Homework SourseYou choose 25 advertisements from a large amount of advertis
You choose 25 advertisements, from a large amount of advertisments, for apartments at random and calculate that their mean monthly rent is $508, and the standard deviation is $78.
1. What would the standard error of the mean be?
2. Construct a 95% confidence interval for the mean monthly rent of all advertised one bedroom apartments.
3. Now, conduct a 90% confidence interval, would the margin of error be smaller or larger than the 95%? Explain why.
Solution
1.
SE = s/sqrt(n)
= 78/sqrt(25)
= 15.6 [ANSWER]
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2.
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 = 508
t(alpha/2) = critical t for the confidence interval = 2.063898562
s = sample standard deviation = 78
n = sample size = 25
df = n - 1 = 24
Thus,
Margin of Error E = 32.19681756
Lower bound = 475.8031824
Upper bound = 540.1968176
Thus, the confidence interval is
( 475.8031824 , 540.1968176 ) [ANSWER]
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3.
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 = 0.05
X = sample mean = 508
t(alpha/2) = critical t for the confidence interval = 1.71088208
s = sample standard deviation = 78
n = sample size = 25
df = n - 1 = 24
Thus,
Margin of Error E = 26.68976045
Lower bound = 481.3102396
Upper bound = 534.6897604
Thus, the confidence interval is
( 481.3102396 , 534.6897604 ) [ANSWER]
As we can see, this has a SMALLER MARGIN OF ERROR than the 95% confidence inetrval. This is so because larger confidence levels give a higher critical t score, which increase the margin of error.
