# Question

For large n, the sampling distribution of S is some-times approximated with a normal distribution having the mean σ and the variance σ2/2n (see Exercise 8.28 on page 250). Show that this approximation leads to the following (1 – σ) 100% large- sample confidence interval for σ:

## Answer to relevant Questions

If x1 and x2 are the values of a random sample of size 2 from a population having a uniform density with α = 0 and β = θ, find k so that Is a (1 – α) 100% confidence interval for θ when (a) α ≤ 1/2; (b) α > 1/2. If a sample constitutes an appreciable portion of a population, that is, more than 5 percent of the population according to the rule of thumb given on page 239, the formulas of Theorems 11.1 and 11.2 must be modified by ...Independent random samples of sizes n1 = 16 and n2 = 25 from normal populations with σ1 = 4.8 and σ2 = 3.5 have the means 1 = 18.2 and 2 = 23.4. Find a 90% confidence interval for µ1 – µ2. Among 100 fish caught in a certain lake, 18 were inedible as a result of chemical pollution. Construct a 99% confidence interval for the corresponding true proportion. With reference to Exercise 11.32, construct a 90% confidence interval for the standard deviation of the population sampled, that is, for the percentage of impurities in the given brand of peanut butter.Post your question

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