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Question 1

Lower Limit Upper Limit
p = 0.25 x0 = 0.8 x0 = 0.92 0.03980 
n = 24 n = 25 0.02967 
Confidence Level S(x0) = 2 S(x0) = 10
0.9305 P[X ≤ x0] = 0.0398011873 P[X ≤ x0] = 0.9703300881 0.0296699119
93.0528900722 LowerLimit = 0.8 Upper Limit = 0.92
Treatment 2 x - x0 Sign x - x0 Sign
0.76 -0.04 - -0.16 -
0.79 -0.01 - -0.13 -
0.8 0 tie -0.12 -
0.81 0.01 + -0.11 -
0.81 0.01 + -0.11 -
0.82 0.02 + -0.1 -
0.89 0.09 + -0.03 -
0.9 0.1 + -0.02 -
0.9 0.1 + -0.02 -
0.91 0.11 + -0.01 -
0.93 0.13 + 0.01 +
0.93 0.13 + 0.01 +
0.94 0.14 + 0.02 +
0.94 0.14 + 0.02 +
0.94 0.14 + 0.02 +
0.95 0.15 + 0.03 +
0.95 0.15 + 0.03 +
0.95 0.15 + 0.03 +
0.96 0.16 + 0.04 +
0.97 0.17 + 0.05 +
0.97 0.17 + 0.05 +
0.97 0.17 + 0.05 +
0.97 0.17 + 0.05 +
0.99 0.19 + 0.07 +
0.99 0.19 + 0.07 +
T1 T2 1 0.78 0.76
1 0.78 0.76 2 0.79 0.79
2 0.79 0.79 3 0.87 0.8
3 0.87 0.8 4 0.91 0.81
4 0.91 0.81 5 0.92 0.81
5 0.92 0.81 6 0.92 0.82
6 0.92 0.82 7 0.92 0.89
7 0.92 0.89 8 0.93 0.9
8 0.93 0.9 9 0.94 0.9
9 0.94 0.9 10 0.95 0.91
10 0.95 0.91 11 0.95 0.93
11 0.95 0.93 12 0.95 0.93
12 0.95 0.93 13 0.95 0.94
13 0.95 0.94 14 0.96 0.94
14 0.96 0.94 15 0.96 0.94
15 0.96 0.94 16 0.96 0.95
16 0.96 0.95 17 0.97 0.95
17 0.97 0.95 18 0.97 0.95
18 0.97 0.95 19 0.97 0.96
19 0.97 0.96 20 0.99 0.97
20 0.99 0.97 21 0.99 0.97
21 0.99 0.97 22 0.99 0.97
22 0.99 0.97 23 0.99 0.97
23 0.99 0.97 24 0.99 0.99
24 0.99 0.99 25 0.99 0.99
25 0.99 0.99

Question 2

Empirical cdf
Treatment 1 Soil Compressibility, X Rank Sorted values F̂(x(i))
x i x(i)
0.78 1 0.78 0.0200
0.79 2 0.79 0.0600
0.87 3 0.87 0.1000
0.91 4 0.91 0.1400
0.92 5 0.92 0.1800
0.92 6 0.92 0.2200
0.92 7 0.92 0.2600
0.93 8 0.93 0.3000
0.94 9 0.94 0.3400 <--1 more than 0.6600
0.95 10 0.95 0.3800
0.95 11 0.95 0.4200
0.95 12 0.95 0.4600
0.95 13 0.95 0.5000
0.96 14 0.96 0.5400
0.96 15 0.96 0.5800
0.96 16 0.96 0.6200
0.97 17 0.97 0.6600
0.97 18 0.97 0.7000
0.97 19 0.97 0.7400
0.99 20 0.99 0.7800
0.99 21 0.99 0.8200
0.99 22 0.99 0.8600
0.99 23 0.99 0.9000
0.99 24 0.99 0.9400
0.99 25 0.99 0.9800

