Answer:
1) Z = -1.8592
2) t = 1.7139
3) p = 0.0062
Explanation:
1) Area: (93.7+100)/2 = 193.7/2 = 96.85
Z = invNorm(0.9685,0,1) = 1.8592, but since the test is one-tailed at the lower tail, Z is actually -1.8592
2) Area: (90+100)/2 = 190/2= 95 where df=n-1=24-1=23
t = invT(0.95,23) = 1.7139
3) Since the population standard deviation is known, our test statistic is z=-2.5, making the p-value for the test p=normalcdf(-1e99,-2.5,0,1) = 0.0062