Download Extra Midterm 2 Review Solution

Question 1: (Don’t Worry about this question)
It is desired to test the claim that a steady diet of wolfbane will cause a
lycanthrope (werewolf) to lose 10 lbs. over 5 months. A random sample of 49
lycanthropes was taken yielding an average weight loss over 5 months of 12.5
pounds, with S=7 lbs. Identify the critical value suitable for conducting a
two-tail test of the hypothesis at the 2% level.
a.
b.
c.
d.
e.
2.06
2.58
1.96
1.65
2.33
t0.01,48
Question 2:
A type I error is always made when:
a.
b.
c.
d.
The null hypothesis is rejected when it is true
The null hypothesis is not rejected when it is false
The research hypothesis is rejected when it is true
The research hypothesis is not rejected when it is false
Question 3:
The level of significance is:
a.
b.
c.
d.
The probability of rejecting the null hypothesis when the null hypothesis is true
The magnitude of the sample size
Symbolized by the Greek Letter Alpha
A and C
Question 4:
It is desired to test the claim that a steady diet of wolfbane will cause an 18-year-old lycanthrope
werewolf to lose exactly 10 lbs. over 5 months. A random sample of 49 lycanthropes was taken, yielding
an average weight loss over 5 months of 12.5 lbs. with S = 7 lbs. Let Alpha = .02. What is the calculated
statistical value (t value) suitable for testing the above hypothesis?
a.
b.
c.
d.
e.
12.5
7*12.5
2.5
-2.5
.35
t= (12.5-10)/(7/SQRT(49))
Question 5:
The standard deviation of a large population is 20. To test:
H(0): MU = 4 vs. H(A): MU > 4
At level of significant 0.05, a sample of size 100 will be taken. You will reject H(0) if:
a. Sample mean >= 7.3
because a = 0.05, so z= 1.645 = (X – 4)/(20/SQRT(100)). As a result, X= 7.29. If the sample mean >=
7.3, the z-value will be bigger than the z-critical value (or the z-value is in the rejection region)
b. Sample mean >= 7.3 or sample mean <= .8
c. Sample mean >= 7.8 or sample mean <= .2
d. None of the above
Question 6:
Suppose a t-test for the null hypothesis that H(0): MU = 0 vs. H(A): MU differs from 0 is carried out and
we find that t (critical point) = 1.8. The descriptive significant level of the test is
a.
b.
c.
d.
e.
The type 1 error probability of the test
The probability of getting a t-value >= 1.8
The probability of getting a t-value >= 1.8 or <= -1.8
The type 1 error probability of the test
None of these
Question 7
Suppose a random sample of size 25 is selected from a population with mean MU, the value of which is
unknown. The sample statistics are XBAR = 6.4, s=10. Test
H(0): MU = 10
H(A): MU < 10 using ALPHA = .05
Which of the following statements do you know if correct?
a. A type 1 error has been committed.
b. H(0) is rejected
Because the t(calc) = (6.4 – 10)/(10/5) = -1.8 < the t(alpha) = -1.711, we reject the H(0) only.
However, since the true value of MU is not known, we do not know if a type 1 error has been
commited
c. H(0) is not rejected
d. Statements (a) and (b) are correct.
e. None of the above
Question 8
The FMA company has designed a new type of 16 lbs bowling ball. The company knows that the average
man who bowls in a scratch league with the company’s old ball has a bowling average of 155. The
variance of these averages is 100. The company asks a random sample of 100 men bowling in scratch
leagues to bowl for five weeks with their new ball. The mean of bowling averages for these men with
the new ball is 170. There is no reason to believe the variance is any different with the new ball. Test the
null hypothesis that the new ball does not improve a bowler’s average at the 5% level of significance.
