Simple statistics problem, using the critical value method

The health of the bear population in Yellowstone National Park is monitored by periodic measurements taken from anesthetized bears. A sample of 48 bears has a mean weight of 192.7 lb with standard deviation 15.7 lb.

a) At ∝ = .01, can it be concluded that the average weight of a bear in Yellowstone National Park is different from 200 lb? Use the critical value method.

And also...

b) Find the p-value

If you could also explain how you came to the conclusion of your answer that would be much appreciated. Thanks!

H0: μ = 200
Ha: μ ≠ 200
Sample mean = 192.7
Standard deviation = 15.7
Standard error of mean = s / √ n
Standard error of mean = 15.7 / √ 48
SE = 15.7/6.9282
Standard error of mean 2.2661
t = (xbar- μ ) / SE
t = (192.7-200) / 2.2661
t = -3.2214

Critical t with 47 degrees of freedom at .01 level = -2.685
Compare the absolute calculated and critical values.
Calculated value > critical value, so reject the null hypothesis.
It can be concluded that the average weight of a bear in Yellowstone National Park is different from 200 lb.

b)
p-value = P( |t| > 3.2214) = .00232
(requires a statistical calculator or a software program to find the exact p-value)