The lifetime of special light-bulbs is normally distributed. The lifetime of special light-bulbs is normally distributed. A sample of 81 bulbs has produced a mean of 738 hours and a standard deviation of 38.2 hours.  

Test a hypothesis $_0 ∶  = 747.5$ versus $-1 ∶  ≠ 747.5,$ at the significance levels of $_1 = 0.05$ and $_2 = 0.01.$ Find the p-value of the test, and use the p-value to verify your answers.

This is how I approached it but not sure if this was right:

$z=\frac{738-747.5}{38.2/√81} = -2.23.$

p-value = $P(z=-2.23)$ --> (1-0.9871) = 0.01

Reject at 0.05 because it is higher than p-value.
Accept at 0.01 because == p-value.

I am extremely unsure about this so any help/verification would be great. Thank you for reading!
You calculated z correctly, however, I noticed when I calculated, I got z=-2.238, so I'd go round to z=-2.24 (rather than z=-2.23), though it doesn't change any results.

You also calculated the p-value correctly. I don't know how many decimals you're required to give, but it may be useful to give one more (as then you can see $p&gt;\alpha_2$).

When it comes to testing hypotheses, if $p&lt;\alpha$ then you reject the null hypothesis, $H_o$, and accept the alternative hypothesis, $H_1$. 

So, your answers are all correct. Good job.
You calculated z correctly, however, I noticed when I calculated, I got z=-2.238, so I'd go round to z=-2.24 (rather than z=-2.23), though it doesn't change any results.

You also calculated the p-value correctly. I don't know how many decimals you're required to give, but it may be useful to give one more (as then you can see $p&gt;\alpha_2$).

When it comes to testing hypotheses, if $p&lt;\alpha$ then you reject the null hypothesis, $H_o$, and accept the alternative hypothesis, $H_1$. 

So, your answers are all correct. Good job.