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 $1.96 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 = 0.43 for the mean price per 100 pounds of watermelon. (Enter a number. 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 (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 limit $

upper limit $

margin of error $

Answer #1

Answer)

As the population s.d is known here we can use standard normal z table to estimate the answers

A)

Mean = 6.88

S.d = 1.96

N = 41

Critical value z from z table for 90% confidence level is 1.645

Margin of error (MOE) = z*s.d/√n = 1.645*1.96/√41 = 0.50353544308

Margin of error (MOE) = 0.5

Lower limit = 6.88 - 0.5 = 6.38

Upper limit = 6.88 + 0.5 = 7.38

B)

Moe = z*s.d/√n

0.43 = 1.645*1.96/√n

N = 57

C)

1 ton = 2000 pounds

15 ton = 15*2000 pounds

We have calculated the interval for 100 pounds

So we need to multiply that interval by 15*2000/100 = 300

Lower limit = 6.38*300 = 1914

Upper limit = 7.38*300 = 2214

Margin of error (MOE) = 0.5*300 = 150

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that σ is known to be $1.90 per 100 pounds. (a) Find a 90%
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