What price do farmers get for their watermelon crops? in the third week of july, a random sample of 37 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. assume that ? is known to be $1.94 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.) lower limit $ upper limit $ margin of error $
(b) find the sample size necessary for a 90% confidence level with maximal error of estimate e = 0.25 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 $

(a) 90% confidence interval for the population mean price (per 100 pounds) is ($6.34, $7.42)

Margin of error is $0.54

(b) The sample size is 172

(c) 90% confidence interval for the population mean cash value is ($1902.51, $2225.51)

Step-by-step explanation:

Confidence interval = mean + or - (t×sd)/√n

Mean = $6.88 per 100 pounds, sd = $1.94 per 100 pounds, n = 37, degree of freedom = n-1 = 37-1 = 36

t-value corresponding to 36 degrees of freedom and 90% confidence level is 1.688

(a) Lower limit = $6.88 - (1.688×$1.94)/√37 = $6.88 - $0.54 = $6.34

Upper limit = $6.88 + (1.688×$1.94)/√37 = $6.88 + $0.54 = $7.42

90% confidence interval is ($6.34, $7.42)

Margin of error (E) = (t×sd)/√n = (1.688×$1.94)/√37 = $0.54

(b) √n = (t×sd)/E = (1.688×$1.94)/0.25 = 13.10

n = 13.10^2 = 172

1 ton = 2000 pounds

15 tons = 15×2000 = 3000 pounds = 300 hundred pounds

Mean = $6.88×300 = $2064, sd = $1.94×300 = $582

Lower limit= $2064 - (1.688×$582)/√37 = $2064 - $161.51 = $1902.49

Upper limit= $2064 + (1.688×$582)/√37 = $2064+$161.51 = $2225.51

90% confidence interval is ($1902.49, $2225.51)

Margin of error = (1.688×$582)/√37 = $161.51