Question

32 out of 60 randomly chosen students travelling to school by bus. Find 99% confidence interval for the true proportion of students travel to school by bus. If 1000 students chosen at random, what can you conclude from the confidence interval that you found?

Answer 1

Ans) - Given,

n= 60 (Randomly chosen students)

no. of students travel by bus = 32

let, 'p' be the probability of students travel by bus

so, p = [no. of students travel by bus]/n

p = 32/60

[p = 0.53]

99% confidence level

We have formula for calculating confidence interval

Where, p: probability of favourable outcome (here p=0.53)

n: sample size (here n=60)

z: corresponding ‘z’ value at 99% confidence level

Calculating ‘z’:

Take, α = 99% = 0.99

(1- α)/2 = (1-0.99)/2 = 0.005

Now, 1 – [(1- α)/2] = 1-0.005 = 0.995

Hence, the value of ‘z’ statistic from z table corresponding to 0.995 will be the value of ‘z’

So, from z table

[z = 2.58]

hence, confidence interval (C.I.)

Hence, Required confidence interval (C.I.) is (0.53+0.166 , 0.53-0.166).

C.I. = (0.696 , 0.364).

NOW IF 1000 STUDENTS CHOSEN AT RANDOM, THEN

New n=1000

Remaining values will be same i.e.

p = 0.53, z=2.58 (at 99% confidence level)

So, put these values in confidence interval formula to find new C.I.

New confidence interval (C.I.) at n=1000 is (0.53+0.0405 , 0.53-0.0405)

C.I. = (0.5705 , 0.4895).

Conclusion: When the sample size is increased (i.e. from 60 to 1000 randomly selected student), the confidence interval with increased sample size became narrow i.e. the width of confidence interval reduced with increased sample size.