Question

A larger orchard has averaged 140 pounds of apples per tree per year. A new fertilizer is tested to try to increase yield. Forty trees are randomly selected, and the mean and standard deviation of the yield are I= 143.2 and 8 9.4. Do the data indicate a significant increase in yield? Use a .05. OH :p < 140 vs. H : > 140 Test Statistic z = 2.1530 > Critical z = 1.645 We can reject the null hypothesis, and conclude that there is a sufficient evidence that > 140. OH : u < 140 vs. Hy :u > 140 . Test Statistic z 0.3404 < Critical z = 1.645 We can reject the null hypothesis, and conclude that there is a sufficient evidence that u > 140. OH, u < 140 vs. H : > 140 Test Statistic z = 2.1530 > Critical z = 1.645 We fail to reject the null hypothesis, and conclude that there is not a sufficient evidence that u > 140. OH : M S 140 vs. H: > 140 Test Statistic 2 = 0.3404 Critical z = 1.645 We fail to reject the null hypothesis, and conclude that there is not a sufficient evidence that > 140.

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ <= 140
Alternative Hypothesis, Ha: μ > 140

Rejection Region
This is right tailed test, for α = 0.05
Critical value of z is 1.645.
Hence reject H0 if z > 1.645

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (143.2 - 140)/(9.4/sqrt(40))
z = 2.153

Reject the null hypothesis

Option A) is conclusion