Farmers and Watermelon Crops

Question

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 42 farming regions gave a sample mean of x = $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. What is the margin of error? (Round your answers to two decimal places.)

lower limit	$
upper limit	$
margin of error	$

Answer 1

First calculate the Standard Error Of The Mean:

σ_{x-bar =} σ/√n or 1.94/√42 equivalent to 0.299.

The Sample Mean is the center of a Confidence Interval, so half of the interval (90%/2 or 45%) is directly to the right of the Sample Mean and half is directly to the left.

In a Table Of Proportions Of Area Under The Standard Normal Curve, find the area that most closely approximates 0.45 and set z_c equal to the corresponding z-score of 1.64.

Substitute x-bar=6.88 and σ_x-bar =1.94/√42 into x-bar ± z_cσ_x-bar to obtain the upper and lower limit of the Confidence Interval Boundary Formulas. The term z_cσ_x-bar is commonly called the Margin Of Error or E and is here equal to 1.64(1.94)/√42 ≈ 0.49.

(Upper/Lower) Limit equals 6.88 (±) 1.64(1.94)/√42, giving 7.37 and 6.39.

The sample gives 90% confidence that the true population mean of the price per hundred pounds of watermelon lies on the interval between 6.39 and 7.37.

Answer 2

On calculator Stat, Tests, ZInterval, Stat (not Data), enter sigma = 1.94, xbar = 6.88, n = 40, C-Level = 0.9, Calculate.

Lower limit = 6.3755

Upper Limit = 7.3845

Margin of Error = (7.3845-6.3755)/2 = 0.5045