What price do farmers get for their watermelon crops? In the third week of July,

Question

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 42 farming regions gave a sample mean of X - $6.88 per 100 pounds of watermelon. Assume that is known to be $1.98 per 100 pounds. (a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop, what is the margin of error? (Round your answers to two decimal places.) ower limit upper limit margin of error (b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.45 for the mean price per 100 pounds of watermelon. (Round up to the nearest whole number.) farming regions (c) A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop, what is the margin of error? Hint: 1 ton is 2000 pounds. (Round your answers to two decimal places.) lower limit upper limit margin of error Need Help? ReadMaster t Master It Submit Answer Save Progress Practice Another Version

Explanation / Answer

a)

Lower Limit = Mean - ME = 6.37746

Upper Limit = Mean + ME = 7.38254

ME(Margin of error) = 0.50254

b)

Required sample size = 52

c)

Lower Limit = Mean - ME 2007.59040

Upper Limit = Mean + ME 2120.40960

Margin of error = 56.40960

CI for 90% n 42 mean 6.88 z-value of 90% CI 1.6449 std. dev. 1.98 SE = std.dev./sqrt(n) 0.30552 ME = z*SE 0.50254 Lower Limit = Mean - ME 6.37746 Upper Limit = Mean + ME 7.38254 90% CI (6.3775 , 7.3825 )

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