Empirical or Actual Distribution

0.78 0.79 0.87 0.91 0.92 0.92 0.92 0.93 0.94 0.95 0.95 0.95 0.95 0.96 0.96 0.96 0.97 0.97 0.97 0.99 0.99 0.99 0.99 0.99 0.99 0.02 0.06 0.1 0.14000000000000001 0.18 0.22 0.26 0.3 0.34 0.38 0.42 0.46 0.5 0.54 0.57999999999999996 0.62 0.66 0.7 0.74 0.78 0.82 0.86 0.9 0.94 0.98

Soil Compressibility, x

F̂(x(i))

Question 3

9 0 Lower Limit Upper Limit
Lower Limit Upper Limit 9 1.1 p = 0.75 x0 = 0.93 x0 = 0.99
p = 0.75 x0 = 0.95 x0 = 0.99 n = 23 n = 23 0.96
n = 21 n = 19 19 0 Confidence Level S(x0) = 10 S(x0) = 23
Confidence Level S(x0) = 9 S(x0) = 19 19 1.1 0.9988 P[X ≤ x0] = 0.0012431137 P[X ≤ x0] = 1
0.9983 P[X ≤ x0] = 0.0016870791 P[X ≤ x0] = 1 99.8756886342 LowerLimit = 0.93 Upper Limit = 0.99
99.8312920899 LowerLimit = 0.95 Upper Limit = 0.99
Binomial Distribution
Binomial Distribution Golfer Standard Ball, X (Yards) x - x0 Sign x - x0 Sign x0 S(x0) Probability
Golfer Standard Ball, X (Yards) x - x0 Sign x - x0 Sign x0 S(x0) Probability 0.76 -0.17 - -0.23 - 0.76 1 0
0.78 -0.17 - -0.21 - 0.78 1 0 0.79 -0.14 - -0.2 - 0.79 2 0
0.79 -0.16 - -0.2 - 0.79 2 0 0.8 -0.13 - -0.19 - 0.8 3 0.0000000001
0.87 -0.08 - -0.12 - 0.87 3 0.0000000001 0.81 -0.12 - -0.18 - 0.81 4 0.000000001
0.91 -0.04 - -0.08 - 0.91 4 0.000000001 0.81 -0.12 - -0.18 - 0.81 5 0.0000000124
0.92 -0.03 - -0.07 - 0.92 5 0.0000000124 0.82 -0.11 - -0.17 - 0.82 6 0.0000001271
0.92 -0.03 - -0.07 - 0.92 6 0.0000001271 0.89 -0.04 - -0.1 - 0.89 7 0.0000010608
0.92 -0.03 - -0.07 - 0.92 7 0.0000010608 0.9 -0.03 - -0.09 - 0.9 8 0.0000073635
0.93 -0.02 - -0.06 - 0.93 8 0.0000073635 0.9 -0.03 - -0.09 - 0.9 9 0.0000430789
0.94 -0.01 - -0.05 - 0.94 9 0.0000430789 0.91 -0.02 - -0.08 - 0.91 10 0.0002145124
0.95 0 tie -0.04 - 0.95 10 0.0002145124 0.93 0 tie -0.06 - 0.93 11 0.0009158314
0.95 0 tie -0.04 - 0.95 11 0.0009158314 0.93 0 tie -0.06 - 0.93 12 0.0033704481
0.95 0 tie -0.04 - 0.95 12 0.0033704481 0.94 0.01 + -0.05 - 0.94 13 0.010734298
0.95 0 tie -0.04 - 0.95 13 0.010734298 0.94 0.01 + -0.05 - 0.94 14 0.0296699119
0.96 0.01 + -0.03 - 0.96 14 0.0296699119 0.94 0.01 + -0.05 - 0.94 15 0.0713282627
0.96 0.01 + -0.03 - 0.96 15 0.0713282627 0.95 0.02 + -0.04 - 0.95 16 0.1494376704
0.96 0.01 + -0.03 - 0.96 16 0.1494376704 0.95 0.02 + -0.04 - 0.95 17 0.2734937885
0.97 0.02 + -0.02 - 0.97 17 0.2734937885 0.95 0.02 + -0.04 - 0.95 18 0.4389019459
