In the absence of auxiliary information, I recommend simple random sampling. It is safe to ignore the fact that you are sampling from a finite population, since the population size $N = 200,000$ is huge, compared to the sample size $n$. Essentially we assume that draws are independent.
Each call can be classified into k = 7 categories. With the simple random sampling design, the number of appearances $a_1, a_2, \ldots a_k$ follow a multinomial distribution, with true proportions $\pi_j$, $j = 1\ldots k$. These will be estimated by the sample proportions $p_j$.
Thompson, 1987, studied sample size determination for estimating a simultaneous $1-\alpha$ confidence set for the $P_j$ with multinomial data. The individual intervals each have the same half-length $2 d$. That is, each interval is of the form:
$$
\left[\,p_j -d,\, p_j +d\,\right]
$$
The analyst chooses $d$ and a protection level $\alpha$, which will be the maximum probability that one or more of the $k$ intervals does not contain its corresponding population proportion; i.e.:
$$
\text{Pr}(|\,P_j - p_j\,| > d\text{, for any j}) \le \alpha.
$$
Equivalently, the probability that all the intervals cover the true $P_j$ is greater than $1-\alpha$.
Thompson proved that the worst case configuration of the $P_j$ in this setup, the one requiring maximum $n$, is one in which the true proportions are equal: $P_j = 1/k$. He provided a simple table for choosing $n$, or, if resources limit $n$, for balancing the choices of $\alpha$ and $d$. These results apply as long as $k$ exceeds a certain minimum value that varied with $\alpha$.
alpha (d^2 x n) min k n if d = 0.05
----------------------------------------
.50 .44129 4 177
.40 .50729 4 203
.30 .60123 3 241
.20 .74739 3 299
.10 1.00635 3 403
.05 1.27359 3 510
.025 1.55963 2 624
.02 1.65872 2 664
.01 1.96986 2 788
.005 2.28514 2 915
.001 3.02892 2 1212
.0005 3.33530 2 1342
.0001 4.11209 2 1645
You can see that $k=7$ is covered by the table, since 7>4. For each $\alpha$ under consideration, choose $d$, then solve for $n$ by dividing the value in column 2 by $d^2$. Alternatively,if your resources or time limit you to a maximum sample size $n_{\text{max}}$, use column 2 solve for $d$. If you were to choose $\alpha =0.20$ and $d = 0.05$, for example, the corresponding value of $n = 0.74739/0.05^2 = 298.954$, rounded up to 299 in column 4.
Other methods for computing simultaneous confidence intervals
Sison and Glaz (1995) proposed two methods for finding simultaneous confidence intervals and recommended their Method 1. May and Williams (2000) published a SAS macro to calculate a version of Sison and Glaz's Method 1. Hou et al. (2003) presented still another method.
References
Hou, Chia-Ding, Jengtung Chiang, and John Jen Tai. 2003. A family of simultaneous confidence intervals for multinomial proportions. Computational Statistics & Data Analysis 43, no. 1: 29-45.
May, Warren L, and William D Johnson. 2000. Constructing two-sided simultaneous confidence intervals for multinomial proportions for small counts in a large number of cells. Journal of Statistical Software 5, no. 6: 1-24.
preprint: http://www.jstatsoft.org/v05/i06/paper
Sison, Cristina P, and Joseph Glaz. 1995. Simultaneous confidence intervals and sample size determination for multinomial proportions. Journal of the American Statistical Association 90, no. 429: 366-369.
http://140.112.142.232/~purplewoo/Literature/!Methodology/!Distribution_SampleSize/SimultConfidIntervJASA.pdf
Thompson, Steven K. 1987. Sample size for estimating multinomial proportions. The American Statistician 41, no. 1: 42-46.
