2 A farmer is worried about the sweetness of his oranges fro
Solution
As the level of significance is not given, we assume it is 0.05.
Formulating the null and alternative hypotheses,
Ho: u >= 0.027
Ha: u < 0.027
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 = 0.025
uo = hypothesized mean = 0.027
n = sample size = 30
s = standard deviation = 0.003
Thus, z = (X - uo) * sqrt(n) / s = -3.651483717
Also, the p value is
p = 0.000130365
Comparing |z| > 1.645 (or, p < 0.05), we REJECT THE NULL HYPOTHESIS.
Thus, there is significant evidence that the sugar content of the said grove is less than the sugar content of the other groves, at 0.027 lb per orange. [CONCLUSION]
