Chapter 10.6, Problem 21E

To determine

To find:

To perfom the test statistic for the following scenerio,

“A school principal claims that the number of students who are tardy in her school does not vary from month to month. Using a 0.05 level of significance, test a teacher’s claim that the number of tardy studens does vary by the month”.

Answer to Problem 21E

Solution:

(a) $H_0:p_1=p_2=\mathrm{...}=p_{10}$

$H_a$: There is some difference amongst the probabilities.

(b) The chi-square test for this goodness of fit and the level of siginificance is $\alpha =0.05$.

(c) The test statistic value is ${\chi }^2=15.56$.

(d) There is no sufficient evidence to support the teacher’s claim that the number of tardy students does vary by the month.

Explanation of Solution

The following table represents the random sample of number of students who are tardy in her school is,

Tardy Students
	Aug.	Sept.	Oct.	Nov.	Dec.	Jan.	Feb.	Mar.	Apr.	May
Number	7	18	16	5	8	12	15	18	11	15

Step 1: To state the null and alternative hypotheses.

When starting the hypotheses to be used we take the null hypothesis to be that the proportion of students who are tardy in her school does not vary from month to month.

Null hypothesis: $H_0$: The proportion of of students who are tardy in her school does not vary from month to month.

Alternative hypothesis: $H_a$: The proportion of of students who are tardy in her school does vary from month to month.

Mathematically, we can write the null and alternative hypothesis as follows,

$H_0:p_1=p_2=\mathrm{...}=p_{10}$

$H_a$: There is some difference amongst the probabilities.

Step 2: To determine whether the observed values of tardy students match with the expected values of tardy students. Since we assume that the necessary condition that have been met. So, use the chi-square test for this goodness of fit and the level of siginificance is $\alpha =0.05$.

Step 3: To calculate the test statistic.

Let‘s calculate the expected value for each month of the year. Since we are assuming that the number of tardy students does not vary for each month, then the probability will be same for every day, so expected number of students for each month is calculated as follows,

$\begin{array}{rll}n& =& 7+18+16+5+8+12+15+18+11+15\\ & =& 125\end{array}$

And,

$\begin{array}{rll}E\left(x_i\right)& =& n{p}_i\\ & =& 125.\frac{1}{10}\\ & =& 12.5\end{array}$

Formula for calculating test statistic for a Chi-Square test for goodness of fit:

The test statistic for a chi-square test for goodness of fit is given below,

${\chi }^2={\displaystyle \sum _{i=1}^n\frac{{\left(O_i-E_i\right)}^2}{E_i}}$

Where O_i, is the observed frequency for the i^th possible outcome and E_i is the expected frequency for the i^th possible outcome and n is the sample size.

From the above table represent the value of n is 125.

The following table represents the test statistic, ${\chi }^2$, for a chi- square test for goodness of fit is given below,

Month	Observed Values	Expected values	$O_i-E_i$	${\left(O_i-E_i\right)}^2$	$\frac{{\left(O_i-E_i\right)}^2}{E_i}$
Aug.	7	12.5	-5.5	30.25	2.42
Sept.	18	12.5	5.5	30.25	2.42
Oct.	16	12.5	3.5	12.25	0.98
Nov.	5	12.5	-7.5	56.25	4.5
Dec.	8	12.5	-4.5	20.25	1.62
Jan.	12	12.5	-0.5	0.25	0.02
Feb.	15	12.5	2.5	6.25	0.5
Mar.	18	12.5	5.5	30.25	2.42
Apr.	11	12.5	-1.5	2.25	0.18
May	15	12.5	2.5	6.25	0.5
					${\chi }^2={\displaystyle \sum _{i=1}^n\frac{{\left(O_i-E_i\right)}^2}{E_i}}=15.56$

Step 4: Degrees of freedom in a Chi-square test for goodness of fit:

In a chi-square test for goodness of fit the number of degrees of freedom for the chi-square distribution of the test statistic is given by,

$df=k-1$

Where k is the number of possible outcomes for each trial.

Rejection Region for Chi-Square Test for Association:

Reject the null hypothesis, $H_0$, if ${\chi }^2\ge {\chi }_{\alpha }^2$.

From the given information k = 10.

Substitute the above values in the formula of degrees of freedom to get the following,

$\begin{array}{rll}df& =& k-1\\ & =& 10-1\\ & =& 9\end{array}$

Conclusion:

Use the level of significance of $\alpha =0.05$ to find the critical value ${\chi }_{0.05}^2$ of the chi-square distribution with df = 9 in order to make a decision about the null hypothesis.

Use the “Area to the right of the critical value ${\chi }^2$” table to get the following,

${\chi }_{0.05}^2=\begin{array}{l}16.9190\hfill \end{array}$

The test statistic value is ${\chi }^2=15.56$.

By using the above condition we fail to reject the null hypothesis.

Since $15.56<16.9190\iff {\chi }^2<{\chi }_{0.05}^2$.

So, there is no sufficient evidence to support the teacher’s claim that the number of tardy students does vary by the month.

Final statement:

(a) $H_0:p_1=p_2=\mathrm{...}=p_{10}$

$H_a$: There is some difference amongst the probabilities.

(b) The chi-square test for this goodness of fit and the level of siginificance is $\alpha =0.05$.

(c) The test statistic value is ${\chi }^2=15.56$.

(d) There is no sufficient evidence to support the teacher’s claim that the number of tardy students does vary by the month.