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.
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 $P\{Z &gt; 2.17\}\approx 0.015$ 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.
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 $P\{Z &gt; 2.17\}\approx 0.015$ 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.