STA 2023 Lecture Notes - Lecture 27: Interval Estimation, Sampling Distribution, General Social Survey
Document Summary
We are 95% confident that the mean dollar amount of all loans at the bank is between . 07 and. . 93: now, find a 90% confidence interval for the mean time required by all college graduates. 900 2. 947 x 120/sqrt(16) = (811. 59, 988. 41) df = 16 1 = 15 t = 2. 947. Interval estimation for binomial data: proportion = number of successes/total sample size, p = population proportion (between 0 and 1, p = sample proportion (unbiased and efficient p-hat) Sampling distribution of (cid:1868) : (cid:862)(cid:373)ea(cid:374)(cid:863) of distri(cid:271)utio(cid:374) (cid:449)ill (cid:271)e p, standard error will be (cid:1868) = (cid:1868)(cid:1869, will be approximately normal for large samples (check (cid:1868) (cid:883)5, (cid:1869) (cid:883)5). (cid:1868) /2 (cid:3043)(cid:3044) (cid:1869) = 1 - (cid:1868) . Note: when n is large, (cid:1868) can approximate p on the standard error. (cid:1868) = x/n. A survey showed that among 785 randomly selected people who completed 4 years of college, 18. 3% smoke.
