A past survey of 1,068,000 students taking a standardized test revealed that 9.6% of the students were planning on studying engineering in college. In a recent survey of 1,4/6,000 SILBBES Ly students were planning to study engineering. Construct a 95% confidence interval for the difference between proportions p, -P, by using the following inequality. Assume the samples are random and independent. A P9, P z+ (d- a) > d- t> The confidence interval is
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- Let n1=100, X1=60, n2=100, and X2=80. a. Determine the zstat, based on the difference p1−p2, and the p-value. b.Determine a conclusion c. Construct a 95% confidence interval estimate of the difference between the two population proportions.11From a random sample of 100 women, 15 have dangerous reading blood sugar levels higher than 300 mg/dL. The 95% confidence interval for the proportion of all women with dangerous reading levels is O a. 0.15+ (1.96) √0.1275 O b. O d. 0.15 0.15 (1.6449) √0.0013 (1.6449) √0.1275 0.15 (1.96) √0.0013
- A past survey of 1,068,000 students taking a standardized test revealed that 9.6% of the students were planning on studying engineering in college. In a recent survey of 1,476,000 students taking the SAT, 9.2% of the students were planning to study engineering. Construct a 90% confidence interval for the difference between proportions p, - P2 by using the following inequality. Assume the samples are random and independent. P191 + Zc P191 P292 P292A data set includes data from student evaluations of courses. The summary statistics are nequals=8585, x overbarxequals=3.423.42, sequals=0.540.54. Use a 0.010.01 significance level to test the claim that the population of student course evaluations has a mean equal to 3.503.50. Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.Suppose a study reported that the average person watched 3.34 hours of television per day. A random sample of 15 people gave the number of hours of television watched per day shown. At the 10% significance level, do the data provide sufficient evidence to conclude that the amount of television watched per day last year by the average person differed from the value reported in the study? (Note: x = 4.400 hours and s= 1.262 hours.) Set up the hypotheses for the one-mean t-test. HoiH Ha: H 2.3 4.0 4.5 4.9 5.0 4.5 3.3 2.0 5.7 5.9 6.2 3.8 5.9 3.9 4.1In 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 past survey of 1,068,000 students taking a standardized test revealed that 9.8% of the students were planning on studying engineering in college. In a recent survey of 1,476,000 students taking the SAT, 9.2% of the students were planning to study engineering. Construct a 90% confidence interval for the difference between proportions p, -p, by using the following inequality. Assume the samples are random and independent. A A P191 P292 + Zc n1 P11 P292The coach of a very popular men’s basketball team claims that the average distance the fans travel to the campus to watch a game is 35 miles. The team members feel otherwise. A sample of 16 fans who travel to games was randomly selected and yielded a mean of M= 36 miles and s= 5 miles. Test the coach’s claim at the 5% (.05) level of significance. one-tailed or two-tailed test: State the hypotheses: df= tα or t value for the critical region = sM = t (test statistic)= Decision:5.A random sample of 362 married couples found that 286 had two or more personality preferences in common. In another random sample of 550 married couples, it was found that only 30 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common. Find a 90% confidence interval for p1 – p2. (Use 3 decimal places.) lower limit upper limitA 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)In a random sample of males, it was found that 29 write with their left hands and 212 do not. In a random sample of females, it was found that 70 write with their left hands and 444 do not. Use a 0.01 significance level to test the claim that the rate of left-handedness among males is less than that among females. Complete parts (a) through (c) below.SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th…StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C…StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. 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