Question

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 40 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 = 0.27 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

Solution-A:

n=40

sigma=1.90

xbar=6.88

Margin of error in excel ,MOE

=CONFIDENCE.NORM(0.1,1.9,40)

=0.494141

=0.49

90% lower limit of mean=xbar-MOE

= 6.88-0.494141

=6.385859

=6.39

90% upper limit of mean=xbar+MOE

= 6.88+0.494141

=7.374141

=7.37

Margin of error=0.49

90% lower limit=6.39

90% upper limit=7.37

Solution-b:

n=(Z*sigma/MOE)^2

=(1.645*1.90/0.27)^2

=134.0021

=134

Required sample size,n=134

n=134

Solution-c:

15 tons=15*2000=30000 pounds

xbar=is 6.88 for 100 pounds

for 30000 pounds

=30000*6.88/100

sd is

1.99 ---100 pounds

?---30000 pounds

=30000*1.99/100

margin of error=1.645*30000*1.99/100

=982.065

=982.07$

90% lower limit=xbar-MOE= 2064-982.065=1081.935=$1081.94

90% upper limit=xbar+MOE= 2064+982.065=3046.065=$3046.07

Margin of error=$982.07

90% lower limit mean=$1081.94

90% upper limit mean=$3046.07

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