Question

2500 randomly selected Ryerson students were asked to fill out a survey on whether they exercise regularly. 2000 students gave a positive answer. a) Calculate a 95% two-sided confidence interval on the proportion of students who regularly exercise. b) Indicate how the length of confidence interval changes if we increase the level of confidence and keep other values constant? Justify your answer.

Answer 1

Let x be the number of students exercide regularly.

Given : x = 2000, n = 2500, $1596208631039_blob.png$ = x /n = 0.8

a) Confidence level = 0.95

Therefore α = 1 - 0.95 = 0.05 , 1 - (α/2) = 0.9750

So we have to find z score corresponding to area 0.9750 on z score table

So z = 1.96

95% confidence interval is given by :

f*(1-P) p£ * n

= $0.8\pm 1.96*{\sqrt{\frac{0.8*(1-0.8)}{2500}}}$

= $0.8\pm 0.01568$

( 0.7843 and 0.8157 )

b) Width of confidence interval = 2*E

E is margin of error = $\textup{Critical value }*\frac{S}{\sqrt{n}}$

Critical value is depends on the level of confidence , As level of confidence increases, critical value also increases.

Therefore margin of error increases and it leads to increase in width of the confidence interval.

So if we increase the level of confidence and keep other values constant , then length of confidence interval also increases.