Description

Patrick Gardocki
9-20
a)
α = P(X¯ ≤ 4.85) + P(X >¯ 5.15) = P(X¯0√.−255 ≤ 4.085√.25−5) + P(X¯0√.−255
8 8 8

> 5.015√.25−5) = P(Z ≤ −1.7) + P(Z > 1.7) =

8

0.04457 + (1 − 0.95543) = 0.08913
pnorm(-1.7)+(1-pnorm(1.7))
## [1] 0.08913093
b)
Power
P(Z ≤ −2.83) = 0.71566 − 0.00233 = 0.7119
pnorm(0.566)-(pnorm(-2.83))
## [1] 0.7119757
9-23
a) x¯ = 5.2
zo
P-value=2(1 − Φ(|2.26|)) = 2(1 − 0.988) = 0.0238
2*(1-pnorm(2.26))
## [1] 0.02382125

P(−2.83 ≤ Z ≤ 0.566) = P(Z ≤ 0.566) −
b) x¯ = 4.7
zo
P-value=2(1 − Φ(| − 3.39|)) = 2(1 − 0.999) = 0.00069
2*(1-pnorm(3.39))
## [1] 0.0006989262
c) x¯ = 5.1
zo
P-value=2(1 − Φ(|1.131|)) = 2(1 − 0.870) = 0.258
2*(1-pnorm(1.131))
## [1] 0.2580551 9-43(a)(b)
a)
if z0 < −zα/2 or z0 > zα/2, reject Ho. α = 0.05, σ = 0.9 zo
− z0.025 = −1.96 < −0.95 ∴ fails to reject null hypothesis
b)
z0.
9-64 (a)(b)(e)!!!!!!!!!!!!!!!!!!!!!!!!
a)
if t0 > tα,n−1, reject Ho. α = 0.01, t0.01,19 = 2.539, x¯ = 26.04, s = 4.78 to
t0.01,19 = 2.539 > 0.97 ∴ fails to reject null hypothesis. Insufficient evidence to show that rainfall is greater than 25 acre-feet.
!!!!!!!!!!!!!!!!!!!!!!P-value=2(1 − Φ(|2.539|)) = 8
b)
data = c(18,30.7,19.8,27.1,22.3,18.8,31.8,23.4,21.2,27.9,31.9,27.1,25.0,24.7,26.9,21.8, qqnorm(data) qqline(data)
29.2,34.8,26.7,31
Normal Q−Q Plot

Theoretical Quantiles
The plot suggests that the data follows a normal distribution.
e)
µ → 23.326 ≤ µ Since the lower limit is less than the mean diameter, no evidence that
true mean rainfall is greater.
qt(.01,19)
## [1] -2.539483 9-98
a)
if z0 < −zα/2 or z0 > zα/2, reject Ho. α = 0.05, x = 117, n = 484 zo
−z0.025 = −1.96 > −11.36 ∴ null hypothesis is rejected and proportion of students planning graduate studies is not 0.5 with α = 0.05. P-value = 2(1 − Φ(11.36)) ≈ 0
b)
Using a two-sided confidence interval, if the confidence interval does not contain 0.5, then the proportion of students planning graduate studies is not 0.5 with α = 0.05.
pˆ
q q
0.2420.280
10-4(a)(b)(c)
a)
Ho : µ1 − µ2 = 0
H1 : µ1 − µ2 ̸= 0
Reject Ho if zo < −zα/2 or zo > zα/2 σ1 = 0.02, σ2 = 0.025, x¯1 = 16.015, x¯2 = 16.005
zo → −1.96 < 0.702 < 1.96 ∴ fail to reject null hypothesis. There is no good proof
that the machines fill volumes are different. P-Value= 2(1 − Φ(0.702)) = 0.4826
test1= c(16.03, 16.01, 16.04, 15.96, 16.05, 15.98, 16.05, 16.02, 16.02, 15.99) test2= c( 16.02, 16.03,15.97, 16.04, 15.96, 16.02,16.01, 16.01, 15.99, 16.00) mean(test1)
## [1] 16.015
mean(test2)
## [1] 16.005
b)
q q
(16.015→ −0.01789 ≤ µ1 − µ2 ≤ 0.03789 Since the interval contains 0, there is no difference between the means with a 95% confidence interval.
c)

