





















Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
The calculations for the statistical analysis of two independent samples, including the means, standard deviations, and hypothesis testing using z-scores and t-tests. The data is presented in two sets, with each set containing sample sizes, sample means, sample standard deviations, and degrees of freedom.
Typology: Study notes
1 / 29
This page cannot be seen from the preview
Don't miss anything!
Q1..
Hypothesis Test:
1
x = 5.30; 2
x = 5.10; s 1 =2.20; s 2 =2.644.
n 1 =75≥30 and n 2 =80≥30, so CLT applies.
Use
2
2
2
1
2
1
1 2 1 2
0
n
s
n
s
x x
z
, α = 0.05.
H 0 : μ 1 – μ 2 ≤ 0; Ha: μ 1 – μ 2 > 0
Reject H 0 if Z 0 >Zα = Z0.05 = 1.645.
2 2
2
2
2
1
2
1
1 2 1 2
0
n
s
n
s
x x
z
not larger than the mean of population 2.
95% Confidence Interval
Use
2
2
2
1
2
1
1 2 / 2 n
s
n
s
x x Z
1 2
2 2
2
2
2
1
2
1
n
s
n
s
2
2
2
1
2
1
1 2 / 2
n
s
n
s
x x Z
. The 95% for μ 1 - μ 2 is given by (-
0.5639. 0.9639)
α=0.
Q2..
Hypothesis Test:
1
2
1 1 1
2
2
2 2 2
2
2
2
1
, unknown, independent random samples
1
x = 430; 2
x = 405; s 1 =20; s 2 =24.
n 1 =25<30 and n 2 =20<30, so CLT does not apply.
Use
2
2
2
1
2
1
1 2 1 2
0
n
s
n
s
x x
t
, α = 0.05,
2 2 2
2
2
2
2
2
1
2
1
2
1
2
2
2
2
1
2
1
.
2 2
2
2
2
1
2
1
1 2 1 2
0
n
s
n
s
x x
t
equal to the mean of population 2.
95% CI:
2
2
2
1
2
1
1 2 / 2 , *
df
2 1
x x
2 2
2
2
2
1
2
1
n
s
n
s
2
2
2
1
2
1
2 1 / 2
n
s
n
s
x x t
. The 95% for μ 1 - μ 2 is given by (-
38.5278, -11.4722)
α/2=0.
Q4..
a.:
1
2
1 1 1
2
2
2 2 2
2
2
2
1
, unknown , samples independent;
1
x (^) = 2.5, 2
x (^) = 3.5,
2
1
s = 2*2=4,
2
2
s = 1.5*1.5=2.25;
n 1 = 15, n 2 = 15, so CLT doesn’t apply, so you must assume a normal distribution.
2
2
2
1
2
1
1 2 1 2
0
n
s
n
s
x x
t
; α=0.05. df*=int(25.9644)=
H 0 : μ 1 ≤ μ 2 ; Ha: μ 1 > μ 2
Reject H 0 if t 0 > tα,df* = t0.05,25≈ 1.7081.
2
2
2
1
2
1
1 2 1 2
0
n
s
n
s
x x
t
likely to increase sales of DVDs as are on-line discounts at the 5% significance level.
b. 95% CI for μ 1 – μ 2
1. First three points are the same as in part a.
2
2
2
1
2
1
1 2 / 2 , *
df
; α=0.05. df*=int(25.9644)=
1 2
x x
3.
/ 2 , * 0. 025 , 25
df
2
2
2
1
2
1
n
s
n
s
2
2
2
1
2
1
1 2 / 2 ,
n
s
n
s
x x t df
or (-2.3294, 0.3294). The 95% CI for
the difference in population means, μ 1 – μ 2 , is (-2.3294, 0.3294).
α=0.
Q5..
a.:
samples are independent random samples
X 1 = 70, X 2 = 75, p 1 = X 1 /n 1 = 70/100=0.7, p 2 = X 2 /n 2 = 75/100 = 0.
n 1 p 1 = 70>5, n 1 (1-p 1 ) = 30>5, n 2 p 2 = 75>5, n 2 (1-p 2 ) = 25>5, n 1 =100≥ 30, n 2 =100≥ 30 so CLT applies
Use
1 2
1 2 1 2
0
n n
p p
p p
Z
p p
;
1 2
1 2
n n
p p α=0.05.
-Zα = -Z0.05 = -1.
1 2
1 2 1 2
0
n n
p p
p p
Z
p p
proportion 2 at the 5% significance level.
b:
Use ^ ^
2
2 2
1
1 1
1 2 / 2
n
p p
n
p p
p p Z
;
1 2
1 2
n n
p p α=0.05.