H(0): MU <= 155
H(A): MU > 155
ALPHA = 0.05
MU = 155
SIGMA = 10
Z (STATISTIC) = (170-155)/(10/SQRT(100) = 15
Z (Critical, ALPHA = 0.05, ONE TAILED) = 1.645
Since Z(Stat) > Z(Critical), reject H(0). Conclude that the new bowling ball does improve a boelwer’s
average at the 5% level
Question 9
A reading coordinator in a large public school system suspects that the poor readers may test lower in IQ
than children whose reading in satisfactory. He draws a random sample of 30 fifth grade students who
are poor readers. Historically fifth grade students in the school system had an average IQ of 105. The
sample of 30 has XBAR = 101.5 and S = 1.42. Test the appropriate hypothesis at the 5% level.
H(0): MU >= 105
H(A): MU < 105
ALPHA = 0.05
Xbar = 101.5
S = 1.42
t (STATISTIC) = (101.5-105)/(1.42/SQRT(30) = -2.465
t (Critical, ALPHA = 0.05, ONE TAILED, df=29) = -1.699
Since t(Stat) < t(Critical), reject H(0). Conclude that the sample evidence is strong enough to suggest that
poor readers test significantly lower in IQ
Question 10
If we would reject a null hypothesis at the 5% level, we would also reject it at the 1% level
a. True
b. False Because rejecting the H(0) at the 5 % level does not mean to reject the H(0) at 1 % level
Question 11
The decision to use a one-sided or two-sided test is usually made after the data is analyzed.
a. True
b. False (Need to make the decision before analyzing the data)
Question 12
The calculated nitrogen content of pure benzanilide is 7.10%. Five repeat analyzes of “representative”
samples yielded values of 7.11%, 7.08%, 7.06%, 7.06% and 7.04%. Using an ALPHA level of size 5%, can
we conclude that the experimental mean differs from the expected value? Assume that the measured
are approximately normal distributed? (Hint: the sample standard deviation is 0.0265)
a.
b.
c.
d.
Yes, because the null hypothesis is rejected
No, because the null hypothesis is rejected
Yes, because the alternative hypothesis is rejected
No, because the alternative is not accepted
t-statistic = (7.07 – 7.10)/(0.0265/SQRT(5)) = 2.53
t-critical (a/2 = 0.025, df = 4) = 2.776
Since t-statistic < t-critical, the null hypothesis is not rejected
Question 13 (I just fixed the error from the answer c)
Suppose in a sample of 25 people, the mean height XBAR was observed to be 70 inches. Suppose the
population standard deviation (sigma) is 3. Construct a 95% confidence interval for the population mean
MU
a.
b.
c.
d.
(50.5,71.2)
(66.8,80.4)
(68.8,71.2)
(69.7,77.1)
Question 14
Which of the following assumption are needed to test a hypothesis about mean MU in a population with
known Variance using the mean of the data X(1) … X(N) and normal tables:
I.
II.
III.
The data are a random sample
The population distribution is normal
The sample size is large
a.
b.
c.
d.
e.
I, II and III
I and either II or III
II and III
Only II
None of these
Question 15
In testing H(0) : MU <= 20 vs. H(A): MU >20 when ALPHA = 0.05, n= 25, S2 = 16 and Xbar = 22, one should
conclude that:
a. MU > 20 with 5% chance of error
t = 2.5 > 1.711, reject the H(0)
b. MU >20 with 5% confidence
c. MU = 20 with 95% confidence
d. MU = 20 with 95 % chance of error
Question 16
If the P-value for your test statistic satisfies P > 0.25, then:
a.
b.
c.
d.
You would not reject H(0)
You would reject H(0) for ALPHA = .05
You would reject H(0) for ALPHA = .10
None of these
because a = 0.01, 0.05, 0.1 ….
