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MATH110 MODULE 8 EXAM | MATH 110 MODULE 8EXAM
QUESTIONS AND ANSWERS- PORTAGE LEARNING
Suppose we have independent random samples of size n 1
= 420 and n 2
= 510. The
proportions of success in thetwo 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
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:
Answer: B.
So, the interval is (.-0.1333, -0.03326).
90% confidence corresponds to z=1.645.
n 1
=70, n 2
=84, s 1
=4.6, s 2
=5.7, - x̄ 1
=30, x̄- 2
Answer: 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.
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 40 full-timelibrary workers and finds that they
re-shelve an average of 185 books per hour with a standard deviation of 17.1books
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 ofbooks 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.
B.
C.
D.
- List the values you would insert into that equation.
- State the final answer to the problem
H 0 : μ 1 - μ 2
= 0 H
1
: μ 1
- μ 2
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
We now find the z-score:
- Consider the following dependent random samples
Observations 1 2 3 4 5 6
x-values 8.1 7.6 8.3 8.4 7.9 7.
y-values 8.4 8.4 8.5 8.9 8.1 7.
a)
Determine the difference between each set of points, x i
b)
Do hypothesis testing to see if μ d
< 0 at the α = .025.
Since we are testing whether or not μd < 0, then our null and alternate
hypothesis will be set as follows:
H
0
: μ d
= 0 H
1
: μ d
n=6. This is a left-tailed test. Note that for t. 025 =
-2.571 for 6-1 = 5. We find the mean in the usual
way:
The sample standard deviation is given by:
Then using the mean, d = -.4333, and the standard deviation, s d
= .2422, that
we found above:
Since t < t. 025 , we reject the null hypothesis.
- 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) 254 276 276 265 271 273 268 281
y-time (after) 265 269 277 279 266 273 275 279
Find the 95 % confidence interval for mean of the differences, μ d
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
Note that n=8. We will define , d i
= x i
- y i . After doing the appropriate
calculations, we find that d =-2.375s d
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
Answer: D.
