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 bar = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.90 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 (in dollars). What is the margin of error (in dollars)? (For each answer, enter a number. 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 = 0.31 for the mean price per 100 pounds of watermelon. (Enter a number. 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 (in dollars). What is the margin of error (in dollars)? Hint: 1 ton is 2000 pounds. (For each answer, enter a number. Round your answers to two decimal places.)

lower limt

upper limit

margin of error

Answer 1

Given,

a random sample of 41 farming regions gave a sample mean of x bar = $6.88 per 100 pounds of watermelon. Assume that is known to be $1.90 per 100 pounds

i.e

Sample size : n= 41

Sample mean : = 6.88

Population Standard deviation : = 1.90

(a)

Formula for Confidence Interval for population Mean When Population Standard deviation is known

for 90% Confidence level = (100-90)/100 = 0.10

/2 = 0.10/2 = 0.05

90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop (in dollars)

Margin of error

Lower limit : 6.39

Upper limit : 7.37

margin of error : 0.49

(b)

Formula for sample size necessary for a confidence interval with given margin of error : E

Given,

E = 0.31

for 90% Confidence level = (100-90)/100 = 0.10

/2 = 0.10/2 = 0.05

Population Standard deviation : = 1.90

Sample Size necessary = 102

(c)

A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop (in dollars)

Let Y be the Population mean cash value per 15 tons of watermelon

Let X be the Population mean cash value 100 pounds of watermelon

1 ton is 2000 pounds

Therefore

15 tons is 15x2000=30000 pounds i.e 300 x 100 pounds

Therefore Y =300 X

90% confidence interval for the population mean cash value of this crop (in dollars) of 15 tons of watermelon

= 300 x 90% confidence interval for the population mean cash value of crop (in dollars) of 100 pounds of watermelon

= 300 * (6.3919,7.3681) = (1917.57 , 2210.43)

Margin of error = 300 * 0.4881 = 146.43

lower limit : 1917.57

upper limit : 2210.43

margin of error: 146.43