Chapter 17, Problem 17CAP

1.

To determine

Check whether to retain or reject the null hypothesis for a chi-square goodness-of-fit test.

Answer to Problem 17CAP

The decision is to reject the null hypothesis.

The distribution of proportions is different from the proportions stated in the null hypothesis.

Explanation of Solution

Calculation:

A sample of 120 students is considered to ask the rate their performance for one of four video games. The given data is,

	Video Games
	McStats	Tic-Tac-Stats	Silly Stats	Super Stats
Frequency Observed	30	30	30	30

Table 1

And the expected frequencies are 70%, 10%, 10%, and 10% respectively.

The formula for total participants is,

$N={\displaystyle \sum _{i=1}^n{f}_{o_i}}$

In the formula, $f_o$ denotes the observed frequencies for each category.

The formula for expected frequency is,

$f_e=Np$

In the formula, N denotes the total participants and p denotes the proportion for each level of categorical variable.

Test statistic:

The formula for chi-square test statistic is,

${\chi }_{\text{obt}}^2={\displaystyle \sum \frac{{\left(f_o-f_e\right)}^2}{f_e}}$

In the formula, $f_o$ denotes the observed frequency and $f_e$ denotes the expected frequency.

The formula for degrees of freedom for chi-square test is,

$df=k-1$

In the formula, k denotes the number of levels of categorical variable.

Null hypothesis:

$H_0:$ The distribution of proportions (7:1:1:1) is the same as expected.

$\begin{array}{c}{p}_{\text{McStats}}=70\%\\ p_{\text{Tic-Tac-Stats}}=10\%\\ p_{\text{Silly Stats}}=10\%\\ p_{\text{Super Stats}}=10\%\end{array}$

Alternative hypothesis:

$H_1:$ The distribution of proportions is different from the proportions stated in the null hypothesis.

Expected frequency:

Substitute the corresponding values of total frequency and proportion for each level of categorical variable in the test statistic formula.

	Video Games
	McStats	Tic-Tac-Stats	Silly Stats	Super Stats
$f_o$	30	30	30	30
$f_e$	$120\times 0.70=84$	$120\times 0.10=12$	$120\times 0.10=12$	$120\times 0.10=12$

Table 2

Substitute, $k=4$ in degrees of freedom formula.

$\begin{array}{c}df=4-1\\ =3\end{array}$

Critical value:

Consider the significance level is $\alpha =0.05$ and the degrees of freedom are 3.

From the Appendix C: Table C.7 Critical values for Chi-square:

• Locate the value 3 in the degrees of freedom (df) column.
• Locate the 0.05 in level of significance row.
• The intersecting value that corresponds to the 3 with level of significance 0.05 is 7.81.

The critical value for $df=3$ at a 0.05 level of significance is 7.81.

Test statistic value:

Substitute the corresponding values of observed and expected frequencies for each level of categorical variable in the test statistic formula.

$\begin{array}{c}{\chi }_{\text{obt}}^2=\frac{{\left(30-84\right)}^2}{84}+\frac{{\left(30-12\right)}^2}{12}+\frac{{\left(30-12\right)}^2}{12}+\frac{{\left(30-12\right)}^2}{12}\\ =\frac{2,916}{84}+\frac{324}{12}+\frac{324}{12}+\frac{324}{12}\\ =34.71+27+27+27\\ =115.71\end{array}$

The value of test statistic is 115.71.

Conclusion:

The test statistic value is 115.71.

The critical value is 7.81.

The test statistic value is greater than the critical value.

The null hypothesis is rejected.

The distribution of proportions is different from the proportions stated in the null hypothesis.

2.

To determine

Check whether to retain or reject the null hypothesis for a chi-square goodness-of-fit test.

Answer to Problem 17CAP

The decision is to retain the null hypothesis.

The distribution of proportions (2.5:2.5:2.5:2.5) is the same as expected.

Explanation of Solution

Calculation:

The given expected frequencies are 25%, 25%, 25%, and 25% respectively.

Null hypothesis:

$H_0:$ The distribution of proportions (2.5:2.5:2.5:2.5) is the same as expected.

$\begin{array}{c}{p}_{\text{McStats}}=25\%\\ p_{\text{Tic-Tac-Stats}}=25\%\\ p_{\text{Silly Stats}}=25\%\\ p_{\text{Super Stats}}=25\%\end{array}$

Alternative hypothesis:

$H_1:$ The distribution of proportions is different from the proportions stated in the null hypothesis.

Expected frequency:

Substitute the corresponding values of total frequency and proportion for each level of categorical variable in the test statistic formula.

	Video Games
	McStats	Tic-Tac-Stats	Silly Stats	Super Stats
$f_o$	30	30	30	30
$f_e$	$120\times 0.25=30$	$120\times 0.25=30$	$120\times 0.25=30$	$120\times 0.25=30$

Table 3

Test statistic value:

Substitute the corresponding values of observed and expected frequencies for each level of categorical variable in the test statistic formula.

$\begin{array}{c}{\chi }_{\text{obt}}^2=\frac{{\left(30-30\right)}^2}{30}+\frac{{\left(30-30\right)}^2}{30}+\frac{{\left(30-30\right)}^2}{30}+\frac{{\left(30-30\right)}^2}{30}\\ =\frac{0}{30}+\frac{0}{30}+\frac{0}{30}+\frac{0}{30}\\ =0+0+0+0\\ =0\end{array}$

The value of test statistic is 0.

Conclusion:

The test statistic value is 0.

The critical value is 7.81.

The test statistic value is less than the critical value.

The null hypothesis is retained.

The distribution of proportions (2.5:2.5:2.5:2.5) is the same as expected.