Self assessment exam B - Confidence Intervals - answers Name: __________________________ 1. True-false test. _True_____ For 10 degrees of freedom, P(t >-1.5) = 0.9177463368 _False_____ s is an unbiased estimator of . _True_____ For 10 degrees of freedom, t 0.05 1.812461102 __False____ z 0.05 2.576 __True____ pˆ is an unbiased estimator of p __True____ For 10 degrees of freedom the ordinate of the t-distribution for t=1.35 is 0.1549. 2. Complete: a) According to the _Central Limit Theorem__________ as n increases the sampling distribution of the mean approaches a normal distribution. b) In a sample of size 26 if 5 conditions must be satisfied, then we have ___21____ degrees of freedom. c) If samples of size n=9 are taken from a population with mean µ =12 and standard deviation then the sampling distribution of the mean has a mean x = 12_____ and a standard deviation x = 4____________. d) As n increases the student’s t distribution approaches the standard normal ___distribution (the z-distribution). e) As n increases the standard error of the mean _decreases___________. 3. The times taken to complete a statistics test by all students are normally distributed with a mean of 55 minutes and a standard deviation of 9 minutes. Find the probability that the mean time taken to complete this test by a class of 36 students would be between 50 and 58 minutes. Solution: 55, 9 A sample of n = 36 students is observed. Find P(50 x 58). The distributi on of x has a mean x 55 and a 9 3 s tan dard deviation x 0.5 n 36 6 P(50 x 58) Normcdf (50, 58, 55, 0.5) 0.99999999 -1- 4. According to the Centers for Disease Control and Prevention, the average length of stay that patients to nonfederal short-stay hospitals In the United States was 5.2 days. Assume that this average was based on a random sample of 36 such hospitals stays and that the sample standard deviation was 1.8 days. Construct a 98% confidence Interval for the mean length of all such hospital stays. Show the work. a) 0.02 ________ b ) Formula : x z 2 s n ___________________ n 36, x 5.2, s 1.8 z 0.01 invNormal 0.01 2.326347877 c) Work : x z 2 s n 5.2 ( 2.326) 1.8 36 5.2 0.6978 5.2 0.7 Therefore , 4.5 5.9 is a 98% confidence int erval for d )Answer (give answers rounded to one decimal place ) We are 98% confident that the mean length of all such hospital stays is between ___4.5 and 5.9__ days. 5. In problem 4 how large a sample will be needed to find the confidence interval for the mean within 1/2 day, that is, with a maximum error of 0.5 days? E = 0.5 days. Use 1.8 2 z 2 2.236 1.8 2 n E .5 65 Answer : 65 hospitals 6. A company wants to estimate the mean net weight of its Top Taste cereal boxes. A sample of 16 such boxes produced the mean net weight of 31.98 ounces with a standard deviation of 0.26 ounces. Construct a 95% confidence interval for the mean net weight of all Top Taste cereal boxes. Assume that the net weights of all such cereal boxes have a normal distribution. a) 0.05 _________ b ) Formula : x t 2 s n _______ n 16, deg rees of freedom 15, x 31.98, s 0.26 t 0.025 invT (0.025, 15) 2.131449536) c) Work : s 0.26 x t 31.98 2.131 31.98 0.1386 31.98 0.14 n 16 2 d )Answer (give answers rounded to two decimal place ) We are 95% confident that the mean weight of Top Taste cereal boxes is between __31.84 and 32.12___ ounces. -2- 7. It is said that happy and healthy workers are efficient and productive. A company that manufactures exercising machines wanted to know the percentage of large companies that provide on-site health club facilities. A sample of 240 such companies showed that 96 of them provide such facilities on site. Construct a 90% confidence interval for the percentage (p) of all such companies that provide such site. p̂q̂ a ) 0.10 _________ b ) Formula : p̂ z _____________________ n 2 96 0.40, q̂ 0.60 240 z 0.05 invNorm (0.05) 1.645 n 240, x 96, p̂ c) Work : p̂ z 2 p̂q̂ 0.40(0.60) 0.40 1.645 0.40 0.0520 n 240 0.35 p 0.45 d )Answer (give answers rounded to two decimal place ) We are 90% confident that the percentage (p) of all such companies that provide such site is between ____0.35 and 0.45_________________________________ 9. A sample of 50 men studying statistics shows an average grade of 80 points with a standard deviation of 5 points. A sample of 40 women studying statistics shows an average grade of 85 points with a standard deviation of 6 points. Use the given sample to construct a 98% confidence interval for the difference in average grade between all men and women who study statistics. Men : n1 50, x1 80, s1 5; Women : n 2 40, x 2 85, s1 6, 0.02 Z 0.01 invNorm (0.01) 2.326 s12 s 22 25 36 Formula : x1 x 2 z 80 85 2.326 5 2.75 n1 n 2 50 40 2 7.75 1 2 2.25 Answer: We have a 98% confidence that the average grade of mean is below the average of women by somewhere between 7.75 points and 2.25 points. . -3- 10. If measurements of the specific gravity of a metal can be looked upon as a random sample from a normal population with the standard deviation of 0.025 ounces, what Is the probability that the mean of a random sample of size n=16 will be off by at most 0.01 ounces? This means that we want the probability that x 0.01 0.01 x 0.01 x 0.01 0.01 x 0.01 0.025 0.025 16 n 16 0.01 0.01 z 1.6 z 1.6 0.00625 0.00625 P( 1.6 z 1.6) normalcdf ( 1.6, 1.6) 0.8904 answer: 0.8904__________ -4-