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A state university finds that 115 of a random sample of 200 of its first-year students say

that "being very well-off financially is an important personal goal. If they conduct the

appropriate hypothesis test, is there evidence that a majority of all first-year students

at this university think being very well-off financially is important? Have all the

conditions been met for this situation?

1 Answer

2 votes

Answer:

There is enough evidence to support the claim that the mayority (more than 50%) of the students think that "being very well-off financially" is an important personal goal.

The conditions are met, as this is a randome sample and the number of of positive answers (np=115) and negative answers (nq=85) are both higher than 10.

Explanation:

The conditions to test the hypothesis are met, as this is a randome sample and the number of of positive answers (np=115) and negative answers (nq=85) are higher than 10.

The claim is that the mayority (more than 50%) of the students think that "being very well-off financially" is an important personal goal.

Then, the null and alternative hypothesis are:


H_0: \pi=0.5\\\\H_a:\pi> 0.5

The significance level is assumed to be 0.05.

The sample has a size n=200.

The sample proportion is p=0.575.


p=X/n=115/200=0.575

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.5*0.5}{200}}\\\\\\ \sigma_p=\sqrt{0.00125}=0.0354

Then, we can calculate the z-statistic as:


z=(p-\pi+0.5/n)/(\sigma_p)=(0.575-0.5-0.5/200)/(0.0354)=(0.0725)/(0.0354)=2.0506

This test is a right-tailed test, so the P-value for this test is calculated as:


P-value=P(z>2.0506)=0.0202

As the P-value (0.0202) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the mayority (more than 50%) of the students think that "being very well-off financially" is an important personal goal.

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