Calculate BJT Amplifier amplification
What am I doing wrong?


(1) You're dividing the output peak-to-peak voltage by the input peak voltage.  When you say the input has an amplitude of 0.01V, that is the peak value.  So, your actual voltage gain is half of what you came up with - 76.5 versus 153

(2) I don't believe the factor of 2 for your calculation of \$r_e\$ is appropriate.  If you use the typical

$$r_e = \frac{26mV}{0.48mA} = 54.2 \Omega$$

instead, the calculated gain, ignoring the capacitor impedance, becomes

$$|A_v| \approx \frac{4400}{54.2} = 81.2 $$

which is close to the 76.5 value I believe your actual gain is.

Including the capacitor impedance, the gain is

$$|A_v| = |\frac{4400}{54.2 - j3.18}| = 81.0 $$

so the capacitor impedance is almost insignificant.