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 means the significance level (\( \alpha \)) is 0.02, and the critical value (z*) needs to be found for a two-tailed test.
Look up the z-score that corresponds to a cumulative probability of 0.99 (since 0.01 is split between two tails, 0.01/2 = 0.005 in each tail) in the standard normal distribution table. This z-score is approximately 2.33.
Calculate the standard error (SE) of the mean using the formula: \( SE = \frac{s}{\sqrt{n}} \). Substitute the values: \( SE = \frac{50}{\sqrt{25}} \).
Construct the confidence interval using the formula: \( \bar{x} \pm z^* \times SE \). Substitute the values to find the lower and upper bounds of the confidence interval.
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