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 = $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 (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 = 0.41 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:

Given:

Sample size = n =

Sample mean = per 100 pounds of watermelon.

Population Standard Deviation = per 100 pounds of watermelon.

Part (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)?

Thus c = confidence level = 90% = 0.90

Formula for confidence interval for mean is:

where

Z_c is z critical value for c = 90% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.90) / 2 = 1.90 / 2 = 0.9500

Look in z table for Area = 0.9500 or its closest area and find corresponding z value.

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus Z_c = 1.645

Thus we get:

Thus

Lower limit =

Lower limit = 6.38

Upper limit =

Upper limit = 7.38

Margin of Error =

Part b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.41 for the mean price per 100 pounds of watermelon.

Thus we have c= 90% confidence level , E = Margin of Error =0.41 and Population Standard Deviation =

Formula for sample size for mean

Part 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)?

We are given that: 1 ton = 2000 pounds

then for 15 tons, total pounds = 15 X 2000 = 30000 pounds.

We know for 100 pounds , mean price is $6.88.

30000 pounds is 300 times 100 pounds

So we have to multiply lower and upper limits of 90% confidence interval obtained in part a)

From part a) we have :

Lower limit = 6.38 and Upper limit = 7.38

Then

Lower limit of the population mean cash value of this crop = 300 X 6.38 = 1914.00

and

Upper limit of the population mean cash value of this crop = 300 X 7.38 = 2214.00

Margin of Error = 300 X 0.50 = 150.00