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A sample of 125 tobacco smokers who recently completed a new smoking-cessation p

ID: 3251233 • Letter: A

Question

A sample of 125 tobacco smokers who recently completed a new smoking-cessation program were asked to rate the effectiveness of the program on a scale of 1 to 10, with 10 corresponding to "completely effective" and 1 corresponding to "completely ineffective". The average rating was 4.1 and the standard deviation was 4.0. Construct a 95% confidence interval for the mean score. A sample of 400 racing cars showed that 80 of them cost over $700,000. What is the 99% confidence interval for the true proportion of racing cars that cost over $7000,000? Construct a 99% confidence interval for the population standard deviation sigma if a sample of size 11 has standard deviation s = 15. Science fiction novels average 290 pages in length. The average length of 10 randomly chosen novels written by I. M. Wordy was 365 pages in length with a standard deviation of 50. At alpha = 0.05, are Wordy's novels significantly longer than the average science fiction novel? It has been claimed that at UCLA at least 40% of the students live on campus. In a random sample of 250 students, 90 were found to live on campus. Does the evidence support the claim at alpha = 001? A statistician claims that the standard deviation of the weights of firemen is more than 25 pounds. A sample of 20 randomly chosen firemen had a standard deviation of their weights of 26.2 pounds. Assume the variable is normally distributed At alpha = 0.05, what is the critical value chi^2 for this test? A) 10.851 B) 10.117 C) 30.144 D) 31.410 A recent survey report that in a sample of 300 students who attend two-year colleges, 105 work at least 20 hours per week. Additionally, in a sample of 225 students attending private four-year universities, only 20 students work at least 20 hours per week. What test value of the difference between these two population proportions? A) 6.95 B) 7.61 C) 4.18 D) 2.38

Explanation / Answer

Z for 95% confidence interval is 1.96

Confidence interval is given by (Mean- z* sigma/sqrt(n) , Mean+ z* sigma/sqrt(n) ) = 4.1-1.96*4/sqrt(125) ,4.1+1.96*4/sqrt(125) )

=(3.3987, 4.8013)

Please repost other questions individually.

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