(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.)

This question is incomplete, however, It can be found on search engines. The complete question is as follows :

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 s 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.

Step-by-step explanation:

(c) Given 1 ton = 2,000 pounds

Therefore, 15 ton = 30,000 pounds

Mean of x = 6.88 per 100 pound

Standard deviation = 1.94 per 100 pound

Mean price would be

Standard deviation in price would be

n = 37

For 90% confidence interval,

The margin of error is :

= 157.38

the 90% confidence interval is

or ($1906.62, $2221.38)

the discriminant is 64 + 4*1*12 = 112

this is prime.

step-by-step explanation: