Question

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 44 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $2.00 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

Here

a ) . Mean M = 6.88

S.D = 2

N = 44

Hence standard error =

= 2 /√ 44

= 0.3015

For 90% confidence interval

Z = 1.645

Hence the interval is given as ( M - 1.645×S.E , M + 1.645×S.E )

= { 6.88 - (1.645×0.3015) , 6.88+(1.645×0.3015) }

= ( 6.384 , 7.375 )

= ( 6.38 , 7.38)

Margin of error = 1.645× 0.3015

= 0.4959

= 0.496

b) sample size is given as

N = Z² × S.D² / E²

E = 0.41

= 64.39

= 65 (say)

c) .

1 ton = 2000 pounds

When we consider price per 100 pounds the 15 tons is converted to an amount. = 15 × 2000/100

= 15 × 20

= 300

To find answer we just have to multiply the answers of the part a) with 300 hence the answers are given as

90% confidence interval = ( 6.38×300 , 7.38×300 )

= ( 1914 , 2214 )

Marginal error = 0.4959 × 300

= 148.77