Question

A student believes that less than 50% of students at his college receive financial aid. A random sample of 120 students was taken. Sixty-five percent of the students in the sample receive financial aid. Test the hypothesis at the 2% level of significance. What are the p-value and conclusion?

Answer 1

Answer:

p= 0.9995 and we can conclude that students receiving aid is more than 50% according to the sample results in 2% significance level.

Step-by-step explanation:

$H_{0}$: less than or equal 50% of students at his college receive financial aid

$H_{a}$: more than 50% of students receive financial aid.

According to the nul hypothesis we assume a normal distribution with proportion 50%.

z-score of sample proportion can be calculated using the formula:

z=$\frac{X-M}{\frac{s}{\sqrt{N} } }$ where

• X is the sample proportion (0.65)
• M is the null hypothesis proportion (0.5)
• s is the standard deviation of the sample ($\sqrt{0.5*(1-0.5)}$)
• N is the sample size (120)

then z=$\frac{0.65-0.50}{\frac{\sqrt{0.25}}{\sqrt{120} } }$ ≈ 3,29

thus p= 0.9995 and since p<0.02 (2%), we reject the null hypothesis.