No. You should use student’s t-test.
I’m assuming the population has a normal distribution. The statistic used in the z-test rely on the population standard deviation, so you should use student’s t-test, wich depends on the sample standard deviation instead.
The statistic for student’s t-test is ##frac{barX – mu_0}{S/sqrtn## , wich has a t distribution with ##n-1## degrees of freedon.
Also note that since the t-distribution converges to a normal distribution when ##n## goes to infinite, if you have a large sample, you can perform t-test similar to the z-test, as the test statistic will have normal distribution.