0.97 0.02 + -0.02 - 0.97 18 0.4389019459 0.96 0.03 + -0.03 - 0.96 19 0.6217214884
0.97 0.02 + -0.02 - 0.97 19 0.6217214884 0.97 0.04 + -0.02 - 0.97 20 0.7862590766
0.99 0.04 + 0 tie 0.99 20 0.7862590766 0.97 0.04 + -0.02 - 0.97 21 0.9037859253
0.99 0.04 + 0 tie 0.99 21 0.9037859253 0.97 0.04 + -0.02 - 0.97 22 0.9678914791
0.99 0.04 + 0 tie 0.99 22 0.9678914791 0.97 0.04 + -0.02 - 0.97 23 0.9929762611
0.99 0.04 + 0 tie 0.99 23 0.9929762611 0.99 0.06 + 0 tie 0.99 24 0.9992474565
0.99 0.04 + 0 tie 0.99 24 0.9992474565 0.99 0.06 + 0 tie 0.99 25 1
0.99 0.04 + 0 tie 0.99 25 1
ERROR:#NAME?
0.97
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 6.7501559897209694E-14 2.4655832930875468E-12 5.7621463156465325E-11 9.6769348090219906E-10 1.2434600904498438E-8 1.2710367514046089E-7 1.0608375653475841E-6 7.3635413242456653E-6 4.3078862624668131E-5 2.14512404866696E-4 9.1583144131135549E-4 3.3704480688676658E-3 1.0734297951536577E-2 2.9669911935542369E-2 7.1328262700355097E-2 0.14943767038437911 0.27349378847076988 0.43890194591929105 0.62172148836239294 0.78625907656118488 0.9037859252746081 0.96789147911829332 0.99297626105712666 0.999247456541835 1 9 9 0 1.1000000000000001 19 19 0 1.1000000000000001 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 6.7501559897209694E-14 2.4655832930875468E-12 5.7621463156465325E-11 9.6769348090219906E-10 1.2434600904498438E-8 1.2710367514046089E-7 1.0608375653475841E-6 7.3635413242456653E-6 4.3078862624668131E-5 2.14512404866696E-4 9.1583144131135549E-4 3.3704480688676658E-3 1.0734297951536577E-2 2.9669911935542369E-2 7.1328262700355097E-2 0.14943767038437911 0.27349378847076988 0.43890194591929105 0.62172148836239294 0.78625907656118488 0.9037859252746081 0.96789147911829332 0.99297626105712666 0.999247456541835 1 0.78 0.79 0.87 0.91 0.92 0.92 0.92 0.93 0.94 0.95 0.95 0.95 0.95 0.96 0.96 0.96 0.97 0.97 0.97 0.99 0.99 0.99 0.99 6.7501559897209694E-14 2.4655832930875468E-12 5.7621463156465325E-11 9.6769348090219906E-10 1.2434600904498438E-8 1.2710367514046089E-7 1.0608375653475841E-6 7.3635413242456653E-6 4.3078862624668131E-5 2.14512404866696E-4 9.1583144131135549E-4 3.3704480688676658E-3 1.0734297951536577E-2 2.9669911935542369E-2 7.1328262700355097E-2 0.14943767038437911 0.27349378847076988 0.43890194591929105 0.62172148836239294 0.78625907656118488 0.9037859252746081 0.96789147911829332 0.99297626105712666

S(x0)

