7. Sampling Distributions & Confidence Intervals: Mean
Confidence Intervals for Population Mean
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Books get more and more expensive every semester, but the distribution of their prices is always normal. 25 randomly selected students in your school spent, on average $500 with a standard deviation of $50. Construct a 98% confidence interval for the true spending on books.
A
(476.74, 523.26)
B
(499.90, 500.10)
C
(490.20, 509.80)
D
(488.38, 511.62)
Verified step by step guidance1
Identify the sample mean (\(\bar{x}\)), which is $500, and the sample standard deviation (s), which is $50. The sample size (n) is 25.
Determine the confidence level, which is 98%. This implies that the significance level (\(\alpha\)) is 0.02, and the critical value (z) for a two-tailed test can be found using a standard normal distribution table or calculator.
Calculate the standard error of the mean (SEM) using the formula: \(SEM = \frac{s}{\sqrt{n}}\). Substitute the values: \(s = 50\) and \(n = 25\).
Find the margin of error (ME) using the formula: \(ME = z \times SEM\). Use the critical value (z) obtained from the standard normal distribution for 98% confidence.
Construct the confidence interval using the formula: \(\bar{x} \pm ME\). This will give you the lower and upper bounds of the confidence interval for the true mean spending on books.
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