Chapter 10.1, Problem 10.8E

To determine

To Calculate: the chi-square goodness of fit test and conclusion.

Answer to Problem 10.8E

  $P<0.05\Rightarrow {\text{Reject H}}_0$

There is enough evidence to the claim that the area is made up of 29% mud, 56% sand, and 15% rocks.

Explanation of Solution

Given:

  $n=200$

Formula used:

  $x^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$

Calculation:

Assume

  $\alpha =0.05$

The null hypothesis statement is the population proportions are equal to the given proportions

  $H_0:p_1=56\%=0.56,p_2=29=o.29,p_3=15\%=0.15$

The alternative hypothesis statement is the opposite of the null hypothesis:

  $H_1:{\text{At least one of the p}}_i\text{'}\text{s is different}$

The expected frequencies $E$ are the product of the sample size $n$ and the probabilities $p_i$ .

  $\begin{array}{c}{\text{E}}_1=n{p}_1=200\times 0.56=112\\ {\text{E}}_2=n{p}_2=200\times 0.29=58\\ E_3=n{p}_3=200\times 0.15=30\end{array}$

The value of the test-statistic is

  $\begin{array}{c}{x}^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ =\frac{{\left(128-112\right)}^2}{112}+\frac{{\left(61-58\right)}^2}{58}+\frac{{\left(11-30\right)}^2}{30}\\ =14.4742\end{array}$

The degree of freedom is

  $\begin{array}{c}df=c-1=3-1=2\\ \end{array}$

The P-Value is the probability of getting the value of the test statistic, or value more extreme. The P-value is the number in the column of the chi-square distribution table containing the $x^2$ -value in the row $df=c-1=3-1=2$

  $P<0.001$

If the P-value is less than or equal to the significance level, then the Alternate hypothesis is accepted and null hypothesis is rejected:

  $P<0.05\Rightarrow {\text{Reject H}}_0$

There is enough evidence to the claim that the area is made up of 29% mud, 56% sand, and 15% rocks.