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MATH 110 MODULE 8 EXAM PORTAGE LEARNING
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 = 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.
Answer the following questions:
- Multiple choice: Which equation would you use to solve this problem?
A.
B.
C.
D.
- List the values you would insert into that equation.
- State the final answer to the problem
Your Answer:
- The answer is B.
- The values listed are: n1=420, n2=510, p1=.38, p2=.43, and 99% confidence relate to z=2.58, sample size is greater than 30.
- When using the formula above (B):
Therefore the interval will be : .-0.1333,-0.
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:
A.
B.
C.
D.
- List the values you would insert into that equation.
- State the final answer to the problem
Your Answer:
- The unswer is A
- The values are: n1=89; n2=70;s1=6.3;s2=4.2; x1=36;x2=
- while using the formula mentioned above (A): After calculation I found:
90% of confidence interval between two population means:
Correction number(7,1785 shuld read 7.1788)
99% cobfidence is: (2.8218, 7.
since 0 does not lies between 2.8212 the hypothesis will be rejected. However, there is 99% confident that there is a difference in the mean of the two population.
Note to professor: Please undesrtand due to limited time sametimes I was not able to use proper elements for example I used (U1) instead of ( ) thank you for understanding.
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, x̄ 1 =36, x̄ 2 =
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.
Incorrect value for z. This is not a hypothesis test.
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 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.
The null hypothesis will not be rejected becouse the z score falls between -1.645 and 1.
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)=.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.
Question 4
0 / 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.)
Consider the following dependent random samples Observations 1 2 3 4 5 6 x-values 209 220 214 229 200 205 y-values 210 230 230 238 215 219
a) Determine the difference between each set of points, xi - yi
b) Do hypothesis testing to see if μd < 0 at the α = .01.
Your Answer:
a. The diference between each set of points X1-X2 is:
X1-y1: 209-210=-
X2-y2=220-230=-
x3-y3=214-230=-
x4-y4=229-238=-
x5-y5=200-215=-
x6-y6=205-219=-
b. Ho:Ud=U1-U2=0 & H1: U1-U2<0, wheras; d=0.
t= -20.83-0/2.2718 =-4.
t=-4.7671; t table value =3.3649, therefore we reject the hypothesis 0.01 less.
Since we are testing whether or not μd < 0, then our null and alternate hypothesis will be set as follows: H 0 : μd = 0 H 1 : 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.
C.
D.
- List the values you would insert into that equation.
- State the final answer to the problem
Your Answer:
- The answer is: D
- The values are: n=8 d= -1.125 and Sd=8.7413; t-chart of 95$ is: t=2.
3While using the formula above (D):
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
D.
Quiz Score: 37 out of 50
