92283 (Class 208, 209) EXAM 2 APR. 3, 2014
Class________ Class ID________ Name______________________
1.
Select correct answer for each of the following questions with analysis or simple calculation. (2 5 points)
(A) What do we hope to capture within a confidence interval? a) The unknown confidence level. b) The unknown parameter µ. c) The unknown statistic x . d) The margin of error
(B) Suppose in a test of hypotheses, data are proved to be significant at leve l

= 0.05, but not significant at level

= 0.01. Which of the following is true about the P-value P , associated with this test? a) P > 0.05 b) P < 0.05 c) 0.01 < P < 0.05 d) P < 0.01
(C) You plan to construct a 99% confidence interval for the mean

of a Normal population with (known) standard deviation

. By using a sample size of 400 rather than 100, you can reduce the margin of error by a factor of a) 2 (the new margin of error will be half that of the one based on 100 observations). b) 4 (the new margin of error will be 1/4 that of the one based on 100 observations). c) 16 (the new margin of error will be 1/16 that of the one based on 100 observations). m
2
/m
1
= (Z *

√𝟒𝟎𝟎

(Z *

√𝟏𝟎𝟎

√𝟏𝟎𝟎

√𝟒𝟎𝟎

(D ) Let X be the number of girls in a family of 3 children. What is the probability that the family has either
one or two girls?
X
Probability
0
0.125
1
0.375
2
0.375
3
0.125
Select: a) 0.375 b) 0.375 +0.375 = 0.750
c) 1-0.125 = 0.875 d) 0.500
Using addition rule .
If A and B are disjoint, P(A or B) = P(A) + P(B).
(E) A random phenomenon results in a number in the range 0 to 15. The density curve (continuous probability modal) is shown below. How does the probability P
1
that the number is between 2.5 and 5 compare to the probability P
2
that the number is between 5 and 7.5?
Select the answer: a) P
1
= P
2 b) P
1
> P
2 c) P
1
< P
2 d) not sure
Hint: the probability is an area beneath the density curve defined by x
1
and x
2
1

2 . A medical researcher treats 400 subjects with high cholesterol with a new drug. The average decrease in cholesterol level is = 90 after two months of taking the drug. Assume that the decrease in cholesterol after two months of taking the drug follows a Normal distribution, with unknown mean

and standard deviation

= 30.
(15 points)
(A) A 95% confidence interval for

is:
a) 90 ± 1.96 b) 90 ± 2.94
𝟑𝟎 m = 1.96 x
√𝟒𝟎𝟎
= 1.96 x 1.5 = 2.94
c) 90 ± 3.92 d) 90 ± 58.8
(B) Which of the following would produce a confidence interval with a smaller margin of error than the
95% confidence interval you computed above? a) Give the drug to only 100 subjects rather than 400, since 100 people are easier to manage
and control. b) Give the drug to 500 subjects rather than 400 . c) Compute a 99% confidence interval rather than a 95% confidence interval. The increase in
confidence indicates that we have a better interval
(C) What sample size should be used to estimate μ within ± 7 (margin of error m) with 95% confidence
if σ = 23?
a) n = (1.645∙23/7)
2
= 30 b) n = (1.96∙23/7)
2
= 42 c) n = (1.645∙23/7)
2
= 29 d) n = (1.96∙23/7)
2
= 41
3 . Select correct answer for each of the following questions with simple calculation (15 points)
(A) The population of the scores of all high school seniors that took the SAT-M test last year followed a
Normal distribution, with mean

and standard deviation

= 100. You read a report that says, “On the basis of a SRS of 500 high school seniors that took the SAT-M test this year, a confidence interval for

is 512.00 ± m. If the confidence level is 99%, please calculate the margin of error m .
C.L.= 99% Z*= 2.576
.
M= 2.576 x 100/ √𝟓𝟎𝟎 = 11.52
(B) Suppose we are testing the null hypotheses H
0
:
µ = 20
against H a
:
µ < 20
. Suppose our population is
Normal with standard deviation
σ
=6. A random sample of 9 observations are drawn from the population, and we find the sample mean of these observations is x = 17
B1) The Z test statistic is: b) 4.5. a) 1.5. c) -1.5
Z-test = (17 – 20) / (6/ √𝟗 ) = -3 /2 = -1.5
d) -4.5
B2) The P -value is closest to : a) 0.0668
. b) 0.1336. c) 0.0332. d) 0.3085
2

4. The weights of newborn in the England vary according to the normal distribution with mean 7.0 pounds and stand deviation 1.25 pounds. The government classifies a newborn as having low birth weight if the weight is less than 5.0 pounds . (25 points)
a) What is the probability that a baby chosen at random and the weight More than 5.0 pounds at birth?
Hint : Write out Population distribution N( µ, σ ). Locate X
0 and calculate Z
0
(Z-score), then given the
probability that the question is asked for.
N(7.0, 1.25) Population Normal Distribution
X
0
= 5.0
Z
0
= (5.0 – 7.0) / 1.25 = -1.6
P (Z ≤ -1.6) = 0.0548
= 5.48 %
P (Z ≥ -1.6) = 1 - 0.0548 = 0.9452
= 94.52% b) You chose three babies at random. What is the probability that their average birth weight is more than 5.0 pounds?
σ
Hint : Write out Sampling distribution N( µ,
√𝑛
). Locate X
0 and calculate Z
0
(Z-score), then given the probability that the question asked for.
N(7.0, 1.25/ √𝟑 ) Sampling Mean Normal Distribution
X
0
= 5.0
Z
0
= (5.0 – 7.0) / (1.25 / √𝟑 ) = - 2.77
P (Z ≤ -2.77
) = 0.0028
= 0.28%
P (Z ≥ -2.77
) = 1 - 0.0028 = 0.9972
= 99.72 %
3

Choose 5A or 5B.
5A. Women’s height. The average height of 18-year-old American women is 64.2 inches with standard deviation 2.7 inches. You wonder whether the mean height of this year’s female graduate from your local high school is shorter than the national average. You measure an SRS of 61 female graduates and find that x = 63.0 inches.
(20 points)
5B Suppose the time that it takes a certain large bank to approve a home loan is Normally distributed with mean µ and standard deviation s = 1. The bank advertises that it approve loans in 5 days, on average, but measurements on a random sample of 100 loan applications to this bank gave a mean approval time of x = 5.3 days. Is this evidence that the mean time to approval is actually longer than advertised? To answer this, test the hypotheses (20 + 3 extra points) a) State null hypothesis H
0
and alternative hypothesis H a
H
0
:

= 64.2
H a
:

< 64.2
H
H
0 a
:
:


= 5
> 5 b) Calculate test statistic Z
T x = 63.0, n= 61, σ = 2.7 x = 5.3, n= 100, σ = 1
Z-test = (63.0 – 64.2) /(2.7 / √𝟔𝟏 )
=
−𝟏.𝟐 x7.81 = -3.47
𝟐.𝟕
Z-test = (5.3 – 5.0) /(1 / √𝟏𝟎𝟎 )
=
𝟎.𝟑
𝟎.𝟏
= 3.0
c) Find the p-value
P ( Z ≤ -3.47
) = 0.0003 P ( Z ≤ 3.0
) = 0.9987
P ( Z
≥
3.0
) = 1- 0.9987 = 0.0013
d) Is the test statistical significant at level

= 0.01?
Yes, the test is statistical significant Yes, the test is statistical significant at level

= 0.01 and reject H
0 at level

= 0.01 and reject H
0 because: p-value = 0.0003 < 0.01 because: p-value = 0.0013 < 0.01
4