Question

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 43 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 = 0.45 for the mean price per 100 pounds of watermelon. (Round up to the nearest whole number.)


(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

Solution :

Given that,

= $6.88

= $1.94

n = 43

a ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* (/n)

= 1.645 * (1.94 / 43 )

= 0.49

At 90% confidence interval estimate of the population mean is,

- E < < + E

6.88 - 0.49 < < 6.88 + 0.49

6.37 < < 7.39

Lower limit= 6.37

Upper limit = 7.39

c ) margin of error = E = 0.45

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Sample size = n = ((Z_/2 * ) / E)²

= ((1.645 * 1.94) /0.45)²

= 50.26

= 50

Sample size = 50

c ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* (/n)

= 1.645 * (1.94 / 43 )

= 0.49

Margin of error = E =0.49 *2000 TON = 980

At 90% confidence interval estimate of the population mean is,

- E < < + E

6.88 - 0.50 < < 6.88 + 0.50

6.38 < < 7.38

Lower limit= 6.37 * 2000 =12740 = 12.74 TON

Upper limit = 7.39* 200 = 14780 = 14.78 TON