Answer to Question #158846 in Statistics and Probability for rr

Question

A farmer is trying out a planting technique that he hopes will increase the yield on his pea plants. The average number of pods on one of his pea plants is 145 pods with a standard deviation of 100 pods. This year, after trying his newplanting technique, he takes a random sample of 144 his plants and finds the average number of pods to be 147. Do the data provide sufficient evidence, at α=0.05, to conclude that the new planting technique is effective in increasing yield?

Answer 1

H₀: μ≤145

H₁: μ>145

$Z = \frac{\bar{x}-μ}{σ/ \sqrt{n}} \\ = \frac{147-145}{100/ \sqrt{144}} \\ = 0.24$

At α=0.05 the critical value will be 1.64 (By using e-calculator https://www.calculators.org/math/z-critical-value.php).

The value of the test statistics is 0.24 < 1.64. So, we accept the null hypotheses and conclude that the new planting technique is not effective in increasing yield.