The owner of a private fishing pond says that the average le
Solution
i)
Formulating the null and alternative hypotheses,
Ho: u >= 30
Ha: u < 30
As we can see, this is a left tailed test.
Thus, getting the critical z, as alpha = 0.05 ,
alpha = 0.05
zcrit = - 1.644853627
Getting the test statistic, as
X = sample mean = 27
uo = hypothesized mean = 30
n = sample size = 30
s = standard deviation = 12
Thus, z = (X - uo) * sqrt(n) / s = -1.369306394
Also, the p value is
p = 0.08545176
As |z| < 1.645 or P > 0.05, we FAIL TO REJECT THE NULL HYPOTHESIS.
Thus, there is no significant evidence that the average fish length is less than 30 cm.
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ii)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 27
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 12
n = sample size = 30
Thus,
Margin of Error E = 4.294065945
Lower bound = 22.70593406
Upper bound = 31.29406594
Thus, the confidence interval is
( 22.70593406 , 31.29406594 )
Comment: 30 cm is inside this interval. Thus, there is no significant evidence that the average fish length is less than 30 cm.