Question 17:
Nine men with a genetic condition that causes obesity entered a weight reduction program. After
four months the statistics of weight loss were: XBAR = 11.2, S = 9.0. The researcher wants to test the
hypothesis: The average four-month weight loss in such a program is <= 6 pounds verses the
alternative: > 6 pounds at a 5% significance level. Given the data of our problem, we
a. Reject the hypothesis
b. Reserve judgement about the hypothesis
Because t (calc) = 1.73 < t(crit) = 1.86, do not reject
Question 18:
A fourth grade teacher wants to try a new teaching method which the authors recommend should
only be used with particularly bright student. The authors offer a short test which they feel can be
used as a guide to decide whether the method should be used. The believe that the average score
for a class on this test should significant exceed 73 and indicate that the national standard deviation
for the test is 10. The teacher’s students take the test and exhibit the following scores: 91, 91, 94,
63, 61, 40, 73, 75, 83, 84, 70, 71, 88, 93, 92, 91, 87, 83, 74, 80.
Should the teacher use the new method? Why? (USE ALPHA = 0.05)
a. Yes
b. No
c. Don’t know
A:
H(0): MU <= 73
H(1): MU > 73
SIGMA = 10
XBAR = 79.2
Z = (79.25 - 73)/(10/SQRT(20)) = 2.79
Z(critical) = 1.645;
with ALPHA = .05 and using a onetail
decision rule.
Reject H(0) and conclude that the average score for this class is
significantly higher than 73. Therefore, she should use the new
method.
Question 19:
A sample of size 16 is taken from a normal population. The 99% confidence interval is set up with
XBAR = 30 and s = 20. The value you would find in the table to completed the information necessary
to obtain the interval would be (Hint: Find the critical value)
a.
b.
c.
d.
e.
2.602
2.326
2.576
2.947
2.921
t(alpha/2 = 0.005, df = 15)
Question 20:
For student’s distribution, 90% of the area lies between t = -1.89 and t= 1.89 if the degrees of
freedom are:
a. 2
b. 3
c. 6
T(alpha/2 = 0.05, df= 7) = 1.895
d. 7
e.8
Question 21:
T-distributions are spread out ------- than a normal distribution with MU = 0 and Sigma = 1
a. More LONGER TAIL
b. Less
c. Equal
Question 22: (Part d, please change the variance from 0 to 1)
A t-distribution with 30 degree of freedom is most similar a ----- distribution
a.
b.
c.
d.
Normal distribution with mean = 1 and variance = 1
Normal distribution with mean = 0 and variance = squared(sigma)
Normal distribution with mean = 0 and variance 29
Normal distribution with mean = 0 and variance = 1
Question 23
A random sample of size 25 is taken from a population with mean 7 and variance 4. The sample
mean is calculated to be 8. What value of the standard normal random variable (Z-score) corresponds to
the sample mean?
a.
b.
c.
d.
25
1.25
– 1.25
2.5 because z = (8-7)/(2/5)
Question 24:
The height of male college freshmen has a normal distribution with mean 71 and standard deviation
3 inches. If 100 male college freshmen are selected at random, and XBAR is the average of their
heights, then P(70.5 < XBAR < 71.3) is approximately:
a.
b.
c.
d.
.0455
.8413
.4525
.7938
Z2 = (71.3-71)/(3/SQRT(100)) = 1
Table
P(X < 71.3) = .8413
Z1 = (70.5-71)/(3/SQRT(100)) = -1.67
Table
P(X<70.5) = .0475
.8413 – 0.0475 = .7938
Question 25 and 26
Suppose that for a sample of 36 Family Nurse Practitioners from several similar type hospital clinics,
a competency score ranging from 0 to 100 was derived based on performance at the clinic. Suppose
further that the population mean competency score for all FNP’s was 80 and the population
variance was 100.
Question 25: For the sample of 36 FNP’s, what is the probability that the mean competency score
will be between 75 and 80?
a. .4987
b. .1915
c. .5013
d. .2287
Sigma = 10 and mu = 80
P(75 < X < 80) = Z(0) – Z(-3) = .5 – 0.0013 = .4987
Question 26: The middle 99% points for the distribution of the sample mean competency score
described above is (rounded to two decimal places):
a.
b.
c.
d.