Cumulative probability

x0

Question 4

Treatment 1 x¯ 0.9404 w¯ -0.0632666465 l = 0.9655665148
sx = 0.0557135531 sw = 0.0627113054 b = 20.4588310045
n = 25
Rank Height (m) Sorted Data Logged Data Empirical cdf Normal cdf Log Normal cdf Weibull cdf 0 0
i xi xi wi = ln(xi) F̂(x(i)) F(x(i)) F(x(i)) F(x(i)) 100 100
1 0.99 0.78 -0.2484613593 0.02 0.00 0.00 0.01
2 0.96 0.79 -0.2357223335 0.06 0.00 0.00 0.02
3 0.99 0.87 -0.1392620673 0.1 0.10319 0.11279 0.11
4 0.97 0.91 -0.0943106795 0.14 0.29 0.31 0.26
5 0.99 0.92 -0.0833816089 0.18 0.36 0.37 0.31
6 0.92 0.92 -0.0833816089 0.22 0.36 0.37 0.31
7 0.92 0.92 -0.0833816089 0.26 0.36 0.37 0.31
8 0.79 0.93 -0.0725706928 0.3 0.43 0.44 0.37 Normal Distribution Log Normal Distribution
9 0.99 0.94 -0.0618754037 0.34 0.50 0.50885 0.43871 1% 0.10% 0.01% 1% 0.10% 0.01%
10 0.95 0.95 -0.0512932944 0.38 0.57 0.58 0.51 Z 2.326347874 3.0902323062 3.7190164855 Z 2.326347874 3.0902323062 3.7190164855
11 0.94 0.95 -0.0512932944 0.42 0.57 0.58 0.51 V 1.0700091058 1.1125678217 1.1475996225 W 0.0826216656 0.1305258556 0.1699577323
12 0.95 0.95 -0.0512932944 0.46 0.57 0.58 0.51 V 1.0861308102 1.1394274 1.1852547522
13 0.97 0.95 -0.0512932944 0.5 0.57 0.58 0.51
14 0.96 0.96 -0.0408219945 0.54 0.64 0.64 0.59 Weibull Distribution
15 0.96 0.96 -0.0408219945 0.58 0.64 0.64 0.59 1% 0.10% 0.01%
16 0.92 0.96 -0.0408219945 0.62 0.64 0.64 0.59 99% 100% 100%
17 0.99 0.97 -0.0304592075 0.66 0.70 0.70 0.67 V 1.0404009706 1.061225926 1.0762537753
18 0.91 0.97 -0.0304592075 0.7 0.70 0.70 0.67
19 0.93 0.97 -0.0304592075 0.74 0.70 0.70 0.67
20 0.78 0.99 -0.0100503359 0.78 0.81 0.80 0.81
21 0.87 0.99 -0.0100503359 0.82 0.81 0.80 0.81
22 0.95 0.99 -0.0100503359 0.86 0.81 0.80 0.81
23 0.99 0.99 -0.0100503359 0.9 0.81 0.80 0.81
24 0.95 0.99 -0.0100503359 0.94 0.81 0.80 0.81
25 0.97 0.99 -0.0100503359 0.98 0.81 0.80 0.81
1 0.99 0.76
2 0.96 0.79
3 0.99 0.8
4 0.97 0.81
5 0.99 0.81
6 0.92 0.82
7 0.92 0.89
8 0.79 0.9
9 0.99 0.9
10 0.95 0.91
11 0.94 0.93
12 0.95 0.93
13 0.97 0.94
14 0.96 0.94
15 0.96 0.94
16 0.92 0.95
17 0.99 0.95
18 0.91 0.95
19 0.93 0.96
20 0.78 0.97
21 0.87 0.97
22 0.95 0.97
23 0.99 0.97
24 0.95 0.99
25 0.97 0.99

In constructing offshore and coastal defence systems one should design to an appropriate wave height. This requires a statistical analysis of extreme waves. The following data set consists of the highest sea waves in the upper Adriatic sea in Venice, Italy. The data include 18 independent storms recorded in a period of 13 months between 2005 and 2006.

1. How is this data distributed? Choose between a normal, log normal and weibull distribution. 2. Use your above choice to answer the following question. If the authorities in Venice are prepared to live with a. a 1%, b. a 0.1% or c. a 0.01% risk of flooding, what wave height should their coastal defences be designed to withstand?

