How to correctly compute p-value for hypothesis test? Let the random variable X represent the amount of money carried by a student in class. To determine μ , the mean amount of money carried by a student, an experiment was conducted in class. The class was divided into four groups, and random samples from each group was taken. From Groups 1 and 2, random samples of size n=5 were taken, while from Groups 3 and 4, random samples of size n=10 were taken. Some sample results for the four groups are shown below.
Group Average
1     46.60
2     6.80
3     16.00
4     13.90

Your instructor is interested in testing the following hypothesis about the mean amount of money carried by a student: 
H0: μ ≥ 28.79 
H1: μ < 28.79 
Assume that the amount of money carried by a student follows a Normal distribution, and s2 = 4929.1. Using the average results for Group 2 from the above table, what is the p-value for this test?
I'm getting the answer as .24, while the correct answer is .2611. What did I do wrong? 
I first calculated z-score as (6.80-28.79)/sqrt(4929.1/5) = -0.7. 
P(z<-0.7) = 0.24
 A: To expand on my comment:
If you know the population standard deviation, you can do a z-test, because $\frac{\bar x - \mu_0}{\sigma/\sqrt n}$ will have a standard normal distribution when $H_0$ is true.
However, normally you don't know the population standard deviation, so you must estimate it from the sample, giving you $\frac{\bar x - \mu_0}{s/\sqrt n}$. Because the sample standard deviation can be larger or smaller than $\sigma$, the distribution of the test statistic is both more sharply peaked and heavier tailed than normal - specifically, it has a Student's $t$ distribution. There are many t-distributions, described by the degrees of freedom. 
The more observations you use to estimate the standard deviation, the more accurately (in a particular sense) it tends to estimate $\sigma$. The degrees of freedom of the t-distribution correspond to the degrees of freedom used in calculating $s$, the estimate of $\sigma$.
Because in your one sample test, you estimated one mean parameter, you lose one degree of freedom, leaving you with $n-1=4$ degrees of freedom in your estimate of $\sigma$, and hence your $t$ distribution under the null hypothesis has 4 degrees of freedom

So you have to find the one-tailed p-value for a $t_4$ statistic of $-0.70037$. In R that comes out like so:
 pt(-0.70037,df=4)
[1] 0.2611462

Some further reading on Wikipedia:
One sample t-test
t-distribution