β = Φ 1.96 − p 0. 0.04 0. −Φ −1.96 − p 0. 0.04 0. = Φ(1.96−3.95)−Φ(−1.96−3.95) = 0.02329
+ +
10-14
a)
q sp =
to
Degree of Freedom= n1 +n2 −2 = 12+16−2 = 26 P-value: 2(P(t > 1.842)) = 0.077 → 2(0.025) < 0.077 <
q
2(0.05) Pooled Standard Deviation: 1. This is a two sided test because Ho : µ1 − µ2 = 0 b)
Since 2(0.025) < 0.077 < 2(0.05), fail to reject the null hypothesis for α = 0.05 and 0.01. c)
The sample standard deviations are only slightly different so it can be assumed that the sample variance is also similar.
d)
P-value: P(t < −1.842) = 0.0326, 0.025 < 0.0326 < 0.05 ∴ P-value is less than α = 0.05, and null hypothesis is rejected.
pnorm(-1.8428)
## [1] 0.03267911 10-24
a)
H0 : µ1 − µ2 = 0 H1 : µ1 − µ2 < 0
v
t0.05,23 = 1.714, x¯1 = 290, x¯2 = 321, s1 = 12, s2 = 22
t
Null Hypothesis is rejected if to < −tα,v. Since -4.64 < -1.714, null hypothesis is rejected and supplier 2 makes gears with higher mean impact strength.
b)
H0 : µ1 − µ2 = 25 H1 : µ1 − µ>25 v =≈ 23
t0.05,23 = 1.714, x¯1 = 290, x¯2 = 321, s1 = 12, s2 = 22, ∆0 = 25
t0 = (321 p−12290)2 22−225 = 0.898
10 + 16
Null Hypothesis is rejected if to > tα,v. Since 0.898 < 1.714, fail to reject null hypothesis and no evidence that supplier 2 makes gears with mean impact strength 25 ft-lb higher than supplier 1. c)
t0.025,25 = 2.0598
q q
17.242 ≤ µ2 − µ1 ≤ 44.758
Interval does not contain 0, this means that supplier 2 makes gears with higher impact strength.
10-52
a)
data = c(-1,2,2,3,-5,3,6,1,2,-1,-2,-2) qqnorm(data) qqline(data)
Normal Q−Q Plot

Theoretical Quantiles
mean(data)
## [1] 0.6666667
sd(data)
## [1] 2.964436
qt(.975,11)
## [1] 2.200985
x = c(17,16,21,14,18,24,16,14,21,23,13,18) y= c(18,14,19,11,23,21,10,13,19,24,15,20)
t.test(x, y, paired = TRUE, alternative = “two.sided”)
##
## Paired t-test ##
## data: x and y
## t = 0.77904, df = 11, p-value = 0.4524
## alternative hypothesis: true mean difference is not equal to 0 ## 95 percent confidence interval: ## -1.216846 2.550179 ## sample estimates:
## mean difference
## 0.6666667
The data seems to follow a normal distribution.
b)
0. → −1.216 ≤ µd ≤ 2.55
Since interval contains 0, can not determine if either language is preferable.
10-88
a)
Ho : p1 = p2, H1 = p1 ̸= p2 n1 = 500, n2 = 400, x1 = 385, x2 = 267, pˆ1 = 0.77, pˆ2 = 0.6675, pˆ = = 0.724
zo = p 0.724(10.−770−.724)0.6675( + ) = 3.42
Ho rejected if zo < −zα/2 or zo > zα/2, 3.42 > 1.96 ∴ null hypothesis is rejected and there is a difference in support between counties.
P-value: 2(1 − P(Z < 3.42)) = 0.000626
2*(1-pnorm(3.42))
## [1] 0.0006262114
b)
q q
→
0.0434 ≤ p1 − p2 ≤ 0.1616
Proportion lies between 0.0434 and 0.1616 with 95% confidence. The interval does not contain 0, implying that there is a difference between counties.