Question 5

Normal distribution 1.9946144065107027E-3 3.4719511676040561E-3 0.10318567499442155 0.29265389242431683 0.35712334358271425 0.35712334358271425 0.35712334358271425 0.42596003408929839 0.49713578595635516 0.56840307198099116 0.56840307198099116 0.56840307198099116 0.56840307198099116 0.63750568236318528 0.63750568236318528 0.63750568236318528 0.70239074951872404 0.70239074951872404 0.70239074951872404 0.81333903164542154 0.81333903164542154 0.81333903164542154 0.81333903164542154 0.81333903164542154 0.81333903164542154 0.02 0.06 0.1 0.14000000000000001 0.18 0.22 0.26 0.3 0.34 0.38 0.42 0.46 0.5 0.54 0.57999999999999996 0.62 0.66 0.7 0.74 0.78 0.82 0.86 0.9 0.94 0.98 Log normal 1.572 8411656437209E-3 2.9798216583316311E-3 0.11278879327157607 0.31028915986364353 0.37419804574198606 0.37419804574198606 0.37419804574198606 0.44102809292573514 0.50884976044926244 0.57570904826505309 0.57570904826505309 0.57570904826505309 0.57570904826505309 0.63979256990040978 0.63979256990040978 0.63979256990040978 0.69956517191320911 0.69956517191320911 0.69956517191320911 0.80194581336837722 0.80194581336837722 0.80194581336837722 0.80194581336837722 0.80194581336837722 0.80194581336837722 0.02 0.06 0.1 0.14000000000000001 0.18 0.22 0.26 0.3 0.34 0.38 0.42 0.46 0.5 0.54 0.57999999999999996 0.62 0.66 0.7 0.74 0.78 0.82 0.86 0.9 0.94 0.98 weibull 1.2617283562724513E-2 1.6343127307424932E-2 0.11180993884885504 0.25726956300369097 0.3106084489455041 0.3106084489455041 0.3106084489455041 0.37124884911095646 0.43871042397574234 0.51184197787896446 0.51184197787896446 0.51184197787896446 0.51184197787896446 0.58870356441213967 0.58870356441213967 0.58870356441213967 0.666 5479129178894 0.6665479129178894 0.6665479129178894 0.81126275399865133 0.81126275399865133 0.81126275399865133 0.81126275399865133 0.81126275399865133 0.81126275399865133 0.02 0.06 0.1 0.14000000000000001 0.18 0.22 0.26 0.3 0.34 0.38 0.42 0.46 0.5 0.54 0.57999999999999996 0.62 0.66 0.7 0.74 0.78 0.82 0.86 0.9 0.94 0.98

