n=16

X = 14

sigma = 4

null: mean >= 15

alternative: mean < 15

For part a)

Find the z-score (one-tail) for a p-value of 0.10. On the ti-83, invNorm(0.10) gives -1.2815.

Find the z-score

Compare the z-score to -1. If it's less than -1, that means the observation of 14 minutes is extreme, so you reject the null and accept the alternative.

For part b) Find the P-value of the z-score of -1 and compare it to the significance level.

I hope this helps.