Question

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 41 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. Assume that σ 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 = 0.41 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	$

Answer 1

given data are:-

sample size (n) = 41

sample mean () = 6.88 per 100 pounds

population sd () = 1.94 per 100 pounds.

a). for 90% confidence interval we get:-

[ from z table for 90% confidence interval ]

so, margin of error be:-

the 90% confidence interval be:-

so, the required answers are as follows:-

lower limit	$ 6.38 per 100 pounds
upper limit	$ 7.38 per 100 pounds
margin of error	$ 0.50 per 100 pounds

b).sample size for 90% confidence interval with marginal error (E) = 0.41 per 100 pounds be:-

so, the required sample size = 61

c). 15 tons = (15 *2000) pounds

= 30000 pounds

= 300 * (100 pounds) [ because here all the information are given and calculated on basis of 100 pounds]

so, the margin of error be:-

[ because from a , the margin of error was = $ 0.50 per 100 pounds]

the 90% confidence interval be:-

so, the required answers are as follows:-

lower limit	$ 1914
upper limit	$ 2214
margin of error	$ 150

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