(a) here sample mean = 6.88 pounds per 100
standard deviation == $ 1.92 pounds per 100
sample size = n = 44
standard error of sample means = se0 = /sqrt(n) = 1.92/sqrt(44) = 0.2895
90% confidence interval = +- Zcritical se0 = 6.88 +- 1.645 * 0.2895
= (6.404 , 7.356)
(b) Here maximum margin of estimate E = 0.41
so margin of error = critical test statistic * standard error of sample mean
0.41 = 1.645 * 1.92/sqrt(n)
sqrt(n) = 59.34 or 60
(c) Here 1 ton = 2000 pounds
total crop = 15 tonnes =30000 pounds
so here expected mean cash value of this croup = 300 * $ 6.88 = $ 2064
standad deviation of population mean cash value = $ 1.92 * 300 = $ 576
Lower limit = 2064 - 1.645 * 576/sqrt(300) = $ 2009.29
Upper limit = 2064 + 1.645 * 576/sqrt(300) = $ 2118.71
margin of error = $ 109.92
Here 90% confidence interval =
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