Q1. Last year, the average yield of soy beans in Iowa was 300 bushels per acre.

Question

Q1. Last year, the average yield of soy beans in Iowa was 300 bushels per acre. Weather conditions were unusual this year and soy bean investors suspected that the yield would be different. A random sample of 50 plots of land selected this year had a mean yield of 311 bushels per acre. From past experience, the population standard deviation is known to be 45.4 pounds. At a significance level of .10, is there evidence that the average yield has changed?

a. Ho: Ha: Should you be using z or t? (Explain.)

b. What is the value of the test statistic? (Show your work.)

c. What is the p-value?

d. Statistical decision: Did you find the evidence the problem was looking for? ***Use the information in the problem to construct a 90% confidence interval estimate for the population mean. (Show work)

e. What is the margin of error of your estimate?

f. Is the H0 mean included in the confidence interval?

g. Explain why this is really giving you the same information as your statistical decision.

Explanation / Answer

a)
H0: mu = 300
Ha: mu not equal to 300

we are using z because sample size is greater than 30
b)
x = 311 , n = 50 , s = 45.4

z = ( x - mean) / (s /sqrt(n))
= ( 311 - 300) / ( 45.4 / sqrt(50))
= 1.7132
c)
p value is calculated using z = 1.7132
p value = 0.0866

d)
There is sufficient evidence that the average yield has changed

z value at 90% CI = 1.645
CI= mean + /- z* ( s/sqrt(n))
= 300 + /- 1.645 * ( 45.4 / sqrt(50))
= (289.438, 310.562)

e)
ME = z* ( s/sqrt(n))
= 1.645 * ( 45.4 / sqrt(50))
= 10.562
f)
Yes , H0 mean is included in CI

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