Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Answer:
Following are the answer to the given question:
Step-by-step explanation:
Following are the null and alternative Hypothesis:
[tex]Null \ Hypothesis: \mu = 4\\\\Alternative \ Hypothesis: \mu < 4[/tex]
Rejection Zone
This is left tailed test, for α = 0.2 and df = 39
[tex]critical\ value(t) = -0.851\\\\reject\ H_0 \ when = t < -0.851[/tex]
Testing statistic:
[tex]t = \frac{(\bar{x} - \mu)}{( \frac{s}{\sqrt{n}})}[/tex]
[tex]= \frac{1.5 - 4}{\frac{5}{\sqrt{40}}}\\\\= \frac{-2.5}{\frac{5}{6.32}}\\\\= \frac{-2.5}{0.79}\\\\ = -3.162[/tex]
[tex]P-value = 0.0015[/tex]
P-value < 0.2, reject the null hypothesis.
Approach to the Rejection Zone:
The null hypothesis is rejected since the test statistic t is outside of the critical range value. Only at the 0.200 significance mark, there's significant evidence that its new shrub's growth rate is less than 4 cm per week.
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.