# Confusion in Confidence Interval and Hypothesis Testing

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?

• 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. Commented Feb 4, 2021 at 13:21
• 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 Commented Feb 4, 2021 at 13:49
• 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. Commented Feb 4, 2021 at 15:03
• 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? Commented Feb 5, 2021 at 5:59
• 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? Commented Feb 8, 2021 at 12:05

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.