MATH 110 Exam 8 Statistics Questions and Answers/ 100%Verified/ Portage Learning

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Exam Page 1 Suppose we have independent random samples of size n1 = 420 and n2 = 510. The proportions of success in the two samples are p1= .38 and p2 = .43. Find the 99% confidence interval for the... difference in the two population proportions. n1=420 n2=510 p1=.38 p2=.43 z=2.58 P1 - P2 ± z √p1 (1 - p1) + p2 (1 - p2) n1 n2 .38 - .43 ± 2.58 √.38 (1 - .38) + .43 (1 - .43) 420 510 -.05 ± 2.58(.03227) So the interval is ( -.1332566 .0332566 ) Instructor Comments Very good. Answer Key Suppose we have independent random samples of size n1 = 420 and n2 = 510. The proportions of success in the two samples are p1= .38 and p2 = .43. Find the 99% confidence interval for the difference in the two population proportions. From table 6.1, we see that 99% confidence corresponds to z=2.58. Notice that the sample sizes are each greater than 30, so we may use eqn. 8.2: So, the interval is (.-.1333,.03326). Exam Page 2 In certain hospital, nurses are required to constantly make rounds to check in on all of the patients. The nursing supervisor would like to know if there is a difference between the number of rounds completed per shift by the nurses on the day shift compared to the nurses on the night shift. So, the nursing supervisor checks the records of 61 day shift nurses and finds that they complete an average (a mean) of 39 rounds per shift with a standard deviation of 6.1 rounds per shift. The nursing supervisor also checks the records of 49 night shift nurses and finds that they complete an average (a mean) of 29 rounds per shift with a standard deviation of 5.2 rounds per shift. a) Find the 98% confidence interval for estimating the difference in the population means (μ1 - μ2). b) Can you be 98% confident that there is a difference in the means of the two populations? a) z=2.33 n1=61 n2=49 xbar1 = 39 xbar2 =29 s1= 6.1 s2= 5.2 ( xbar1 - xbar2) - z √s1 2 + s2 2 <u1 - u2 < ( xbar1 - xbar2) + z √s1 2 + s2 2 n1 n2 n1 n2 ( 39 - 29) - 2.33 √6.1 2 + 5.2 2 <u1 - u2 < ( 39 - 29) + 2.33 √6.1 2 + 5.2 2 61 49 61 49 8.92313 <u1- u2< 12.56259 -2.0 points Instructor Comments You have a math error in your computations. Answer Key In certain hospital, nurses are required to constantly make rounds to check in on all of the patients. The nursing supervisor would like to know if there is a difference between the number of rounds completed per shift by the nurses on the day shift compared to the nurses on the night shift. So, the nursing supervisor checks the records of 61 day shift nurses and finds that they complete an average (a mean) of 39 rounds per shift with a standard deviation of 6.1 rounds per shift. The nursing supervisor also checks the records of 49 night shift nurses and finds that they complete an average (a mean) of 29 rounds per shift with a standard deviation of 5.2 rounds per shift. a) Find the 98% confidence interval for estimating the difference in the population means (�1 - �2). b) Can you be 98% confident that there is a difference in the means of the two populations? When we look back at table 6.1, we see that 98% confidence corresponds to z=2.33. If we say that the day shift nurses corresponds to population 1 and the night shift nurses corresponds to population 2, then: n1=61, n2=49, s1=6.1, s2=5.2, x̄1=39, x̄2=29 a) We will use eqn. 8.1: b) Since the entire confidence interval is positive, we can be 98 % sure that there is a difference in the means of the two populations. [Show More]

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