Sample standard deviation is measured to be 121 lets

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Unformatted text preview: tα/2,n−1 s / n ¯ where tα/2,n−1 is t-value with n − 1 degrees of freedom and α/2 upper tail probability. Notice now the margin of error is a function of both x and s ¯ Utku Suleymanoglu (UMich) Interval Estimation 11 / 16 Interval Estimation of Population Mean Example Similar example: Suppose you have a normally distributed population with variance σ 2 unknown. You have a sample of 14 observations, whose sample average is x = 7. Sample ¯ standard deviation is measured to be 1.21. Let’s build a 95% confidence interval for µ. We need to get the margin of error. Let’s calculate tα/2,n−1 is first. α = 0.05, so α/2 = 0.025. n = 14, so n − 1 = 13. Let’s look up the table: √ 1. We have tα/2,n−1 = 2.160. So margin of error is tα/2,n−1 s / n = 2.160 × √21 . It is 14 0.6985. So the confidence interval: (7 − 0.6985, 7 + 0.6985) = (6.3015, 7.6985) Food for thought: suppose the 1.21 was σ not s . What would happen to the width of our interval estimate? Why? Utku Suleymanoglu (UMich) Interval Estimation 12 / 16 Interval Estimation of Population Mean Case 3: Population Not Normally Distributed, σ unknown Above results relied on the population values having a normal distribution. If this is violated, sampling distribution of x cannot be normal or t-distribution. So the confidence ¯ intervals cannot be constructed as described. Luckily, CLT comes to rescue. For large enough n, x is approximately normally ¯ distributed, so we can build confidence intervals. CI for µ, Case 3: Any Population, σ unknown If population has unknown dis...
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