Consider a normal distribution $N(\mu, \sigma^{2})$ where $\sigma^2 = 9$ and we draw 100 samples from this distribution. Now, we set up a hypothesis testing problem: $H_{0}:\mu = 0$ against $H_{a}:\mu \ne 0$. We reject null if $|mean| > K$ with 0.05 as the size of the test. It is required to obtain the $95$% confidence interval of $\mu$.

In this problem, I don't understand why have they provided us with all this extra information about hypothesis testing ($H_0$ and $H_a$) when the confident interval could be just computed by using the formula :

$\bar{X}_{n} \pm \frac{\sigma}{\sqrt{n}}*Z_{0.05}$

What is the use of this extra information about hypothesis testing when the CI could just be obtained without it? They also have provided four options to choose from:

  1. $(-0.488, 0.688)$
  2. $(-1.96, 1.96)$
  3. $(0.422, 1.598)$
  4. $(0.588, 1.96)$

EDIT.1 If I instead specify the problem as follows:

Consider a normal distribution $N(\mu, \sigma^{2})$ where $\sigma^2 = 9$ and we draw 100 samples from this distribution. It is required to obtain the $95$% confidence interval of $\mu$.

Here, I have omitted the hypothesis specification from the problem. Now, my doubt was that, we can form the confidence interval even without that information, then what was the need for that?

  • 1
    $\begingroup$ Please can you clarify which bit of the problem statement do you consider extra (and perhaps unnecessary as you seems to imply) information? The formula required $\sigma$ and $n$, which is given in the statement, and the formula is only applicable when the problem setting as specified in the question is specified. $\endgroup$
    – B.Liu
    Feb 4, 2021 at 13:21
  • $\begingroup$ Why is it so that formula is only applicable when the problem setting is specified as in it is specified? If I have $X$ following $N(\mu, \sigma^{2})$ and consider a sample of size $n$. Now, $\bar{X}_{n}$ follows normal $N(\mu, \frac{\sigma^{2}}{n})$ . If I have to make a CI of $\mu$ with size $100(1-\alpha)$% then I can just write: $P(|\frac{\bar{X}_{n}-\mu}{\sqrt{\frac{\sigma^{2}}{n}}}| <k) = 1-\alpha$. From here, I can use adjust the terms for $\mu$ to get an interval with probability $100(1-\alpha)$%. Here, I have never used the hypothesis specification. This probability is true in general $\endgroup$
    – userNoOne
    Feb 4, 2021 at 13:49
  • 1
    $\begingroup$ 1) to let you know that it is a two-sided equal tail test and 2) they want you to use $K$ in your answer. $\endgroup$
    – John L
    Feb 4, 2021 at 15:03
  • $\begingroup$ Alright, if I use their rejection region as a starting point to form the CI, then here is what I will get: $ P(| \bar{X}_{n} | >K | H_{0}) = \alpha $ which implies $ P(| \bar{X}_{n} | <=K | H_{0}) = 1-\alpha $. Using this probability, i can possibly make the CI for $\mu$. But if you notice, then this in this probability statement under null hypotheis, the left hand side of inequality does not depend on $\mu$ under null. How am i supposed to form the CI for $\mu$ in this case? $\endgroup$
    – userNoOne
    Feb 5, 2021 at 5:59
  • 1
    $\begingroup$ In your second option, I don't know $\bar{X}$. I think you need either this or a null hypothesis in order to determine where to center your CI. Without having $\mu = 0$ or $\bar{X}$, where would you center your CI? $\endgroup$ Feb 8, 2021 at 12:05

1 Answer 1


So, I found out the correct Approach for this problem. First, I can compute the width of the interval which will be $\frac{2\sigma}{\sqrt{n}} Z_{0.025}$. It will be equal to $1.176$ after taking the value of $Z_{0.025} = 1.96$.

Now, out of the four options given, Only one of the option has this limit. Hence, that would be an answer. I also understood why the hypothesis specification was necessary. It was because usually the rejection region is the complement of the confidence interval which means the value of $\mu$ under the null hypothesis cannot be in the interval. Hence, First two options where we have 0 inside the interval cannot be the option. This is just an additional information. This problem just could have been solved by obtaining width of the interval.


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