A random sample of 29 lunch orders at Noodles and Company showed a mean bill of
ID: 3387304 • Letter: A
Question
A random sample of 29 lunch orders at Noodles and Company showed a mean bill of $9.6 with a standard deviation of $5.73. Find the 95 percent confidence interval for the mean bill of all lunch orders. (Round your answers to 4 decimal places.)
The 95% confidence interval is from _________ to ___________
A random sample of 29 lunch orders at Noodles and Company showed a mean bill of $9.6 with a standard deviation of $5.73. Find the 95 percent confidence interval for the mean bill of all lunch orders. (Round your answers to 4 decimal places.)
The 95% confidence interval is from _________ to ___________
Explanation / Answer
Note that
Margin of Error E = t(alpha/2) * s / sqrt(n)
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 9.6
t(alpha/2) = critical t for the confidence interval = 2.048407142
s = sample standard deviation = 5.73
n = sample size = 29
df = n - 1 = 28
Thus,
Margin of Error E = 2.179575434
Lower bound = 7.420424566
Upper bound = 11.77957543
Thus, the confidence interval is
( 7.4204 , 11.7796 ) [ANSWER]
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