Chapter 14, Problem 14.48SE

To determine

To identify:

The value of $p$ and degrees of freedom of $x^2$ .

Answer to Problem 14.48SE

The value of P is $0.04$ and null hypothesis is rejected and the data provide sufficient evidence to indicate there is a significant difference between actual and experimental observations.

Explanation of Solution

Given information:

The problem experience consists of $4$ tails was repeated $100$ times.

The number of success gets as:

No of successes	No of times obtained
$0$	$11$
$1$	$17$
$2$	$42$
$3$	$21$
$4$	$9$

Calculation:

As know that,

  $p=\frac{\overline{x}}{n}$

Where,

  $n=$ number of trials.

  $n=4$

Now, for estimating $p$ ,

So, consider

  $\begin{array}{l}\overline{x}={\displaystyle \sum \frac{x_i{f}_i}{f_i}}\\ \overline{x}=\frac{200}{100}\\ \overline{x}=2\end{array}$

Put the value of $\left(\overline{x}=2,n=4\right)$ in the equation of $p$

  $\begin{array}{l}p=\frac{\overline{x}}{n}\\ p=\frac{2}{4}\\ p=0.5\end{array}$

Now, the expected values for a binomial experiment of $4$ trails are,

  $Exp=N\times \left(n_{C_{xi}}{p}^{x_i}{q}^{n-x_i}\right)$

Are $0.0625,0.25,0.375,0.25,0.0625$ respectively.

Now, for testing whether the data provide sufficient evidence to indicate there is any significance difference between actual and the experimental observations or not.

The null hypothesis is,

  $H_0:$ There is sufficient evidence to indicate there is a significance difference between actual observation and experimental observations.

The alternative hypothesis is,

  $H_1:$ There is sufficient evidence to indicate there is a significance difference between actual observation and experimental observations.

In case test statistic $x^2$ is too large, so there is a rejection of null hypothesis, $H_0$ .

Obs. count	Exp. count	Obs.-exp	${\left(Obs-\exp \right)}^2$	$\frac{{\left(Obs-\exp \right)}^2}{\exp }$
$11$	$6.25$	$4.75$	$22.5625$	$3.61$
$17$	$25$	$-8$	$64$	$2.56$
$42$	$37.5$	$4.5$	$20.25$	$0.54$
$21$	$25$	$-4$	$16$	$0.64$
$9$	$6.25$	$2.75$	$7.5625$	$1.21$

The chi-square statistic is,

  $\begin{array}{l}{x}^2={\displaystyle \sum \frac{{\left(Obs-\exp \right)}^2}{\exp }}\\ =8.5555\\ \approx 8.56\end{array}$

Test statistic has $\left(k-1-1\right)$ degrees of freedom,

  $\begin{array}{l}df=\left(k-1-1\right)\\ df=\left(k-2\right)\\ df=\left(5-2\right)\\ df=3\end{array}$

The P-value can be determined as,

  $\begin{array}{l}P\left(x^2>8.56\right)=1-P\left(x^2>8.56\right)\\ P\left(x^2>8.56\right)=1-0.96\\ P\left(x^2>8.56\right)=0.04\end{array}$

Thus, the value of P is $0.04$ which is smaller than the $0.05=\alpha$ .

So, there is rejection of the null hypothesis because of low $p-$ value.

Thus, the data provide sufficient evidence to indicate there is a significant difference between actual and experimental observations.