Chapter 10.P, Problem 10P

To determine

To find:

To write is there sufficient evidence to conclude at $\alpha =0.10$ that the makeup of your institution has significantly changed between Year 1 and Year 2.

Answer to Problem 10P

Solution:

There is no sufficient evidence to conclude that the makeup of your institution has significantly changed between Year 1 and Year 2.

Explanation of Solution

Consider two years that are not consecutive from which to collect data.

That is choosing the current academic year for Year 2 and the academic year four years earlier for Year 1.

Suppose the 2015 for year 1 and 2019 for year 2.

For year 1:

The number of students who were enrolled at your institution for each classification during year 1 is given below,

	2015
Number of Freshmen Enrolled	15
Number of Sophomores Enrolled	18
Number of Juniors Enrolled	21
Number of Seniors Enrolled	73
Total Number of Students Enrolled	127

Let $p_1,p_2,p_{3}and{p}_4$ be the probabilities for freshmen, sophomores, junior and senior respectively.

Mathematically, we can write the null and alternative hypothesis for four different students as follows,

$H_0:p_1=0.1181,p_2=0.1417,p_3=0.1654,and{p}_4=0.5748$

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

Let‘s calculate the expected value for year 1. Since we are assuming that the proportion percentages stated for the student in year 1 are not incorrect is calculated as follows,

$\begin{array}{rll}n& =& 15+18+21+73\\ & =& 127\end{array}$

And,

$\begin{array}{l}E\left(x_1\right)=n{p}_1=127\times 0.1181=14.9987\\ E\left(x_2\right)=n{p}_2=127\times 0.1417=17.9959\\ E\left(x_3\right)=n{p}_3=127\times 0.1654=21.0058\\ E\left(x_4\right)=n{p}_4=127\times 0.5748=72.9996\end{array}$

For year 2:

The number of students who were enrolled at your institution for each classification during year 2 is given below,

	2015
Number of Freshmen Enrolled	24
Number of Sophomores Enrolled	20
Number of Juniors Enrolled	19
Number of Seniors Enrolled	20

Let $p_1,p_2,p_{3}and{p}_4$ be the probabilities for freshmen, sophomores, junior and senior respectively.

Mathematically, we can write the null and alternative hypothesis for four different students as follows,

	2019
Percentage of Freshmen Enrolled	28.92%
Percentage of Sophomores Enrolled	24.10%
Percentage of Juniors Enrolled	22.89%
Percentage of Seniors Enrolled	24.10%

$H_0:p_1=0.2892,p_2=0.2410,p_3=0.2289,and{p}_4=0.2410$

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

Let‘s calculate the expected value for year 2. Since we are assuming that the proportion percentages stated for the student in year 2 are not incorrect is calculated as follows,

$\begin{array}{rll}n& =& 24+20+19+20\\ & =& 83\end{array}$

And,

$\begin{array}{l}E\left(x_1\right)=n{p}_1=83\times 0.2892=24.0036\\ E\left(x_2\right)=n{p}_2=83\times 0.2410=20.003\\ E\left(x_3\right)=n{p}_3=83\times 0.2289=18.9987\\ E\left(x_4\right)=n{p}_4=83\times 0.2410=20.003\end{array}$

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 for year 1 is 127 and for year 2 is 83.

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

Year 1
Category	Observed Values	Expected values	$\left(O_i-E_i\right)$	${\left(O_i-E_i\right)}^2$	$\frac{{\left(O_i-E_i\right)}^2}{E_i}$
Freshman	15	14.9987	0.0013	$1.69\times {10}^{-6}$	$1.1268\times {10}^{-7}$
Sophomore	18	17.9959	0.0041	$1.681\times {10}^{-5}$	$9.341\times {10}^{-7}$
Junior	21	21.0058	-0.0058	$3.364\times {10}^{-5}$	$1.6015\times {10}^{-6}$
Senior	73	72.9996	0.0004	$1.6\times {10}^{-7}$	$2.1918\times {10}^{-9}$
					${\chi }^2={\displaystyle \sum _{i=1}^n\frac{{\left(O_i-E_i\right)}^2}{E_i}}=2.6504\times {10}^{-6}$
Year 2
Freshman	24	24.0036	-0.0036	$1.296\times {10}^{-5}$	$5.39919\times {10}^{-7}$
Sophomore	20	20.003	-0.003	$9\times {10}^{-6}$	$4.49933\times {10}^{-7}$
Junior	19	18.9987	0.0013	$1.69\times {10}^{-6}$	$8.89535\times {10}^{-8}$
Senior	20	20.003	-0.003	$9\times {10}^{-6}$	$4.49933\times {10}^{-7}$
					${\chi }^2={\displaystyle \sum _{i=1}^n\frac{{\left(O_i-E_i\right)}^2}{E_i}}=1.52874\times {10}^{-6}$

The calculated test statistic for year 1 and for year 2 is $2.6504\times {10}^{-6}$ and $1.52874\times {10}^{-6}$ respectively.

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 = 4.

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

$\begin{array}{rll}df& =& k-1\\ & =& 4-1\\ & =& 3\end{array}$

Conclusion:

Use the level of significance of $\alpha =0.10$ to find the critical value ${\chi }_{0.10}^2$ of the chi-square distribution with df = 3 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,

${\chi }_{0.10}^2=6.2514$

Year 1:

The test statistic value is ${\chi }^2=2.6504\times {10}^{-6}$.

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

Since $2.6504\times {10}^{-6}<6.2514\iff {\chi }^2<{\chi }_{0.10}^2$.

Year 2:

The test statistic value is ${\chi }^2=1.52874\times {10}^{-6}$.

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

Since $1.52874\times {10}^{-6}<6.2514\iff {\chi }^2<{\chi }_{0.10}^2$.

The null hypothesis will be accepted for both year 1 and year 2.

Hence the proportion percentage for the students does not vary for both year 1 and year 2.

So, there is no sufficient evidence to conclude that the makeup of your institution has significantly changed between Year 1 and Year 2. Since the proportion percentage will not vary between two years. That is, percentage of students in every classification in year 1will be equal to percentage of students in every classification in year 2.

Final statement:

There is no sufficient evidence to conclude that the makeup of your institution has significantly changed between Year 1 and Year 2.