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9. Solutions

Solutions for the exercises in Chapter 8.

ExerciseGroupNo.QuestionAnswer
8.1I1-a)The sample size31
   The range of the sample values469 cm
   The median267 cm
   The mean261.9 cm
   The total value8118 cm
   The sample variance15636.5
   The sample standard deviation.125.0 cm
   The sample coefficient of variation.0.48
8.1II1From this data calculate the mean and the variance.Mean(x) = 23.2 cm
Var(x) = 12.5
  2Choose an adequate class interval and build up a table with the length frequencies distributionClass interval chosen -1 cm Because it is simple, and gives an adequate number of classes (between 15 and 30)
  3-a)The sample mean and sample variance. Compare this results with the ones obtained in 1Sample mean = 23.2 cm
Sample variance = 12.5
The values of the statistics are equal.
Since the class interval used was the same as the resolution of the original measurements, no information was lost.
  3-b)Three statistics of locationMean = 23.2 cm
Median = 22.5 cm
Mode = 20 cm
  3-c)Three statistics of dispersionInter Quartile Range = 5.5 cm
Variance = 12.49
CV = 0.15
  3-d)Number of individuals with a length less than 20 cm41
  3-e)Percentage of individuals with a length equal to or greater than 20 cm79%
  3-f)Percentage of observations between 23 and 25 cm15%
  3-g)The value that corresponds to a length equal to or greater than 45% of all the observations21.5 cm
  3-h)The value that corresponds to a length less than 21% of all the observations26.5 cm
  3-i)The quantile of order 96%30.5 cm
8.2I1-a)Relative frequencies of number of boats by fishing gearPurse-seines: 0.3
Trawls: 0.21
Handlines: 0.17
Longlines: 0.23
Trammel nets: 0.09
Purse-seines: 0.3
  1-b)Percentage of vessels that operate handlines17%
  1-c)Percentage of vessels that operate trammel nets9%
  2-a)Relative frequency of boats not operating trammel nets0.91
  2-b)Percentage of boats operating purse seines or longlines53%
  2-c)Relative frequency of boats that do not operate with handlines, nor trammel nets, nor longlines0.51
  2-d)Proportion of the total fleet that are small boats without engine0.25
  2-e)Proportion of the total fleet that are small boats with engine0.37
  2-f)Proportion of the total fleet that are small boats0.62
  2-g)Check that the proportion of 2.f) is the sum of 2.d) plus 2.e)0.62 = 0.25+0.37
  3-a)Proportion of the small boats without engine that operate handlines0.34
  3-b)Proportion of the total fleet that are small boats without engine0.25
  3-c)Proportion of the total fleet that are small boats without engine operating handlines0.08
  3-d)Check that the proportion of 3.c) is the product of 3.a) times 3.b)0.08 = 0.34×0.25
  4-a)Percentage of small boats without engine operating handlines or longlines65%
  4-b)Percentage of purse seiners operating purse seines100%
  4-c)Relative frequency of vessels that are not purse-seiners.81%
  4-d)Percentage of small boats with engine that operate trawls4%
  4-e)Percentage of the fleet that fishes with traps0%
8.2II1-a)Probability that the boat operates handlines0.17
  1-b)Probability that the boat operates Trammel nets0.09
  1-c)Probability that the boat does not operate trammel nets0.91
  1-d)Probability that the boat operates purse-seines or longlines.0.53
  1-e)Probability that the boat does not operate handlines, nor longlines nor trammel nets0.51
  2-a)Probability that the boat will be a small boat without engine0.245
  2-b)Probability that the boat will be a small boat with engine0.374
  2-c)Probability that the boat will be a small boat0.62
  2-d)Show that the probability 2.c) is equal to the sum of probability 2.a) plus the probability 2.b)0.619 = 0.245+0.374
  3-a)Probability of the boat being a purse seiner0.189
  3-b)Probability of the boat being a stern trawler0.192
  3-c)Probability of the boat not being a purse seiner nor a stern trawler0.619
  3-d)Show that probability 3.c) is equal to probability 2.c)Shown from the comparison of the values
  4-a)If we choose a boat from the small boats without engine, what is the probability that she operates with handline?0.338
  4-b)If we choose a boat out of the total fleet what is the probability that she is a small boat without engine?0.245