Question 6

cdf of the normal distribution
Empirical cdf Method 1 Method 2
Sea State F̂(x(i)) x F(x(i)) x F(x(i))
1 0.0277777778 0.82 0.015345582 0.87 ERROR:#DIV/0!
2 0.0833333333 1.54 1 1.33 ERROR:#DIV/0!
3 0.1388888889 1.59 1 1.4 ERROR:#DIV/0!
4 0.1944444444 1.92 1 1.69 ERROR:#DIV/0!
5 0.25 2.23 1 1.82 ERROR:#DIV/0!
6 0.3055555556 2.55 1 2.42 ERROR:#DIV/0!
7 0.3611111111 4.09 1 3.46 ERROR:#DIV/0!
Sea State 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 8 0.4166666667 5.5 1 5.3 ERROR:#DIV/0!
Method 1 0.78 0.79 0.87 0.91 0.92 0.92 0.92 0.93 0.94 0.95 0.95 0.95 0.95 0.96 0.96 0.96 0.97 0.97 0.97 0.99 0.99 0.99 0.99 0.99 0.99 9 0.4722222222 5.79 1 5.87 ERROR:#DIV/0!
Method 2 0.76 0.79 0.8 0.81 0.81 0.82 0.89 0.9 0.9 0.91 0.93 0.93 0.94 0.94 0.94 0.95 0.95 0.95 0.96 0.97 0.97 0.97 0.97 0.99 0.99 10 0.5277777778 5.79 1 5.87 ERROR:#DIV/0!
11 0.5833333333 5.91 1 6.44 ERROR:#DIV/0!
Mooring Method 1 12 0.6388888889 7.38 1 7.41 ERROR:#DIV/0!
Sample mean, x̅ = 0.9404 13 0.6944444444 7.99 1 8.26 ERROR:#DIV/0!
Sample variance, s2 = 0.00310 0.00012416 14 0.75 8.98 1 8.88 ERROR:#DIV/0!
Sample s. deviation, s = 0.05571 0.0111427106 15 0.8055555556 9.62 1 9.77 ERROR:#DIV/0!
Sample size, n = 25 16 0.8611111111 9.96 1 9.82 ERROR:#DIV/0!
17 0.9166666667 10.75 1 10.32 ERROR:#DIV/0!
a = 0.1 18 0.9722222222 10.83 1 11.2 ERROR:#DIV/0!
1 - a = 0.9 0.00012416
tn-1,a/2 = 1.7108820799
Critical Point
(tn-1,a/2) x (s / √n) = 0.0190638639
90% confidecnce level
Lower Limit 0.9213361361
Upper Limit 0.9594638639
Mooring Method 2
Sample mean, x̅ =
Sample variance, s2 =
Sample s. deviation, s =
Sample size, n =
a =
1 - a =
tn-1,a/2 =
(tn-1,a/2) x (s / √n) =
Lower Limit
Upper Limit

An experiment was carried out on scale models in a wave tank to investigate how the choice of mooring method affected the bending stress produced in a device used to generate electricity from wave power at sea. The model system was subjected to the same sample of 18 sea states with each of the two mooring methods. The resulting data (root mean square bending moment in Newton -meters) are shown below. Construct 90% confidence intervals for the true mean bending stress associated with each type of mooring. What needs to be assumed and how well do the data meet these assumptions?

Probability Plot

Method 1 1.5345581951694647E-2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2.7777777777777776E-2 8.3333333333333329 E-2 0.1388888888888889 0.19444444444444445 0.25 0.30555555555555558 0.3611111111111111 0.41666666666666669 0.47222222222222221 0.52777777777777779 0.58333333333333337 0.63888888888888884 0.69444444444444442 0.75 0.80555555555555558 0.86111111111111116 0.91666666666666663 0.97222222222222221 Method 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2.7777777777777776E-2 8.3333333333333329E-2 0.1388888888888889 0.19444444444444445 0.25 0.30555555555555558 0.3611111111111111 0.41666666666666669 0.47222222222222221 0.52777777777777779 0.58333333333333337 0.63888888888888884 0.69444444444444442 0.75 0.80555555555555558 0.86111111111111116 0.91666666666666663 0.97222222222222221

F(x(i)) assuming X is normally distributed

F̂(x(i))

The 90% confidence intervals for the two mooring methods do overlap. There is therefore no evidence to suggest that the true (population) mean bending moments associated with each method are significantly different from each other. The methods produce bending moment measurements which on the average are not different from each other. This conclusion requires the assumption that the bending moment measurements from the two different test methods are both normally distributed as the sample size is to small to invoke the central limit theorem. The probability plot alongside suggests this assumption is not reasonable.