1 2
2
2 2
1
1 1
n
p p
n
p p
0. 05 1. 96 * 0. 0630 0. 05 0.
2
2 2
1
1 1
1 2 / 2
n
p p
n
p p
p p Z . The 95% CI for
the difference in the two population proportions is (-0.1736, 0.0736).
Note: The 95% CI for (p 2 – p 1 ) is (-0.0736, 0.1736), which requires changes to fformulas in Steps 1, 2 & 5.
c: The population standard error for the difference in sample proportions is:
2
2 2
1
1 1
n n
. A
CI makes no assumptions about π 1 and π 2 , so estimate them by p 1 and p 2 , respectively. If π 1 = π 2 = π 2 , as in the
“=” part of H 0 , then π 1 (1- π 1 ) = π 2 (1- π 2 ) = π (1- π), so common factoring gives:
α=0.
b. Classical Hypothesis Test:
Use
0
s
x
Z
H 0 : μ ≤ 50 (km); H a : μ > 50 (km);
(^) Reject H 0 if Z 0 > Zα = Z0.10 = 1.
0
s
x
Z
drive on a punctured tire for an average of more than 50km.
c. Hypothesis Test with p-value approach:
Steps 1 and 2 are the same as in b.
drive on a punctured tire for an average of more than 50km.
d. In general, you cannot use the results from a 90% CI to make a conclusion about the engineers’ one-
tailed claim, since the CI is two-tailed and the test is 1-tailed. Though both use the same standard
error, the CI has a smaller critical value.
α = 0.
Q8. Since the random samples come from the same 11 days, the test of difference in means should use paired
samples, since the day of sale is likely to influence volume of sales.
1. Assume sales in the two locations are normally distributed (so the difference is normally distributed),
assume α=0.05, and assume test is H 0 : μA = μB
/ 139 / 11 12. 6364
d d n i ;^ 97,629.654 6
(^222)
2
n
d n d
s
i
d
n=11<30, so CLT does not apply
Use
s n
d
t
d
d
/
0
with α=0.
Location
A
Location
B
Difference
di=Ai-Bi di
2
$444 $233 (^211 )
200 299 -99 9801
(^167 800) -633 400689
300 780 -480 230400
(^478 127 351 )
400 250 150 22500
(^250 340) -90 8100
(^600 370 230 )
(^501 230 271 )
(^350 300 50 )
300 400 -100 10000
∑di=-139 ∑di
2 = 978053
2. H 0 : μA = μB , HA: μA ≠ μB, or, H 0 : μd = 0 , HA: μd ≠ 0 where μd = μA – μB 3. Reject H 0 if t 0 >tα/2, n-2 =t0.025, 10 =2.2281 or t 0 <-tα/2= -2. -
t α/ = - 2.2281 t α/ = 2.
4. -0.
97629.65455/ 11
/
0
s n
d
t
d
d
5. Do not reject H 0 since -tα/2 < t 0 < tα/2 (-2.2281< -0.1341 < 2.2281). Do not reject the claim, at the 5%
level, that average daily sales are the same in the two locations.
α/2=0.
the population average service time for customers is the same in both the drive-through service and the
regular walk-in service, which is the same conclusion as the hypothesis test.
.
Q10..
1. Same as Q9. 2. H 0 : μ 1 ≥ μ 2 , Ha: μ 1 < μ 2 3. Reject H 0 if t 0 < -tα, df = -t0.05,25 = -1.
-t α = -1.
4.
1 2
1 2 1 2
0
n n
s
x x
t
p
5. Do not reject H 0 since t 0 > - tα/2 (-1.5492 > -1.7081). Therefore the evidence suggests you reject the
manager’s claim, at the 5% level, that the drive-through service is more efficient.
b. 95% CI does not change. The answer is the same as in Q9.
c. You can use the (2-sided) CI as a (2-tailed) test on means - both use the same critical values and standard
errors.
Q11..
a.
1
2
1 1 1
2
2
2 2 2
2
2
2
1
unknown, 2 independent random samples.
3. 2 1
x ; 2 2
x ; s 1 =4; s 2 =1.5.
n 1 =16<30 and n 2 =0<30, so CLT does not apply.
2
2
2
1
2
1
2 1 / 2 , n
s
n
s
x x t df
, α = 0.05 , df=int(20.9790)=
1 2
/ 2 , * 0. 025 , 20
df
2
2
2
1
2
1
n
s
n
s
2
2
2
1
2
1
1 2 / 2 ,
n
s
n
s
x x t df The 95% CI for the difference in
population average internet use between boys and girls is (-1.1322, 3.5322).
b.
