Question

1. For a random sample of 256 owners of medium-sized cars, it was found that their average monthly car insurance premium for comprehensive cover was R356. Assume that the population standard deviation is R44 per month and that insurance premiums are normally distributed.

1. Find the 95% confidence interval for the average monthly comprehensive car insurance premium paid by all owners of medium-sized cars. Interpret the results.                                                                                                                                           [9 Marks]
2. Find the 90% confidence interval for the same problem. Interpret the result.

[9 Marks]

3. Compare the two intervals, what can be observed and concluded from the interval as we decrease the confidence level/percentage?                             [2 Marks]

Answer 1

a)

sample mean, xbar = 356
sample standard deviation, σ = 44
sample size, n = 256

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 44/sqrt(256)
ME = 5.39

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (356 - 1.96 * 44/sqrt(256) , 356 + 1.96 * 44/sqrt(256))
CI = (350.61 , 361.39)

b)

sample mean, xbar = 356
sample standard deviation, σ = 44
sample size, n = 256

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

ME = zc * σ/sqrt(n)
ME = 1.64 * 44/sqrt(256)
ME = 4.51

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (356 - 1.64 * 44/sqrt(256) , 356 + 1.64 * 44/sqrt(256))
CI = (351.49 , 360.51)

c)

As we decrease the confidence level th einterval would be narrower