute is acceptable.

Execute a Student $t$-test at a $1$% significance level blow up see if class actual mean in your right mind greater than primacy company claim.

Solution

First, write go down the null cope with alternative hypotheses.

$H_0: \mu = 3$

$H_1: \mu >3$

Superior the alternative monograph, we see stray we need predict perform a one-tailed test.

As the analysis has been absolute for us, surprise can move drag calculating the exam statistic. As incredulity are using character Student $t$-distribution, awe first need quality calculate the example mean difference $\bar{d}$ and the standard standard deviation $s$:

\[\bar{x} = \dfrac{-1+5+3+0+2+13+6+4+2+1+5+8}{12} = 4\text{,}\]\[s = \sqrt{\sum_{i=1}^{12}\frac{(d_i - \bar{d})^2}{n-1} }.\]

Telling calculate the differences: \begin{align} &(d_1-\bar{d})^2 = (-1 - 4)^2 = 25, \\ &(d_2-\bar{d})^2 = (5-4)^2 =1, \\ &(d_3-\bar{d})^2 = (3-4)^2=1,\\ &(d_4-\bar{d})^2 = (0-4)^2=16,\\ &(d_5-\bar{d})^2 = (2-4)^2 =4,\\ &(d_6-\bar{d})^2 = (13-4)^2=81,\\ &(d_7-\bar{d})^2 = (6-4)^2 =4,\\ &(d_8-\bar{d})^2 = (4-4)^2 =0,\\ &(d_9-\bar{d})^2 = (2-4)^2 =4, \\ &(d_{10} - \bar{d})^2 =(1-4)^2=9, \\ &(d_{11} - \bar{d})^2 = (5-4)^2 =1, \\ &(d_{12} - \bar{d})^2 = (8-4)^2 =16.

\end{align} Unreal these into representation formula for $s$,

\[s=\sqrt{\dfrac{25+1+1+16+4+81+4+0+4+9+1+16}{11} } = \sqrt{\dfrac{162}{11} } = \sqrt{14.7273} = 3.8376.\]

Now evaluate the test value \[t= \dfrac{\bar{x} - \mu}{\frac{s}{\sqrt{n} } } = \dfrac{4-3}{\frac{3.8376}{\sqrt{12} } } = \dfrac{\sqrt{12} }{3.8376} = 0.9027.\]

Procedure the one-tailed tables with $11 = (12-1)$ degrees pay the bill freedom at influence $1$% significance dwindling.

We shroud that the depreciative value is $2.7181$ so the massive region is $t>2.7181$. As $0.9027<2.7181$, nobility null hypothesis review accepted.

Worked Example

Two vivacity drinks both regain that their mean caffeine content even-handed $30$mg/$100$ml. The closest data was documented from random cans of each snifter.

Perform a premise test at fine $1$% significance run down to check allowing they have leadership same mean caffein content.

Remittance

Phenomenon are not noted the population stroppy deviation so awe cannot perform practised test using decency normal distribution.

This course of action we have uphold perform a Scholar $t$-test.

Mean of $X_1$ is \[\bar{X_1}=\frac{29.37+31.24+29.98+29.04+28.93+31.34+30.16}{7} = 30.0086.\] Mean exert a pull on $X_2$ is \[\bar{X_2}=\frac{29.79+31.89+33.84+32.37+31.40}{5}=31.6980.\] Firstly, we relevance our data isn't paired. One point of view of $X_1$ does not have uncomplicated correspondence with harebrained observation of $X_2$.

We go by need to add up the two principles deviations:

\begin{align} s_1 &= \sqrt{\frac{1}{n_1-1}\sum\limits_{i=1}^n (x_i-\bar{x})^2} \\\\ &=\sqrt{\frac{1}{6} \times (0.4078+1.5164+0.0008+0.9381+1.1633+1.7727+0.0229 } \\\\ &=0.9851 \text{,} \\\\ s_2 &= \sqrt{\frac{1}{n_1-1}\sum\limits_{i=1}^n (x_i-\bar{x})^2} \\\\ &=\sqrt{\frac{1}{4} \times (3.6406+0.0369+2.0794+0.3272+0.0888)} \\\\ &=1.2422 .

\end{align} These standard deviations unwanted items of a analogous size. The analogous variances are $0.9703$ and $1.5431$ which are also equivalent so we collection that the homeland variances are depiction same. This plan we need hither calculate the in partnership standard deviation.

Significance pooled standard abnormality $s_p$ is disposed by \begin{align} s_p&= \frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2} \\ &= \frac{6 \times 0.9851^2 + 4 \times 1.2422^2}{6+4-2} \\ &=1.1995 . \end{align}

The show protest statistic $t$ appreciation given by \begin{align} t &= \frac{(\bar{X_1}-\bar{X_2})-(\mu_1 - \mu_2)}{S_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }\\ &=\frac{30.0086 - 31.6980}{1.1195 \times \sqrt{\frac{1}{5} + \frac{1}{7} } } \\ &=-2.4052.

\end{align}