# Hypothesis test (p-value)

Suppose I conduct a hypothesis test and $H_0: \mu = 7$ and $H_A: \mu \neq 7$. Let's say the $p$-value is 0.03. Would a 95% confidence interval contain the value $7$? I think it would since alpha (which is 0.05) is higher than the $p$-value.

However, the question that is really bothering me is would I have enough information from this hypothesis test (if the only information we have is the $p$-value and hypotheses) to conclude if the value $6$ is contained in this 95% confidence interval?

I don't think I would since I have no information on the sample mean and variance.

-
Maybe ask this on stats.SE instead? – Dilip Sarwate Jan 25 '12 at 2:41
If you decide to take @Dilip up on his suggestion (a reasonable one), it is best to flag your post and ask that it be migrated instead of, e.g., creating another identical copy over there. Cheers. :) – cardinal Jan 25 '12 at 2:56
I think I'll leave it here. It's probably not too complicated. I'm just a bit rusty on the theory behind hypothesis testing. – helpme123321 Jan 25 '12 at 3:01
The main problem at this point is that the question as stated is a bit ill-posed. Is this for a normal model? What procedure are you using to construct the confidence interval? What test was used to obtain the stated $p$-values and how does it relate to the construction of the CI? Do you know the sample size? These are questions to help you begin the thought process and to consider what information would be relevant to even have a chance at arriving at a concrete answer. – cardinal Jan 25 '12 at 3:07
Unfortunately, the only data I have available are the hypotheses and the p-value, which leads me to conclude I don't have enough information to conclude the second assertion I made. The first one, however, I think I am correct about since the p-value is less than 0.05 – helpme123321 Jan 25 '12 at 3:14

This appears to be a two-sided test. Since the $p$-value $0.03$ is the probability of an observation as extreme as or more extreme than the observed sample mean $\hat{\mu}$, then assuming normal distributions and large samples, $\hat{\mu} = 7 \pm 2.17\sigma$, that is, the observed sample mean $\hat{\mu}$ is larger (or smaller) than the mean $7$ under the null hypothesis by $2.17$ standard deviations. Note that $P\{Z > 2.17\}\approx 0.015$ for a standard normal random variable $Z$. But, under the assumption that the data is a large sample of independent identically distributed normal random variables, the sample mean is normal with standard deviation $\sigma$, its observed value $\hat{\mu}$ is the maximim-likelihood estimator for the mean, and the $95\%$ confidence interval for the mean is the interval $(\hat{\mu} - 1.96\sigma, \hat{\mu} + 1.96\sigma)$ and does not include $7 = \hat{\mu} \pm 2.17\sigma$.
Note that this answer makes several assumptions that might not be applicable in this problem. Other assumptions e.g. small samples, non-normal data, different method for construction of the confidence interval, disbelief in frequentist hypothesis testing and confidence intervals, etc. can lead to different conclusions as to whether the confidence interval contains $7$ or not.