Suppose we have independent random samples of size n1 = 615 and n2 = 505. The proportions of success in the two samples are p1= .53 and p2 = .45. Find the
... [Show More] 90% 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
B
C
D
2. List the values you would insert into that equation.
3.. State the final answer to the problem.
We see that 90% confidence corresponds to z=1.645. Notice that the sample sizes are each greater than 30, so we may use eqn. B:
So, the interval is (.03065,.12935).
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 70 day shift nurses and finds that they complete an average (a mean) of 30 rounds per shift with a standard deviation of 4.6 rounds per shift. The nursing supervisor also checks the records of 84 night shift nurses and finds that they complete an average (a mean) of 25 rounds per shift with a standard deviation of 5.7 rounds per shift.
a) Find the 90% confidence interval for estimating the difference in the population means (µ1 - µ2).
b) ) Can you be 90% 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
We see that 90% confidence corresponds to z=1.645. If we say that the day shift nurses correspond to population 1 and the night shift nurses corresponds to population 2, then: n1=70, n2=84, s1=4.6, s2=5.7, x̄1=30, x̄2=25
We will use eqn:
A.
b) Since the entire confidence interval is positive, we can be 90 % sure that there is a difference in the means of the two populations.
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 40 full-time library workers and finds that they re-shelve an average of 185 books per hour with a standard deviation of 17.1 books per hour. The records of 40 part-time library show that they re-shelve an average of 190 books per hour with a standard deviation of 9.2 books per hour.
Using a level of significance of α=.10, 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:
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 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)=.1/2=.05. and P(Z > z)=.1/2=.05.
In the standard normal table, z=-1.645 and z=1. 645.We will reject the null hypothesis if the z-score is less than -1.645 or the z-score is greater than 1.645.
We now find the z-score:
Since the z-score is between -1.645 and 1.645, we do not reject the null hypothesis. [Show Less]