Chapter 14, Problem 14.35SE

To determine

To identify: The provided data is sufficient evidence to indicate that the probability of finding manganese nodules in the Deepsea earth’s crust is dependent on the magnetic-age classification.

Answer to Problem 14.35SE

The data is sufficient evidence to indicate the equal preference for all polished patches of floor.

Explanation of Solution

Given information:

Sample of $100$ housekeepers viewed five patches of flooring that had received the five polishes $A,B,C,D$ and $E$ .

Polish	$A$	$B$	$C$	$D$	$E$
Frequency	$27$	$17$	$15$	$22$	$19$

Calculation:

The null hypothesis,

  $H_0:$ the data provide sufficient evidence to indicate the equal preference for all polished patches of floor.

The alternative hypothesis,

  $H_1:$ the data provide sufficient evidence to indicate a preference for one or more of the polished patches of floor.

If the test statistic $x^2$ is too large, null hypothesis is rejected $H_0$ .

From the given data,

Expected count is equal for all polished patches of floor $A,B,C,D$ and $E$ .

Since, from the null hypothesis there is equal preference for the five polished patches of floor.

i.e. $\frac{100}{5}=20$ .

ObservedCount   $O_i$	Expected Count   $E_i$	ObservedCount − Expected Count   $\left(O_i-E_i\right)$	${\left(O_i-E_i\right)}^2$	$x^2=\frac{{\left(O_i-E_i\right)}^2}{E_i}$
$27$	$20$	$7$	$49$	$2.45$
$17$	$20$	$-3$	$9$	$0.45$
$15$	$20$	$-5$	$25$	$1.25$
$22$	$20$	$2$	$4$	$0.2$
$19$	$20$	$-1$	$1$	$0.05$

The chi-square statistic is,

  $\begin{array}{l}{x}^2={\displaystyle \sum \frac{{\left(obs-Exp\right)}^2}{Exp}}\\ =4.4\end{array}$

The test statistic has,

  $\begin{array}{l}\left(k-1\right)df\\ =\left(5-1\right)\\ =4df\end{array}$

P-value is,

  $\begin{array}{l}P\left(x^2>4.4\right)\\ =1-P\left(x^2<4.4\right)\\ =1-0.65\\ =0.35\end{array}$

The P-value is larger than the $0.05$ and because of the higher P-value, fail to reject the $H_0$ .

That means data is sufficient evidence to indicate the equal preference for all polished patches of floor.

Now, to find the polish $A$ is superior to the others, if one can reject the hypothesis of no preference for this experiment.

Consider, proportions for polished patches,

Polish	$A$	$B$	$C$	$D$	$E$
Frequency	$0.27$	$0.17$	$0.15$	$0.22$	$0.19$

Yes, the polish $A$ is superior to the others. Since, the sample proportions appear that the polish patches of floor $A$ is the best to the other polish patches of floors.