7. Sampling Distributions & Confidence Intervals: Mean
Confidence Intervals for Population Mean
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You want to purchase one of the new Altima. You randomly select 400 dealerships across the United States and find a mean of $25,000 and sample standard deviation of $2500. Construct and interpret a 94% confidence interval for the true mean price for the new Nissan Altima.
A
(24996.25, 25003.75); We are 94% confident that the true mean price for the new Nissan Altima falls between $24996.25 and $25003.75.
B
(24999.25, 25000.24); We are 94% confident that the true mean price for the new Nissan Altima falls between $24999.25 and $25000.24.
C
(24984.912, 25015.088); We are 94% confident that the true mean price for the new Nissan Altima falls between $24984.912 and $25015.088.
D
(24764.25, 25235.75); We are 94% confident that the true mean price for the new Nissan Altima falls between $24764.25 and $25235.75.
Verified step by step guidance1
Identify the sample mean (\( \bar{x} \)) and sample standard deviation (\( s \)) from the problem. Here, \( \bar{x} = 25000 \) and \( s = 2500 \).
Determine the sample size (\( n \)), which is given as 400 dealerships.
Select the confidence level, which is 94%. This will help you find the critical value (\( z^* \)) from the standard normal distribution table.
Calculate the standard error of the mean (SEM) using the formula: \( \text{SEM} = \frac{s}{\sqrt{n}} \).
Construct the confidence interval using the formula: \( \bar{x} \pm z^* \times \text{SEM} \). This will give you the range within which the true mean price is expected to fall with 94% confidence.
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