A survey of 219 students at a community college found that 81 of them had graduated with some kind of academic honors. Find the 95% confidence interval of the true proportion of community college students that graduated with some kind of academic honors.Write your answer as an interval (smaller number first, then bigger number).

Answer:

Given Data:
The favorable cases (x)=81
The sample size (n)=219
The confidence interval level =95%

→ The sample proportion (p ̂ )=x/n=81/219=0.3699

→ The level of significance (α)=1-confidence interval level=1-0.95=0.05

→ The critical value at α=0.05 is 1.96 [ use z table ]

→ The confidence interval = p ̂±z_c√((p ̂(1-p ̂ ))/n) = 0.3699±1.96√(0.3699(1-0.3699)/219)

=(0.306,0.434)

Adbhutah.com