Mathematics, 06.05.2020 04:44 thedocgalloway

A private and a public university are located in the same city. For the private university, 1042 alumni were surveyed and 655 said that they attended at least one class reunion. For the public university, 796 out of 1318 sampled alumni claimed they have attended at least one class reunion. Is the difference in the sample proportions statistically significant? (Use α=0.05)

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Answer from: lLavenderl

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the proportion of alumni that assist to at least a one class reunion is different for privates university and public university (p-value=0.222).

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the proportion of alumni that assist to at least a one class reunion is different for privates university and public university.

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq 0$

where π1: proportion of private university alunmi that have attended at least one class reunion, and π2: proportion of public university alunmi that have attended at least one class reunion.

The significance level is 0.05.

The sample 1 (private), of size n1=1042 has a proportion of p1=0.6286.

p_1=X_1/n_1=655/1042=0.6286

The sample 2 (public), of size n2=1318 has a proportion of p2=0.6039.

p_2=X_2/n_2=796/1318=0.6039

The difference between proportions is (p1-p2)=0.0247.

p_d=p_1-p_2=0.6286-0.6039=0.0247

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{655+796}{1042+1318}=\dfrac{1451}{2360}=0.6148$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.6148*0.3852}{1042}+\dfrac{0.6148*0.3852}{1318}}\\\\\\s_{p1-p2}=\sqrt{0.00023+0.00018}=\sqrt{0.00041}=0.0202$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.0247-0}{0.0202}=\dfrac{0.0247}{0.0202}=1.222$

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=2\cdot P(t1.222)=0.222$

As the P-value (0.222) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the proportion of alumni that assist to at least a one class reunion is different for privates university and public university.

Answer from: Quest

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Tess weighted two colored metal balls during