Exam 8 Portage Learning Statistics Equation Sheet, Exams of Nursing

Download Exam 8 Portage Learning Statistics Equation Sheet and more Exams Nursing in PDF only on Docsity! Exam 8 Portage Learning Statistics Equation Sheet Question 1 10 / 10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) standard normal table (Links to an external site.) t-table (Links to an external site.) Suppose we have independent random samples of size n1 = 520 and n2 = 450. The proportions of success in the two samples are p1= .65 and p2 = .52. Find the 95% confidence interval for the difference in the two population proportions. Answer the following questions: 1.Multiple choice: Which equation would you use to solve this problem? A. P, - P, + Zz Pi(l=Ps) 5 P2(=Pe) nm nN From table 6.1, we see that 95% confidence corresponds to z=1.96. Notice that the sample sizes are each greater than 30, so we may use eqn. 8.2: B. So, the interval is (.06828,.19172). Question 2 10 / 10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) standard normal table (Links to an external site.) t-table (Links to an external site.) 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 89 day shift nurses and finds that they complete an average (a mean) of 36 rounds per shift with a standard deviation of 6.3 rounds per shift. The nursing supervisor also checks the records of 70 night shift nurses and finds that they complete an average (a mean) of 31 rounds per shift with a standard deviation of 4.2 rounds per shift. a) Find the 99% confidence interval for estimating the difference in the population means (µ1 - µ2). b)Can you be 99% confident that there is a difference in the means of the two populations? Answer the following questions: 1.Multiple choice: Which equation would you use to solve this problem? A. B. C. D. 2. List the values you would insert into that equation. 3.State the final answer to the problem Your Answer: a)A b) n1 =89 n2 = 70 S1 = 6.3 S2 =4.2 X 1^ = 36 X^ 2 = 31 z=2.58 (36-31)-2.58 6.3 2 89 + 4.2 70 < μ 1 − μ 2 < (36-31)+2.58 6.3 2 89 + A. B. C. D. 2. List the values you would insert into that equation. 3.State the final answer to the problem Your Answer: a) I would choose C b) X 1^ = 190 X^ 2 = 180n1 =35 n2 = 35 S1 = 18.3 S2 =14.4 H0 = μ 1 − μ 2 =0 H1 = μ 1 − μ 2 ≠ 0 Two tailed test P(Z<z) = .05/2=.025 and P(Z>z).05/2= .025 Stand Normal T. z= -1.96 and 1.96 so we will reject the Hypo if the z-score is less than -1.96 = 190 − 180 − 0 18.3 2 + 14.4 2 35 35 = 2.54 06 The score is above so we reject the null hypothesis The null hypothesis is that there is no difference between the mean number of books re-shelved by the full-time and part- time workers: H0 : µ1 - µ2 = 0 H1 : µ1 - µ2 ≠0. Since this is a two-tailed test, we must find the z that satisfies: P(Z<z)=.05/2=.025 and P(Z > z)=.05/2=.025. In the standard normal table, z=-1.96 and z=1.96. We will reject the null hypothesis if the z-score is less than -1.96 or the z-score is greater than 1.96. We now find the z-score: Since we are testing whether or not µd < 0, then our null and alternate hypothesis will be set as follows: H0: µd = 0 H1: µd < 0 There are 6 data points. So, n=6. This is a left-tailed test. Note that for t.01 = -3.365 for 6-1 = 5. We find the mean in the usual way: The sample standard deviation is given by: Then using the mean, d = -10.833, and the standard deviation, sd= 5.565, that we found above: Since t < t.01, we reject the null hypothesis. Question 5 10 / 10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet (Links to an external site.) standard normal table (Links to an external site.) t-table (Links to an external site.) A new energy drink is supposed to improve a person’s time in the one mile run. The times, in seconds, of eight runners with and without the drink are given below: Runner 1 2 3 4 5 6 7 8 x-time (before) 267 283 270 265 261 247 250 241 y-time (after) 260 277 281 279 260 242 244 250 Find the 95 % confidence interval for mean of the differences, µd. Answer the following questions: 1.Multiple choice: Which equation would you use to solve this problem? A. B. C. D. 2. List the values you would insert into that equation. 3.State the final answer to the problem Your Answer: 8-1+ 7 95% 2.37 Formula D x^= -1.125-2.37 8.74/8) <md< -1.125+2.37( 8.74/8) -1.125-7.34 <md< -1.125+7.34 Note that n=8. We will define , di = xi - yi. After doing the appropriate calculations, we find that d=-1.125 sd= 8.7413.