2
2
2
1
2
1
1 2 1 2
0
α=0.
α/2=0.
Or to < - tα/2 =- 2.
2
2
2
1
2
1
1 2 1 2
0
boys and girls do have the same average daily internet use.
c. In general, you can use the (2-sided) CI to test the (2-tailed) claim on population means since they both use
the same critical value and standard error.
Q12..
Classical Hypothesis Test:
1.^11.^79 1
x , 2. 50 1
2
x , 1. 95 2
,
2
2
2
1
2
1
1 1 1 1
n n
x x
.
Small n 1 =25, n 2 =25, with known distribution and population variances
Use
2
2
2
1
2
1
1 2 1 2
0
n n
x x
and α=0.
2. H 0 : μ 1 - μ 2 ≤ 2 , Ha: μ 1 - μ 2 > 2 (men’s spending is at least women’s + $2) 3. Reject H 0 if Z 0 > Z α = Z 0. =1.
4.
2 2
2
2
2
1
2
1
1 2 1 2
0
n n
x x
Z
5. Do not reject H 0 since Z 0 < Z α (1.2616 < 1.645). Based on the evidence, reject the claim, at the 5% level,
that men spend, on average, at least $2 more for lunch than women at this restaurant.
p-value approach
1. Same as above. 2. Same as above. 3. Reject H 0 if p-value<α = 0.
α=0.
α=0.
Q13..
a. Hypothesis Test:
1. Let X 1 = Number of Engineers in sample who found a job within 6 months ;
X 2 =Number of Computer Scientists in sample who found a job within 6 months.
X 1 & X 2 come from 2 independent random samples each with a Binomial distribution
X 1 =70; X 2 =80,
1
1
1
n
p ,
2
2
2
n
p ,
1 2
1 2
n n
p p
n 1 p 1 =X 1 =70>5; n 1 (1-p 1 )=30>5; n 2 p 2 =X 2 =80>5; n 2 (1-p 2 )=20>5; n 1 =100≥30; n 2 =100≥30; so the CLT applies
and
2
2 2
1
1 1
1 1 1 2
n n
p p
Use
1 2
1 2 1 2
0
n n
p p
p p
Z
p p
, α=0.
2. H 0 : π 2 ≤ π 1 ; Ha: π 2 > π 1 or H 0 : π 2 – π 1 ≤ 0, Ha: π 2 – π 1 > 0 3. Reject H 0 if Z 0 >Zα = Z0.01 =2.
4.
1 2
1 2 1 2
0
n n
p p
p p
Z
p p
5. Since Z 0 < Zα (1.6330 < 2.3264) do not reject H 0. The evidence does not support, at the 1% level, the
claim that a larger proportion of Computer Scientists than Engineers found a job within 6 months of
graduation.
b. 99% CI
2
2 2
1
1 1
2 1 / 2
n
p p
n
p p
p p Z
2 1
p p
3. Zα/2 = Z0.005 = 2.
2
2 2
1
1 1
n
p p
n
p p
2
2 2
1
1 1
2 1 / 2
n
p p
n
p p
p p Z or (-0.0567,
0.2567). The 99% CI for the difference in the population proportion of Computer Scientists to
Engineers that found a job within 6 months (π 2
α=0.
Q14. Answers to numerically equivalent to Q13 part a. and part b., though concluding statements should
refer to Population 1 and Population 2, not to Engineers and Computer Scientists.
c. For p-value = 0.0332 and Ha: π 2 – π 1 >10 ( Ho: π 2 – π 1 ≤10 and Decision Rule : Reject H 0 if p-value < α )
i) α=0.10, α>p-value, so reject H 0 and accept the claim in Ha (π 2 – π 1 >10)
ii) if α=0.05, α>p-value, so reject H 0 and accept the claim in Ha (π 2 – π 1 >10)
iii) if α=0.01, α<p-value, so do not reject H 0 (π 2 – π 1 ≤10)
Q15..
a.
p 2 = Proportion of Men in sample primarily concerned with weight of laptop.
Assume X 1 & X 2 come from 2 independent random samples; both have a Binomial distribution
p 1 =0.70 ; p 2 =59,
1 2
1 1 2 2
n n
pn p n
p p
n 1 p 1 =0.7481 = 336.7>5; n 1 (1-p 1 )=144.3>5; n 2 p 2 =0.59374=220.66>5; n 2 (1-p 2 )=153.34>5; n 1 =481≥30;
n 2 =374≥30; so CLT applies, and
2
2 2
1
1 1
1 1 1 2
n n
p p
2
2 2
1
1 1
2 1 / 2
n
p p
n
p p
p p Z
, α=0.