  4-c)If we choose a boat out of the total fleet what is the probability that she is a small boat without engine operating with handline?0.083
  4-d)Check that the probability of 4.c) is equal to the product of the probability of 4.a) times the probability of the 4.b).0.083 = 0.338×0.245
8.2III1-a)Probability of X being less than or equal to 180.097
  1-b)Probability of X being greater than 180.903
  1-c)Probability of X being less than 250.986
  1-d)Probability of X being between 18 and 250.889
  2-a)x such that Prob {X ≤ x} = 0.841322.60
  2-b)x such that Prob {X ≥ x} = 0.977216.60
  2-c)x such that Prob {X < x} = 0.998626.58
  2-d)x such that x is the 95% quantile of the distribution of X23.89
  2-e)x such that x is the median of the distribution of X20.60
8.2IV1-a)Probability of the values of Z being between -1 and 10.683
  1-b)Probability of the values of Z being between -2 and 20.954
  1-c)Probability of the values of Z being between -3 and 30.997
  2-a)The z1 value for which the probability that the values of the variable Z will be smaller than z1 is 2.5% (Prob{Z<z1} = 0.025)-1.96
  2-b)The z2 value for which the probability of the variable Z being smaller than z2 is 97.5% (Prob{Z<z2} = 0.975)1.96
  3-a)Probability that the variable Z will be within the interval (z1, z2) given by (Prob{Z<z1} = 0.025) (Prob{Z<z2} = 0.975)0.950
  3-b)Probability that the variable Z will be within the interval (z1, z2) given by (Prob{Z<z1} = 0.004)Prob{Z<z2} = 0.954).0.950
  3-c)Probability that the variable Z will be within the interval (z1, z2) given by (Prob{Z<z1} = 0.012) (Prob{Z<z2} = 0.962)0.950
   Verify that the smallest of these intervals is the interval with symmetrical values, z1 and z2Width 3-a) = 1.96-(-1.96) = 3.92
Width 3-b) =1.68-(-2.65) = 4.33
Width 3-c) = 1.77-(-2.26) = 4.03
8.2V1-a)Prob {t(10) >1.812}0.050
  1-b)Prob {t(19) <1.729}0.950
  1-c)Prob {-1.34 < t(15) < +2.602}0.890
  2-a)a such that Prob {t(8) < a} = 0.951.860
  2-b)a such that Prob {t(26) > a} = 0.99-2.479
  2-c)a such that Prob {-a < t(20) < +a} = 0.952.086
  3-a)a such that Prob {t < a} = 0.95, with 40, 60, 120 and infinite degrees of freedom40 df: a = 1.684
60 df: a = 1.671
120 df: a = 1.658
df: a = 1.645
  3-b)Compare a obtained in 3.a) with a such that Prob {Z < a} = 0.95a (Z) = 1.645 = a (t∞)
8.2VI1Show that the expected value of the random variable X is equal to P
  2Show that the variance of the random variable X is equal to PQ
8.3I1-a)Mean shrimp landing per sampled fishing trip698.7 Kg
  1-b)Variance of the sampled landings137 696
  1-c)Standard deviation of the sampled landings371.1 Kg
  2Estimate of the total amount of shrimp landed in that week6 986.7 Kg
8.3II1-a)Average shrimp landing per fishing trip during that week542 Kg
  1-b)Population variance and modified variance of the landingsσ2 = 127 730
S2 = 141 922
  1-c)Standard deviation of the landings357.4 Kg
  1-d)Total amount of shrimp landed5 420 Kg
  1-e)Proportion of all landings below 400 Kg0.300
  1-f)Relative frequency of landings between 400 and 800 Kg0.400
  2Build at least 10 samples of 3 landings each that could have been selected from that population.Sample no.Land 1Land 2Land 3
15380906
25380230
35380598
4538020
55380435
62305980
7230598435
8230598859
92305981123
10230598711
  3Repeat the calculations done on number 1. a) to d), for each of these samplesSample. no.AverageVarSDTotal
14812076174564813
2256728682702560
33791084413293787
4186930283051860
5324815462863243
6276909883022760
7421340031844210
8562998643165623
96502014164496503
10513632592525130
  4Compare the values of the statistics obtained in the previous item with the values of the corresponding population parameters.The values of these statistics both over-estimate and under-estimate the corresponding population parameters
8.3III1Histogram of the sampling distribution of the estimator, , using an appropriate class interval.Appropriate Class Interval: 50 Kg (approx. 20 classes)

Sampling distribution - Mean landing

8.3III2-a)The 120 values of the estimatorNo.Mean Land
    11481.3
    2256.0
    3378.7
    4186.0
    5324.3
    6465.7
    7553.7
    8416.3
    9558.0
    10680.7
    11488.0
    12626.3
    13767.7
    14855.7
    15718.3
    16455.3
    17262.7
    18401.0
    19542.3
    20630.3
    21493.0
    22385.3
