A statistics instructor randomly selected four bags of oranges, each bag labeled

Question

A statistics instructor randomly selected four bags of oranges, each bag labeled 10 pounds, and weighed the bags. They weighed 9.7,9.7,9.4, and 9.4 pounds. 9D Assume that the distribution o weights is No al. Find a 95% confidence interval for the mean weight of all bags of oranges. Use technology for your calculations. Answer parts a and b below. a. Choose the correct interpretation of the confidence interval below and, if necessary, fill in the answer boxes to complete your choice. A. We are 95% confident the population mean is between and 0 B. We are 95% confident that the sample mean is between and ° C. There is a 95% chance that all intervals will be between and O D. The requirements for constructing a confidence interval are not satisfied (Type integers or decimals rounded to the nearest thousandth as needed. Use ascending order)

Explanation / Answer

a.
TRADITIONAL METHOD
given that,
sample mean, x =9.55
standard deviation, s =0.1732
sample size, n =4
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.1732/ sqrt ( 4) )
= 0.087
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 3 d.f is 3.182
margin of error = 3.182 * 0.087
= 0.276
III.
CI = x ± margin of error
confidence interval = [ 9.55 ± 0.276 ]
= [ 9.274 , 9.826 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =9.55
standard deviation, s =0.1732
sample size, n =4
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 3 d.f is 3.182
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 9.55 ± t a/2 ( 0.1732/ Sqrt ( 4) ]
= [ 9.55-(3.182 * 0.087) , 9.55+(3.182 * 0.087) ]
= [ 9.274 , 9.826 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 9.274 , 9.826 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

b.
we are 95% sure that the interval [ 9.274 , 9.826 ] contains the true population mean

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