Question

Ivy League schools receive many applications, but only some can be accepted. At the schools listed in Table 11.22, two types of applications are accepted: regular and early decision.

We want to know if the number of regular applications accepted follows the same distribution as the number of early applications accepted. State the null and alternative hypotheses, the degrees of freedom and the test statistic, sketch the graph of the p-value, and draw a conclusion about the test of homogeneity.

Short answer

The null hypothesis can be stated as:

\(H_{0}:\) The number of regular applications accepted follows the same distribution as the number of early applications accepted.

And the alternative hypothesis can be stated as:

\(H_{a}:\) The number of regular applications accepted does not follow the same distribution as the number of early applications accepted.

\(df=5\)

\(p-value<\alpha\)

Therefore the decision will be to reject the null hypothesis, \(H_{0}\). Therefore, the null hypothesis gets rejected while alternative is accepted. Hence, it can be concluded that there are sufficient evidence to ensure that the number of regular applications accepted does not follow the same distribution as the number of early applications accepted.

Step-by-step solution

Step 1. Given information

Ivy League schools receive many applications, but only some can be accepted. At the schools listed in the below given table, two types of applications are accepted: regular and early decision;

Application Type Accepted	Brown	Columbia	Cornell	Dartmouth	Penn	Yale
Regular	2115	1792	5306	1734	2685	1245
Early Decision	577	627	1228	444	1195	761

Step 2. Calculation

The null hypothesis can be stated as:

\(H_{0}:\) The number of regular applications accepted follows the same distribution as the number of early applications accepted.

And the alternative hypothesis can be stated as:

\(H_{a}:\) The number of regular applications accepted does not follow the same distribution as the number of early applications accepted.

The degrees of freedom can be calculated by the formula given below;

\(df=\)(number of columns \(-1\)

\(\Rightarrow df=(6-1)\)

\(\Rightarrow df=5)\)

The observed value table is already given. Now let’s calculate the expected frequencies by using the formula shown below;

\(E=\frac{(row total)(column total)}{overall total}\)

Let’s calculate the expected (E) values in excel as shown below;

The test statistic of independence test is given below;

Test statistics \(=\sum_{i\times j}=\frac{(O-E)^{2}}{E}\)

To calculate \(\frac{(O-E)^{2}}{E}\) apply formula \(=((B4-B11)^{2})/B11\) in cell \(B17\) and drag the same formula up to cell F18. After that, take total of columns total and row total. The table of test statistic is shown below;

Hence, the test statistic is calculated as \(430.063\)

The p-value can be calculated in excel by using formula as shown below:

Therefore the p-value is zero.

The p-value sketch is shown below;

We have the alpha value given as \(0.05\). Now, since \(p-value<\alpha\) therefore the decision will be to reject the null hypothesis, \(H_{0}\) . Therefore, the null hypothesis gets rejected while alternative is accepted. Hence, it can be concluded that there are sufficient evidence to ensure that the number of regular applications accepted does not follow the same distribution as the number of early applications accepted.