In the third week of july, a random sample of 40 farming regionsgave a sample me

Question

In the third week of july, a random sample of 40 farming regionsgave a sample mean of = $6.88 per 100 pounds of watermelon.assume that standard deviation is known to be $1.92 per 100 pounds.
a.) find a 90 % confidence interval (1.645) for the populationmean price (per 100 pounds) that farmers in this region get fortheir watermelon crop. what is the margin of error?
b.) find the sample size necessary for a 90% confidence levelwith maximal error of estimate E= 0.3 for the mean price per100 pounds of watermelon.
a.) find a 90 % confidence interval (1.645) for the populationmean price (per 100 pounds) that farmers in this region get fortheir watermelon crop. what is the margin of error?
b.) find the sample size necessary for a 90% confidence levelwith maximal error of estimate E= 0.3 for the mean price per100 pounds of watermelon.

Explanation / Answer

Given n=40, xbar=6.88, s=1.92 (a) =0.1, Z(0.05)=1.645 (check normal tabe) xbar±Z*s/n -->6.88±1.645*1.92/sqrt(40) --> (6.38, 7.40) (b) Given E=0.3 n=(Z*s/E)^2 = (1.645*1.92/0.3)^2 = 110.8388 Take n=111

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