    23523.7
    24665.0
    25753.0
    26615.7
    27331.0
    28472.3
    29560.3
    30423.0
    31610.7
    32698.7
    33561.3
    34840.0
    35702.7
    36790.7
    37378.7
    38501.3
    39308.7
    40447.0
    41588.3
    42676.3
    43539.0
    44276.0
    4583.3
    46221.7
    47363.0
    48451.0
    49313.7
    50206.0
    51344.3
    52485.7
    53573.7
    54436.3
    55151.7
    56293.0
    57381.0
    58243.7
    59431.3
    60519.3
    61382.0
    62660.7
    63523.3
    64611.3
    65578.0
    66385.3
    67523.7
    68665.0
    69753.0
    70615.7
    71508.0
    72646.3
    73787.7
    74875.7
    75738.3
    76453.7
    77595.0
    78683.0
    79545.7
    80733.3
8.3III2-b)Expected value of the estimator542.0 Kg
  2-c)Sampling variance of the estimator33 115.2
  2-d)Error of the estimator182.0 Kg
  3Compare the expected value obtained in 2.b) with the population mean calculated in Group I - 1.a)They have the same value
  4Check the theoretical expression:
     
  5-a)Percentiles of the sampling distribution of the estimator with the following orders:  
i) 1.0%i) 1.0%: 158.2 Kg
ii) 2.5%ii) 2.5%: 205.5 Kg
iii) 3.5%iii) 3.5%: 222.8 Kg
iv) 0.0%iv) 50.0%: 540.7 Kg
v) 95.0%v) 95.0%: 840.8 Kg
vi) 96.0%vi) 96.0%: 856.7 Kg
vii) 97.5%vii) 97.5%: 876.2 Kg
viii) 98.5%viii) 98.5%: 901.0 Kg
  5-b)Four intervals that encompass 95% of all possible sample meansIntervalLower LimitUpper Limit
From minimum value to 95th percentile83.3840.8
From 5th percentile to largest value242.9962.7
From 1st percentile to 96th percentile158.2856.7
From 2.5th percentile to 97.5th percentile205.5876.2
8.3III5-c)The width of the four intervalsIntervalWidth
From minimum value to 95th percentile757.5
From 5th percentile to largest value719.8
From 1st percentile to 96th percentile698.5
From 2.5th percentile to 97.5th percentile670.7
  6The shortest of these intervals that holds 95% of all possible sample meansThe last interval 
  7-a)Probability of getting a sample of 3 landings with an average landing below or equal to 600 Kg62.5% 
  7-b)Probability of getting a sample of 3 landings with an average landing above 600 Kg37.5% 
  7-c)Probability of getting a sample of 3 landings with an average landing between 199 and 953 Kg96.7% 
  8-a)l such that Prob {< l} = 0.95840.8 Kg 
  8-b)l1 and l2 such that Prob {l1<< l2} = 0.95l1=205.5 Kg
l2=876.2 Kg
 
8.4I1-a)Mean 39.29
  1-b)Variance 16.41
  1-c)Standard deviation 4.05
  2-a)Estimate of population mean 39.29
  2-b)The estimator is biased? No
  2-c)Sampling variance of  4.10
  2-d)Error of  2.03
  2-e)A 95% confidence interval of m 35.32–43.26
8.4II1A procedure to select a simple random sample of 4 numbered vesselsSelect a whole random number between 1 and 20. Include the corresponding vessel in the sample. Repeat 4 times. If at any moment the vessel selected was already included in the sample, repeat the random number selection, until a new vessel is selected.
  2-a)An estimator of the total amount of fish landed in the port during this dayN x Sample Mean
  2-b)Sampling distribution of that estimator and the formulae to obtain the expected value and the expected sampling varianceSampling Distribution: 
    ŶN (E[Ŷ], V[Ŷ])
E[Ŷ] = Y
  3-a)Estimate of total landing153.5 Kg
  3-b)Estimate of sampling variance3923.4
  3-c)Estimate of error of the estimate62.6 Kg
  3-d)A 95% confidence interval for the population total landings0–352.8
  4Approximate size of the sample necessary to have an error 10% smaller than the one previously calculated in 3.c)5
8.4III1-a)Proportion of females in the sample0.400
  1-b)Sample variance0.248
E[p]=P
  1-c)Expressions for the expected value of p, and for the sampling variance of p
  1-d)Estimate of the sampling variance of p0.0058
  1-e)Estimate of the error of p0.0761
  1-f)Estimate of the 95% confidence interval for the proportion P applying the binomial distribution0.227–0.594
  1-g)Estimate of the 95% confidence interval for the proportion P applying the normal approximation to the binomial distribution0.251–0.549
  2-a)Estimate of total number of females in the population40
  2-b)Estimate of the sampling variance and of the error of the total numberV[Np]=57.93
SNp=7.61
  2-c)Estimate of the 95% confidence interval for the total number of females of the population applying the binomial distribution23 – 59
