Chapter 8.1, Problem 17BSC

Binomial Prob. Question: A new teaching method claims to improve student engagement. A survey reveals that 60% of students find this method engaging. If 15 students are randomly selected, what is the probability that: a) Exactly 9 students find the method engaging?b) At least 7 students find the method engaging? (2 points = 1 x 2 answers) Provide answers in the yellow cells

In a survey of 2273 adults, 739 say they believe in UFOS. Construct a 95% confidence interval for the population proportion of adults who believe in UFOs. A 95% confidence interval for the population proportion is ( ☐, ☐ ). (Round to three decimal places as needed.)

Find the minimum sample size n needed to estimate μ for the given values of c, σ, and E. C=0.98, σ 6.7, and E = 2 Assume that a preliminary sample has at least 30 members. n = (Round up to the nearest whole number.)

In a survey of 2193 adults in a recent year, 1233 say they have made a New Year's resolution. Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. The 90% confidence interval for the population proportion p is (Round to three decimal places as needed.) J.D) .

Let p be the population proportion for the following condition. Find the point estimates for p and q. In a survey of 1143 adults from country A, 317 said that they were not confident that the food they eat in country A is safe. The point estimate for p, p, is (Round to three decimal places as needed.) ...

(c) Because logistic regression predicts probabilities of outcomes, observations used to build a logistic regression model need not be independent. A. false: all observations must be independent B. true C. false: only observations with the same outcome need to be independent I ANSWERED: A. false: all observations must be independent.  (This was marked wrong but I have no idea why. Isn't this a basic assumption of logistic regression)

Business discuss

Spam filters are built on principles similar to those used in logistic regression. We fit a probability that each message is spam or not spam. We have several variables for each email. Here are a few: to_multiple=1 if there are multiple recipients, winner=1 if the word 'winner' appears in the subject line, format=1 if the email is poorly formatted, re_subj=1 if "re" appears in the subject line. A logistic model was fit to a dataset with the following output:   Estimate SE Z Pr(>|Z|) (Intercept) -0.8161 0.086 -9.4895 0 to_multiple -2.5651 0.3052 -8.4047 0 winner 1.5801 0.3156 5.0067 0 format -0.1528 0.1136 -1.3451 0.1786 re_subj -2.8401 0.363 -7.824 0 (a) Write down the model using the coefficients from the model fit.log_odds(spam) = -0.8161 + -2.5651 + to_multiple  + 1.5801 winner + -0.1528 format + -2.8401 re_subj(b) Suppose we have an observation where to_multiple=0, winner=1, format=0, and re_subj=0. What is the predicted probability that this message is spam?…

Consider an event X comprised of three outcomes whose probabilities are 9/18​, 1/18​,and 6/18. Compute the probability of the complement of the event. Question content area bottom Part 1 A.1/2 B.2/18 C.16/18  D.16/3