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 = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.92 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

Solution :

Given that,

= $6.88

= $1.92

n = 41

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( /n)

=1.645 * (1.92 / 41)

E =0.49

At 90% confidence interval estimate of the population mean is,

- E < < + E

6.88 - 0.49 < < 6.88 + 0.49

6.39< < 7.37

($6.39,$7.37 )

lower limit	$6.39
upper limit	$7.37

margin of error = 0.49