"What will happen to Confidence Interval if you change from using sample SD to population SD? Would the CI you calculated previously increase, decrease or remain?" As I think the replacement of population SD will make the confidence interval more accurate but not sure why. Please help me if you know the answer.

• That depends on tail heaviness and length.
– Carl
Commented May 5, 2020 at 3:24

If the population variance $$\sigma^2$$ is known, then you should use it. Then a 95% CI for the unknown normal population mean $$\mu$$ would be a z interval of the form $$\bar X \pm 1.96\frac{\sigma}{\sqrt{n}}.$$

If you decided to ignore the known value of $$\sigma$$ estimating it by the sample standard deviation $$S,$$ then a 95% t interval would be of the form $$\bar X \pm t^*\frac{S}{\sqrt{n}},$$ where $$t^*$$ cuts probability 0.025 from the upper tail of Student's t distribution with $$n - 1$$ degrees of freedom.

For small $$n,$$ the value $$t^*$$ is noticeably larger than $$1.96,$$ so the CI would tend to be larger on that account. However, there is no way to know whether the estimate $$S$$ is larger or smaller than $$\sigma.$$ For large $$n,$$ you'd have $$t^* \approx 1.96,$$ but but $$S$$ can still be either larger or smaller than $$\sigma.$$

The bottom line is that you don't know the relationship between $$S$$ and $$\sigma,$$ so it is not possible to know whether the z or t interval is longer.

Consider the example below with $$n=50$$ observations from $$\mathsf{Norm}(\mu=100,\sigma=15).$$

set.seed(2020)
x = rnorm(50, 100, 15);  mean(x);  sd(x)
[1] 101.8843
[1] 16.67984


So $$S = 16.68 > \sigma = 15.$$ The 95% t confidence interval $$(97.144, 106.625)$$ from R is shown below. Its length is $$9.481.$$ If you want to check it by hand (within rounding error), then use $$t^* = 2.01$$ and $$S = 16.68.$$

t.test(x)$conf.int [1] 97.14397 106.62469 attr(,"conf.level") [1] 0.95 qt(.975, 49) [1] 2.009575  And the 95% z interval is $$(97.727, 106,042),$$ which happens to be shorter (length $$8.316)$$ because $$\sigma < S.$$ pm = c(-1,1) mean(x) + pm*1.96*15/sqrt(50) [1] 97.72654 106.04212  The expected length of a 95% t confidence inter for a sample of size $$n=50$$ from $$\mathsf{Norm}(\mu=100,\sigma=15)$$ is about $$8.48,$$ based on the simulation below---shorter than for the t interval in the example above. Because the length of the z interval with $$\sigma=15$$ and $$n = 50$$ is not random, we can say that in this case the average 95% t interval is (a little) longer than the z interval. set.seed(2020) len = replicate(10^5, diff(as.numeric(t.test(rnorm(50,100,15))$conf.int)))
mean(len)
[1] 8.481077