What price do farmers get for their watermelon crops? In the third week of July,
ID: 3336075 • Letter: W
Question
What price do farmers get for their watermelon crops? In the third week of July, a random sample of 44 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. Assume that is known to be $1.96 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.)
(b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.39 for the mean price per 100 pounds of watermelon. (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. What is the margin of error? Hint: 1 ton is 2000 pounds. (Round your answers to two decimal places.)
Explanation / Answer
TRADITIONAL METHOD
given that,
standard deviation, =1.96
sample mean, x =6.88
population size (n)=44
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 1.96/ sqrt ( 44) )
= 0.295
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 0.295
= 0.486
III.
CI = x ± margin of error
confidence interval = [ 6.88 ± 0.486 ]
= [ 6.394,7.366 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =1.96
sample mean, x =6.88
population size (n)=44
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 6.88 ± Z a/2 ( 1.96/ Sqrt ( 44) ) ]
= [ 6.88 - 1.645 * (0.295) , 6.88 + 1.645 * (0.295) ]
= [ 6.394,7.366 ]
[ANSWER]
Lower = 6.394
Upper = 7.366
Margin of error = 0.486
B.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.1% LOS is = 1.645 ( From Standard Normal Table )
Standard Deviation ( S.D) = 1.96
ME =0.39
n = ( 1.645*1.96/0.39) ^2
= (3.224/0.39 ) ^2
= 68.346 ~ 69
C.
farm brings 15 tons of watermelon to market = 15 * 2000 = 30000 = 300 hudread pounds
mean = 300 * 6.88 = 2064
sd = 300 * 1.96 = 588
n = 44
I.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 88.644
= 145.82
II.
CI = x ± margin of error
confidence interval = [ 2064 ± 145.82 ]
= [ 1918.18,2209.82 ]
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