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MATH 110 Module 8 Exam
Exam Page 1
Suppose we have independent random samples of size n 1 = 420 and n 2 = 510. The proportions of success in the two samples are p 1 = .38 and p 2 = .43. Find the 99% confidence interval for the difference in the two population proportions.
n 1 =420 n 2 =510 p1=.38 p 2 =.43 z=2.
P 1 - P 2 ± z √p 1 (1 - p 1 ) + p 2 (1 - p 2 )
n 1 n 2
So the interval is ( -.1332566 .0332566 )
Instructor Comments
Suppose we have independent random samples of size n 1 = 420 and n 2 = 510. The proportions of success in the two samples are p 1 = .38 and p 2 = .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).
Very good.
Answer Key
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 (μ 1 - μ 2 ).
b) Can you be 98% confident that there is a difference in the means of the two populations?
a) z=2.
Exam Page 3
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 42 full-time library workers and finds that they re-shelve an average of 113 books per hour with a standard deviation of 8.1 books per hour. The records of 42 part-time library show that they re- shelve an average of 110 books per hour with a standard deviation of 10.1 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?
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:
n 1 =61, n 2 =49, s 1 =6.1, s 2 =5.2, x̄ 1 =39, x̄ 2 =
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.
Z-Score:
- H 0 : u 1 -u 2 =
- H 1 :u 1 -u 2 ≠ - Two tailed test: a=.
- P(Z<z)=.05/2 = .025 P(Z>z)=a/2 = .05/2=.
- z=-1.96 and z=1.
- z=(xbar 1 - xbar 2 ) - - √s 12 + s - √n 1 n
- z = (113- 110) -
Exam Page 4
Consider the following dependent random samples
Observations 1 2 3 4 5 6
x-values 22.0 21.0 18.1 19.6 13.2 17.
y-values 23.1 21.7 18.7 20.7 14.7 16.
a) Determine the difference between each set of points, xi - yi
b) Do hypothesis testing to see if μd < 0 at the α = .05.
H 0 : ud=
H 1 : ud <
n=6 and this is a left tailed test and note for t. 05 = -2.015 for 6-1=
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 the z-score is between -1.96 and 1.96, we do not reject the null hypothesis.
Consider the following dependent random samples
Observations 1 2 3 4 5 6
x-values 22.0 21.0 18.1 19.6 13.2 17.
y-values 23.1 21.7 18.7 20.7 14.7 16.
a) Determine the difference between each set of points, xi - yi
dbar= -1.1 - .7 - .6 - 1.1 - 1.5 + 1.1 = -0.
6
sd=
√ -1.1 -(-0.65))^2 - .7 -(-0.65))^2 - .6 - (-0.65))^2 - 1.1- (-0.65))^2 - 1.5 -(-0.65))^2 + 1.1 (-0.65))^2 =
-6.0 points
Answer Key
y-time (after) 260 277 281 279 260 242 244 250
Find the 95 % confidence interval for mean of the differences, μd.
n=
d= -1.125 sd= 8.
DOF= 8-1=7 t=2.
dbar- t sd <ud < dbar + t sd
√n √n
-1.125- 2.365 8.714 <ud < -1.125+ 2.365 8.
√8 √
-1.125 - 7.31 <ud < -1.125 + 7.
The times, in
y-time (after) 260 277 281 279 260 242 244 250
Find the 95 % confidence interval for mean of the differences, μd.
Note that n=8. We will define , di = xi - yi. After doing the appropriate calculations, we find that
d=-1.125 sd= 8.7413.
When we look at the student’s t chart for 95% confidence (the 95% is found along the bottom row of the chart) and DOF=8-1=7 (the df=7 is found in the leftmost column) we find that t=2.365. Then
-8.434. <ud < 6.
Instructor Comments
Very good.
Answer Key
A new energy drink is supposed to improve a person’s time in the one mile run. 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
