7. 121 points BBUnderStat12 7.1.018.ML My Notes Ask Your Teacher What price do f
ID: 3325597 • Letter: 7
Question
7. 121 points BBUnderStat12 7.1.018.ML My Notes Ask Your Teacher What price do farmers get for their watermelon crops? In the third week of July, a random sample of 43 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. ound your answers a) Find a 90% confidence interval for the population mean price per 100 pounds that a mers in his region ge to two decimal places.) lower limit upper limit or their atermelon crop whats the margin o erro 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. (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.) lower limit upper limit margin of errorExplanation / Answer
7.
a.
TRADITIONAL METHOD
given that,
standard deviation, =1.96
sample mean, x =6.88
population size (n)=43
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 1.96/ sqrt ( 43) )
= 0.299
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 * 0.299
= 0.492
III.
CI = x ± margin of error
confidence interval = [ 6.88 ± 0.492 ]
= [ 6.388,7.372 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =1.96
sample mean, x =6.88
population size (n)=43
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 ( 43) ) ]
= [ 6.88 - 1.645 * (0.299) , 6.88 + 1.645 * (0.299) ]
= [ 6.388,7.372 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [6.388 , 7.372 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 6.88
standard error =0.299
z table value = 1.645
margin of error = 0.492$
confidence interval = [ 6.38 , 7.37 ]$ per 100 pounds
b.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.1% LOS is = 1.64 ( From Standard Normal Table )
Standard Deviation ( S.D) = 1.96
ME =0.41
n = ( 1.64*1.96/0.41) ^2
= (3.21/0.41 ) ^2
= 61.47 ~ 62
c.
not sure Answer what to do
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