2 1
p p
6. Zα/2 = Z0.025 = 1.
2
2 2
1
1 1
n
p p
n
p p
2
2 2
1
1 1
2 1 / 2
n
p p
n
p p
p p Z
. The 95% CI
for the difference in the population proportion of women minus men who primarily care about laptop
weight is (0.0455, 0.1745).
b. Test
1. As above. Use formula
1 2
1 2
0
n n
p p
p p
Z
p p
2. H 0 : π 1 ≤ π 2 ; Ha: π 1 > π 2 3. Reject H 0 if Z 0 >Zα = Z0.05 =1.
4.
1 2
1 2
0
n n
p p
p p
Z
p p
α=0.
Q17..
a.
1. Let X 1 = # of government donors out of sample; X 2 = # of private sector donors out of sample
16 1
1
1
1
n
p ; 28 2
2
2
2
n
p ;
1 2
1 2
n n
p p
n 1 p 1 =X 1 =16>5; n 1 (1-p 1 )=24>5; n 2 p 2 =X 2 =28>5; n 2 (1-p 2 )=52>5; n 1 =40≥30; n 2 =80≥30; CLT applies
α=0.05. Both samples are large, n≥30 and np>5 and n(1-p)>5, so CLT applies
1 2
1 2 1 2
0
n n
p p
p p
Z
p p
with α=0.05.
2. H 0 : π 1 ≤ π 2 , Ha: π 1 > π 2 , or H 0 : π 1 - π 2 ≤ 0, Ha: π 1 - π 2 > 0 3. Reject H 0 if Z 0 > Zα = Z0.05 = 1.
4.
1 2
1 2 1 2
0
n n
p p
p p
Z
p p
5. Do not reject H 0 since Z 0 < Zα (0.5358< 1.645). Reject the claim that government employees donate
more than private sector employees.
b. 95% Confidence Interval:
2
2 2
1
1 1
1 2 / 2
n
p p
n
p p
p p Z
1 2
3. Zα/2= Z0.025 = 1.
4.
2
2 2
1
1 1
n
p p
n
p p
5.
2
2 2
1
1 1
1 2 / 2
n
p p
n
p p
p p Z
So the 95% CI for the difference in the proportion of public sector and private sector employees who
donate is (-0.1343, 0.2343).
c. In general, you cannot use a (2-sided) CI to test this (1-sided) hypothesis, since the CI and Hypothesis Test
use different critical values and different standard errors. In this particular example, however, you would reach
the same conclusion since the CI also leads you to accept, at the 5% significance level, that the proportion of
donors is the same in the two groups.
α=0.
Q19..
1..
Given: 2 independent random samples from normal distributions; population variances unknown (and
unequal).
11
73
11
12 3 6 11 3 8 12 8 33 14 13
n
d
d .
10
10
2045 11 * 6. 6364
1
(^222)
2
n
d n d i
d
n=11<30 so CLT does not apply.
Use
n
d
t
d
d
/
0
(^) , α = 0.05.
H 0 : μs = 5; Ha: μs ≠ 5
-2.2281 t α/2, n- =t α/2, 10 =2.2281.
Reject H 0 if t 0 >tα/2 = 2.2281 or t 0 < - tα/2 = -2.
0545 / 11
6364 5
/
0
n
d
t
d
d
difference in the means of the two populations is equal to 5.
This conclusion would not change at the 1% level of significance, since the critical values would be
larger (in absolute value), so the test statistic would still lie between these critical values.
Sample 1 50 47 44 48 40 36 43 46 72 40 54
Sample 2 38 44 38 37 43 44 31 38 39 54 41
di = x 1 – x 2 12 3 6 11 -3 -8 12 8 33 -14 13 ∑di= 73
di
2 144 9 36 121 9 64 144 64 1089 196 169 ∑di
2 = 2045
α/2=0.
Q20.
a.
2009 2010 d i =x 2010
- x 2009 d i
2
11 12 1 1
22 24 2 4
6 4 -2 4
12 10 -2 4
26 22 -4 16
∑di = -5 ∑di
2 = 29
n
d
d
i
6
4
24
4
29 5
4
1 4 4 4 16 5 * 1
1
2 2
2
n
d n d i
d
b. 95% CI:
d 1 ,^6
2 d
n=5, so CLT does not apply.
Use d^ t / 2 , n 1 d / n
2 n d
in movies watched between 2010 and 2009 is (-4.0415. 2.0415)
c. Based on the 95%CI, we cannot conclude movie goers watched the same number of movies, on
average, at the 5% significance level, since: a) a hypothesis test would use the same critical value and
standard error, and b) the CI includes the value “0”, so it is possible there is no average difference in
movies watched in the two years.