Confidence interval for normal population having both mean and variance..

Question

I'm dealing with this problem:

Let $X$ denote the concentration level of a drug in blood of a certain population of $90$ people and suppose $X \sim \mathcal{N}(\mu,\sigma^2)$ with $\mu= 100$ and $\sigma^2 = 1200$.

1)Let $H_0$ be the following null hypothesis: $H_0$ : the mean drug level of the population is equal to $95$. Calculate a bilateral confidence interval with 95% significancy for the mean drug concentration.

2)calculate the critical value $t^*$ of a $t$ student in the case we want to do a bilateral test for $H_0$ at 1% level. Is it possible to reject

My attempts:

1)The formula for calculating the (bilateral) confidence interval is $$(\bar{X}_n - z_{\alpha /2}\frac{\sigma}{\sqrt{n}},\bar{X}_n + z_{\alpha /2}\frac{\sigma}{\sqrt{n}})$$

But in this case I know the exact mean $\mu$ for the entire population... does this mean $\bar{X}_{90} = \mu = 100?$

So I would proceed saying

$$(100 - z_{0.025}\frac{\sqrt{1200}}{\sqrt{90}},100 + z_{0.025}\frac{\sqrt{1200}}{\sqrt{90}})$$

Does this make sense?

2)Why does the problem talk about $t$ student? I have both the real mean and standard deviation....

Thanks for the help!

Answer 1

(1) If you don't know $\mu$ and do know $\sigma^2 = 1200, \sigma=34.64,$ and if you have $n$ observations from this normal population, then you can find a $95\%$ confidence interval for $\mu$ of the type $\bar X \pm 1.96\sigma/\sqrt{n},$ where $\bar X$ is the sample mean. (Note that this is a confidence level, not a level of significance, which would apply to a z test.)

(2) If you know neither $\mu$ nor $\sigma^2$ but know that the population is normally distributed, then for a sample of size $n = 90.$ you would have the two-sided 95% CI $\bar X \pm 1.987\,S/\sqrt{90},$ where $\bar X$ is the sample mean, $S$ is the sample standard deviation, and $1.987$ cuts probability $0.025 = 2.5\%$ from the upper tail of Student's t distribution with $\nu = n - 1 =89$ degrees of freedom (as found in a printed table of t distributions, or by using software, such as R, as below).

qt(.975, 89)
[1] 1.986979

Note: Some authors might use 1.96 (because the t distribution with 89 DF is nearly normal) or 2 (because it is an integer and 'almost correct') instead of $1.987.$