3 Of 266 employees selected randomly from one company, 13.53% of them commute by carpooling. Find the margin of error for a 90% confidence interval for the proportion of all employees of the company who carpool. a. 0.54 b. 0.53 OC. 0.03 O d. 0.43 Q11 Travelocity would like to test if the average roundtrip airfare between New York and London is less than $1,400. A Type I error would occur if Travelocity concludes that the average airfare is a. $1,400 or more when, in reality, the average airfare is less than $1,400. O b. equal to $1,400 when, in reality, the average airfare is not equal $1,400 or more. c. not equal to $1,400 when, in reality, the average airfare is equal to $1,400. O d. less than $1,400 when, in reality, the average airfare is $1,400 or more. A survey of 300 union members in New York State reveals that 112 favor the Republican candidate for governor. Construct the 98% confidence interval for the proportion of all New York State union members who favor the Republican candidate. 0.316 to 0.430 0.301 to 0.445 0.308 to 0.438 O d. 0.304 to 0.442 O a. b. C. Q12 A sample mean of 137, sample standard deviation of 14.2, and sample size of 20, Hop-132, H1 μ 132, a=0.01 are given. Use the one-mean t-test to perform the required hypothesis test about the mean, μ, of the population from which the sample was drawn. Use the P-value approach. Also, assess the strength of the evidence against the null hypothesis. a. Test statistic: t = 1.57. P-value = 0.15.&; Reject the null hypothesis. b. Test statistic: t = 1.57. P-value = 0.15.&; Do not reject null hypothesis. O c. Test statistic: t = 1.57. P-value = 0.05.&; Do not reject null hypothesis. O d. Test statistic: t = 1.57. P-value = 0.05.&; Reject null hypothesis.

In a survey funded by the UW school of medicine, 750 of 1000 adult Seattle residents said they did not believe they could come down with a sexually transmitted infection (STI). Construct a 95% confidence interval estimage of the proportion of adult Seattle residents who don't believe they can contract an STI. (Use a z score of 1.96 for your computations.) (.728, .772) (.723, .777) (.718, .782) (.713, .878) (.665, .835)

A survey of 240 teachers found that two-thirds of those surveyed do not live in Sydney. Can you conclude statistically that the population proportion who do not live in Sydney is less than ​0.75? Use a​ 95% confidence interval. What is the confidence interval and is 0.75 within the interval?

A survey of 2450 golfers showed that 281 of them are left-handed. Construct a 90% confidence interval for the proportion of golfers that are left-handed. Question 10 options: (0.104, 0.125) (0.369, 0.451) (0.100, 0.130) (0.203, 0.293)

Identify the t-score for a 95% confidence interval if the sample size is 15. t =  the answer isnt 2.131 so what is the answer?

A study is conducted by a physical therapist to determine whether there is a difference in the proportion of men and women in the U.S. that participate in regular sustained physical activity. It is found that 22% of menout of 550 and 19.5% of women out of 450 participated in regular sustained physical activity. Construct a 90% confidence interval for the difference in proportions of men and women that participate in regular sustained and state your conclusion.

A recent poll indicated that 42% of men supported raises taxes to fund additional park services.   The poll also found that 64% of women supported such a tax.  The sample consisted of 29 men and 25 women.   Construct a 90% confidence interval around the difference in proportions.

If n=430 and (p-hat) = 0.67, construct a 90% confidence interval. Give your answers to three decimals