Question 7

Treatment 1 Treatment 2
0.78 0.76
0.79 0.79
0.87 0.8
0.91 0.81
0.92 0.81
0.92 0.82
0.92 0.89
0.93 0.9
0.94 0.9
0.95 0.91
0.95 0.93
0.95 0.93
0.95 0.94
0.96 0.94
0.96 0.94
0.96 0.95
0.97 0.95
0.97 0.95
0.97 0.96
0.99 0.97
0.99 0.97
0.99 0.97
0.99 0.97
0.99 0.99
0.99 0.99
Sample mean, x̅1 = 0.9404 Sample mean, x̅2 = 0.9096
Sample variance, s21 = 0.00310 Sample variance, s22 = 0.00482
Sample s. deviation, s1 = 0.05571 Sample s. deviation, s2 = 0.06943
Sample size, n1 = 25 Sample size, n2= 25
a = 0.01 0.01
1 - a = 0.99 0.99
tn-1,a/2 = 2.7969395048 2.7969395048
Critical Point Critical Point
(tn-1,a/2) x (s / √n) = 0.0311654875 0.038838873 0.0388388731
1-a% Confidence Interval 1-a% Confidence Interval
Lower Limit 0.9092345125 1 0.8707611269
Upper Limit 0.9715654875 1 0.9484388731
DO OVERLAP

The 99% confidence intervals for the two testing methods do overlap. There is therefore no evidence to suggest that the true (population) mean thicknesses associated with each test method are significantly different from each other. The ultrasound technique produce thickness measurements which on the average are not different to those obtained from sectioning. The company can therefore safely introduce the non destructive test procedure and thereby speed up delivery times and remove the scrapping cost associated with the sectioning technique. This conclusion requires the assumption that the thickness measurements from the two different test methods are both normally distributed as the sample size is to small to invoke the central limit theorem.

Question 8

Treatment 1
0.78
0.79
0.87
0.91
0.92
0.92
0.92
0.93
0.94
0.95
0.95
0.95
0.95
0.96
0.96
0.96
0.97
0.97
0.97
0.99
0.99
0.99
0.99
0.99
0.99
Sample mean, x̅1 = 0.9404
Sample variance, s21 = 0.00012
Sample s. deviation, s1 = 0.01114
Sample size, n1 = 25
35.4308390023
a = 0.05
1 - a = 0.95
tn-1,a/2 = 2.0638985616
reduced by 16%
Critical Point L = 2 * critical point L 0 = L * (1-0.16) Answer
(tn-1,a/2) x (s / √n) = 0.0045994849 0.0091989697699446 0.00772713 35.4308390023
1-a% Confidence Interval
Lower Limit 0.9358005151
Upper Limit 0.9449994849

Question 8 (2)

Treatment 1
0.78
0.79
0.87
0.91
0.92
0.92
0.92
0.93
0.94
0.95
0.95
0.95
0.95
0.96
0.96
0.96
0.97
0.97
0.97
0.99
0.99
0.99
0.99
0.99
0.99
Sample mean, x̅1 = 0.9404
Sample variance, s21 = 0.00310
Sample s. deviation, s1 = 0.05571
Sample size, n1 = 25
35.4308390023
a = 0.05
1 - a = 0.95
tn-1,a/2 = 2.0638985616
reduced by 16%
Critical Point L = 2 * critical point L 0 = L * (1-0.16) Answer
(tn-1,a/2) x (s / √n) = 0.0229974244 0.0459948488497231 0.03863567 35.4308390023
1-a% Confidence Interval
Lower Limit 0.9174025756
Upper Limit 0.9633974244

Question 8 (3)

Treatment 1
0.78
0.79
0.87
0.91
0.92
0.92
0.92
0.93
0.94
0.95
0.95
0.95
0.95
0.96
0.96
0.96
0.97
0.97
0.97
0.99
0.99
0.99
0.99
0.99
0.99
Sample mean, x̅1 = 0.9404
Sample variance, s21 = 0.00310
Sample s. deviation, s1 = 0.05571
Sample size, n1 = 25
35.4308390023
a = 0.01
1 - a = 0.99
tn-1,a/2 = 2.7969395048
reduced by 16%
Critical Point L = 2 * critical point L 0 = L * (1-0.16) Answer
(tn-1,a/2) x (s / √n) = 0.0311654875 0.0623309750564699 0.05298133 34.6020761246
1-a% Confidence Interval
Lower Limit 0.9092345125
Upper Limit 0.9715654875