Step-by-step explanation:
Given:
population mean, mu = 57 lb
population standard deviation, sigma = 6.3 lb
Sample mean, x = 60.2 lb
Sample size, n= 45
Null hypothesis x = mu (i.e. sample mean is not greater than usual, using a significance level of 0.05)
Here we have a case where population mean and standard deviation are known (given).
We calculate the z-score
z=(x-mu)/(sigma/sqrt(n))
= (60.2-57)/(6.3/sqrt(45))
= 3.41
P(x>mu) = P(z>3.41) = 0.99968 >> 0.95 (one sided tail)
Therefore we can reject the null hypothesis and accept the alternate hypothesis that fish are larger than usual.