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lOMoARcPSD| MATH 110 exam 8 Portage Learning Statistics MATH 110 exam 8 Portage Learning Statistics
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 n 1 = 520 and n 2 = 450. The proportions of success in the two samples are p 1 = .65 and p 2 = .52. Find the 95% confidence interval for the difference in the two population proportions. Answer the following questions:
- Multiple choice: Which equation would you use to solve this problem? A. B.
- MATH 110
C.
D.
- List the values you would insert into that equation.
- State the final answer to the problem Your Answer: I would choose B P 1 = .65 P 2 = .52 n 1 =520 n 2 = 450 .65-.52 +/- 1. . ( 1 − . )
.
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 (μ 1 - μ 2 ). b) Can you be 99% confident that there is a difference in the means of the two populations? Answer the following questions:
- Multiple choice: Which equation would you use to solve this problem? A. B. C.
μ 1 − μ 2 <7.15 Here the whole confidence Interl is positive so we are 99% confident that we there is a different between the mean of the two populations. When we look back at table 6.1, we see that 99% confidence corresponds to z=2.58. If we say that the day shift nurses corresponds to population 1 and the night shift nurses corresponds to population 2, then: n 1 =89, n 2 =70, s 1 =6.3, s 2 =4.2, x 1 =36, x 2 =31 We will use eqn. 8.1: A. b) Since the entire confidence interval is positive, we can be 99% sure that there is a difference in the means of the two populations.
Question 3
10 / 10 pts You may find the following files helpful in
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 head librarian supervises a number of libraries in a large county. He wants to know if full-time library workers and part-time library workers re-shelve books at the same rate. So, he checks the records of 35 full-time library workers and finds that they re-shelve an average of 190 books per hour with a standard deviation of 18.3 books per hour. The records of 35 part-time library show that they re-shelve an average of 180 books per hour with a standard deviation of 14.4 books per hour. Using a level of significance of α=.05, is there enough evidence to indicate a difference in the mean number of books re-shelved by full-time workers compared to part- time workers? Answer the following questions:
- Multiple choice: Which equation would you use to solve this problem? A.
μ 2 = H 1 = μ 1 − μ 2 ≠ 0 Two tailed test P(Zz).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. = 190 − 180 − 0
2
2 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: H 0 : μ 1 - μ 2 = 0 H 1 : μ 1 - μ 2 ≠0. Since this is a two-tailed test, we must find the z that satisfies: P(Z z)=.05/2=.025.
b) Do hypothesis testing to see if μd < 0 at the α = .01. Your Answer: Ho= Md=0. H1: Md< n=6 left tailed test 6-1= 5 t 0.01= 3.37 d^ = Ed1/n = - 10.8 sd = Sq root of E (d-d^)/ n-1 = 5.56 d= d^/Sq /root of n =- 4.77 t 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:
Formula D x^= -1.125-2.37 8.74/8)
