What price do farmers get for their watermelon crops? In the third week of July,

Question

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 36 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. Assume that is known to be $1.86 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 = 0.25 for the mean price per 100 pounds of watermelon. (Round up to the nearest whole number.) t. 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 $

Explanation / Answer

n=36, mean=6.88, sd=1.86

answer a) 90% confidence interval for sample mean=mean±z(.05)*sd/sqrt(n)

=6.88±1.6445*1.86/sqrt(36)=6.88±0.51=(6.37,7.39)

margin of error=z(.05)*sd/sqrt(n)=1.6445*1.86/sqrt(36)=0.51

answer b) maximal error=0.25 i.e. margin of error=0.25=z(.05)*sd/sqrt(n)

or 1.6445*1.86/sqrt(n)=0.25

or sqrt(n)=1.6445*1.86/0.25=12.24

or n=149.82 (next integer is 150

answer is sample size=150

answer c) 15 tons=15*2000=$30000 then new mean and standard deviation will be as

now mean1=300*mean=300*6.88=2064 (here scale has been changed)

and sd1=300*sd=300*1.86=558

margin of error=z(.05)*sd/sqrt(n)=1.6445*558/sqrt(36)=152.94

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.