Question

(a) Construct and interpret a 95 % confidence interval for the true proportion p of all first-year students at the university who would identify being well-off as an important personal goal.

(b) Explain what the interval in part (a) tells you about whether the national value holds at this university.

Short answer

a. The confidence interval is$(0.59,0.73)$

b. The national value holds at this university as the value lies in the interval

Step-by-step solution

Introduction

A confidence interval is the mean of your estimate in addition to and minus the variation in that estimate. This is the scope of values you anticipate that your estimate should fall between if you re-try your test, within a certain degree of confidence. Confidence, in statistics, is one more method for describing probability.

Explanation Part (a)

The number of students is n = $200$

Number of students in favour of well being x = $132$

population proportion $73\%=0.73$

$\stackrel{-}{p}=\frac{x}{n}=\frac{132}{200}=0.66$

Using,

$CI=\stackrel{-}{p}\pm z_{\alpha /2}\times \sqrt{\frac{\stackrel{-}{p}(1-\stackrel{-}{p})}{n}}$

$CI=0.73\pm 1.96\times \sqrt{\frac{0.73(1-0.73)}{200}}$

$CI=$$(0.59,0.73)$

Explanation Part (b)

The national value holds at this university as the value $0.73$ lies in the interval.