Charles Stevens, the owner of Wilmot Orchards, historically had on average of 18
ID: 3131403 • Letter: C
Question
Charles Stevens, the owner of Wilmot Orchards, historically had on average of 185 apples per tree for his MacIntosh variety. He applies a new fertilizer to his crop and from a random sample of 36 trees, the average yield is 199 apples per tree with a standard deviation of 48 apples per tree. From these data, test the claim at 95% confidence that the fertilizer actually improved his crop yield.
a) What are the Null and Alternative Hypotheses?
b) Prepare the PDF and state the Decision/Rejection Rule for this problem
c) Conduct the test
d) State the Decision and Interpretation
e) What is the Pvalue?
f) Using this same information, prepare the 95% confidence interval for this new crop.
g) Does the 95% confidence interval contain the older average yield (i.e.,185 apples per tree) before the application of this new fertilizer?
If the decision of part i) is to reject the Null Hypothesis, why w/could your confidence interval of part l) include the older average yield of 185 apples per tree? Explain.
Explanation / Answer
a)
Formulating the null and alternative hypotheses,
Ho: u <= 185
Ha: u > 185 [ANSWER]
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b)
As n > 30, we use z distribution.
As we can see, this is a right tailed test.
Thus, getting the critical z, as alpha = 0.05 ,
alpha = 0.05
zcrit = + 1.644853627
Hence, we reject Ho when z > 1.6449. [ANSWER]
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c)
Getting the test statistic, as
X = sample mean = 199
uo = hypothesized mean = 185
n = sample size = 36
s = standard deviation = 48
Thus, z = (X - uo) * sqrt(n) / s = 1.75
Hence, as z > 1.6449, we reject Ho. [DECISION]
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d)
Hence, as z > 1.6449, we reject Ho. [DECISION]
Hence, there is significant evidence that the fertilizer actually improved his crop yield at 0.05 level. [CONCLUSION]
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E)
Also, the p value is, as it is right tailed,
p = 0.040059157 [ANSWER]
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