Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, x overbar​, is found to be 108​, and the sample standard​ deviation, s, is found to be 10. ​(a) Construct a 95​% confidence interval about mu if the sample​ size, n, is 25. ​(b) Construct a 95​% confidence interval about mu if the sample​ size, n, is 12. ​(c) Construct a 70​% confidence interval about mu if the sample​ size, n, is 25. ​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed?

Answer 1

Given X̄ ~ N(108,10)

And we know, CI = X̄ ± Z* (σ/√n)

(a) Since z value for 95% CI is 1.96, 95% CI for mu is X̄ ± 1.96* (σ/√n)

Here, n=25, X̄=108,σ=10

Thus, CI= 108±1.96*(10/√25)

= 108 ± 1.96*2

= 108±3.92

= (104.08,111.92)

(b) Since z value for 95% CI is 1.96, 95% CI for mu is X̄ ± 1.96* (σ/√n)

Here, n=12, X̄=108,σ=10

Thus, CI= 108±1.96*(10/√12)

= 108 ± 1.96*2.886

= 108±5.658

= (102.341,113.658)

(c) z value for 70% CI is 1.036.

Thus, 70% CI for mu is X̄ ± 1.036* (σ/√n)

Here, n=25, X̄=108,σ=10

Thus, CI= 108±1.036*(10/√25)

= 108 ± 1.036*2

= 108± 2.072

= (105.928,110.072).

(d) If the sample size were large(>= 30), then we could have used CLT to calculate the CI for mu even when the population is not normal. But in this case, Since the sample size is small and if the population were non-normal we couldn't have computed the confidence interval for the population mean.