(55.24, 105.76)
(76.12, 83.88)
(56.74, 103.26)
(75.71, 84.29)
Z(ALPHA/2 = 0.005) = 2.576 (Note: 0.99+0.005 = 0.995, z = 2.56 or 2.57 from the normal table)
80 +/- 2.576 * 10/6 = (75.71, 84.29)
Question 27
Suppose that the weight (W) of male patients registered at a diet clinic has the normal distribution
with mean 190 and variance 100. For a sample of size 25 from the clinic, which of following
statements is equivalent to the statement: P(WBAR > 195)
Where WBAR denotes the mean weight of the sample?
a.
b.
c.
d.
P(Z < -2.5)
P(Z < 2.5)
P(Z > 2.5)
A and C because the probability from the left and right tail are the same (according to the
symmetric case)
Question 28
A random sample of size 100 is selected from a population with MU = 0 and Sigma = 20. Find P(XBAR
> 2)
a.
b.
c.
d.
.0228
.3085
.4207
.1587 z= (2-0)/(20/10)= 1 P(X>2) = 1 – P(X<2) = 1- P(Z<1) = 1- .8413 = .1587
Question 29
A sample of 3600 cases is drawn at random from an infinitely large population. The standard
deviation of the population is 10. The standard error of the mean is
a.
b.
c.
d.
2/15
1/6
4/5
10
10/SQRT(3600)
Question 30
As the sample size (n) increases, the mean of a random sample is less likely to be near the mean of
the population
a. True
b. False BECAUSE THE SAMPLE MEAN IS GETTING CLOSER TO THE POPULATION MEAN AS N
INCREASES
c. Don’t know
Question 31
A sample of twenty-five observations is taken from a normal population with variance 9. 90%
confidence limits corresponding to a sample mean of 30 are best represented by:
a.
b.
c.
d.
30 +/- 9
30 +/- .79
30 +/- 4.935
30 +/- .99
Variance = 9
so
Standard deviation (Sigma) = 3
Z (ALPHA/2 = 0.05) = 1.645 (Note 0.9 + 0.05 = .95 from the table, z = 1.64 or 1.65)
30 +/- 1.645 * 3/SQRT(25) = 30 +/- .987
Question 32 (Change part c to .25 instead of 2.5)
Suppose that a sample of size 16 from a normal distribution with mean MU yielded a sample mean
of 3.2 and a sample standard deviation of 4. For 99% confidence level, the lower confidence limit for
a two sided interval for MU is
a.
b.
c.
d.
0.6
.8
.25
3
T(alpha/2 = 0.005, df = 15) = 2.947
3.2 – 2.947*4/SQRT(16) = .253
Question 33
To find the confidence intervals for the mean of a normal distribution, the t distribution is usually
used instead of the standard deviation because
a.
b.
c.
d.
The mean of the population is not known
The t distribution is more efficient
The variance of the population is usually not known
The sample mean is known
Question 34
The size of a confidence interval for a mean is affected by changes in which of the following?
a.
b.
c.
d.
e.
The size of the sample
The confidence coefficient
The variance of the sample
B and c only
A, b and c
Question 35
Breaking strengths (load at breaking point) in lbs. of two brands of fishing line. We take 8
measurements from each brand. Assume the population standard deviation of Trilene X2 (sample mean
= 10.525) and Eagle Claw (sample mean = 8.050) are 0.4 and .6 respectively. Assume normality. Does
Trilene X2 have a higher mean Strength than Eagle Claw? Using a = .01
Key:
1/ the population standard deviation is known
2/ Using 8 samples from each brand. (Meaning the pool independent normal hypothesis test because
we analyze 8 Trilene brands to compare with 8 Eagle brands)
Let u1 = the population mean of Trilene X2
Let u2 = the population mean of Eagle Strength
Step 1:
H(0) : u1 - u2 = 0
H(0) : u1 - u2 > 0
Step 2:
Z(Stat) = ((10.525 – 8.050) – 0) / (SQRT(.42/8 + .62/8)) = 9.71
Z(Critical, alpha = 0.01, one tailed) = 2.33
Because Z(Stat) > Z(Critical), we reject the null hypothesis. There is a strong evidence that Trilene X2 is
stronger than E