  2-d)Estimate of the 95% confidence interval for the total number of females of the population applying the normal approximation to the binomial distribution25 – 55
8.5I
(landing ports)
1Number of vessels to be sampled in each stratumClass (stratum)Sample Size
100-2
200-10
>3004
Total16
  2-a)Average landing per vessel in each categoryClass (stratum)Average Landing
100-2
200-6
>3005
  2-b)Variance between total landings within each stratumClass (stratum)Variance
100-3.84
200-19.18
>30011.56
  3-a)Estimates of mean landing for each stratumClass (stratum)Est. Mean Landing
100-2
200-6
>3005
  3-b)Estimates of expected sampling variance of the estimator of the mean for each stratumClass (stratum)Est. Exp. Var
100-1.54
200-1.54
>3002.31
  3-c)Estimates of error of the estimator of the mean for each stratumClass (stratum)Est. Error
    100-1.24
    200-1.24
    >3001.24
8.5I3-d)Estimates of total landing for each stratumClass (stratum)Est. Total Landing
    100-20
    200-300
    >300100
  3-e)Estimates of expected variance of the estimator of total landing for each stratumClass (stratum)Est. Exp. Var. Total
    100-153.66
    200-3836.88
    >300924.80
  3-f)Estimates of error of the estimator of total landing for each stratumClass (stratum)Est. Exp. Error Total
    100-12.40
    200-61.94
    >30030.41
  4-a)Estimate of mean landing for total fleet5.25 
  4-b)Estimate of expected variance of the estimator of the mean for total fleet0.77 
  4-c)Estimate of error of the estimator of the mean for total fleet0.88 
  4-d)Estimate of total landing for total fleet420 
  4-e)Estimate of expected variance of the estimator of total landing for total fleet4915.34 
  4-f)Estimate of error of the estimator of total landing for total fleet70.11 
8.5II (surveys)1-a)Estimate of index of total biomass of European hake for each stratumStratum                    Est. Index (tonnes)
A1, 100–200 m129.8
A1, 200–300 m263.3
A1, 300–400 m1101.4
A1, 400–500 m289.2
A1, 500–600 m142.5
A2, 100–200 m764.0
A2, 200–300 m759.7
A2, 300–400 m239.7
A2, 400–500 m354.7
A2, 500–600 m319.4
  1-b)Estimate of error of the estimator for each stratumStratumEst. Error (tonnes)
A1, 100–200 m3.0
A1, 200–300 m7.6
A1, 300–400 m43.7
A1, 400–500 m9.9
A1, 500–600 m3.6
A2, 100–200 m20.6
A2, 200–300 m12.3
A2, 300–400 m7.6
A2, 400–500 m8.7
A2, 500–600 m10.2
  1-c)Estimate of coefficient of variation of the estimator for each stratumStratumCV Est.
A1, 100–200 m2.3%
A1, 200–300 m2.9%
A1, 300–400 m4.0%
A1, 400–500 m3.4%
A1, 500–600 m2.5%
A2, 100–200 m2.7%
A2, 200–300 m1.6%
A2, 300–400 m3.2%
A2, 400–500 m2.5%
A2, 500–600 m3.2%
  2-a)Estimate of total biomass of European hake for the total area4 634 tonnes
  2-b)Estimate of error of the estimator54 tonnes
8.5II2-c)Estimate of 95% confidence limits of the total biomass in the total area4 258 tonnes – 4 469 tonnes
  3-a)Proportional allocation of the total 100trawls to the strata areasStratumCV Est.
A1, 100–200m26
A1, 200–300 m10
A1, 300–400 m8
A1, 400–500 m7
A1, 500–600 m11
A2, 100–200 m16
A2, 200–300 m6
A2, 300–400 m4
A2, 400–500 m6
A2, 500–600 m6
Total100
  3-b)Strata allocation that gives the maximum precision in the estimation of the total abundanceStratumNum. Hauls
A1, 100–200m3
A1, 200–300 m8
A1, 300–400 m29
A1, 400–500 m8
A1, 500–600 m3
A2, 100–200 m17
A2, 200–300 m13
A2, 300–400 m5
A2, 400–500 m7
A2, 500–600 m7
Total100
8.6I (clusters)115 clusters selected with equal probabilitiesElementCluster Nš
120
235
311
425
521
627
75
828
94
1013
111
1210
1330
147
1519
8.6I215 clusters selected with probabilities proportional to the cluster sizesElementCluster No.
110
211
320
423
59
625
714
835
924
1028
1112
1229
1327
1434
1522
8.6I315 clusters selected by selecting 3 clusters from each province with probabilities proportional to the sizes of the clusters of each stratumElementProvCluster No.
113
217
311
4210
5211
629
7317
8316
9318
10420
11423
12425
13535
14528
15529
8.6II1-a)Number of clusters in the population23
  1-b)Number of clusters in the sample5
  1-c)Number of elements in cluster 1450
  2-a)Sample mean value per cluster1294.6
  2-b)Sample mean value per element25.9
  2-c)Sample variance between clusters1422.8
  3-a)Estimate of the expected value of the estimatorEstimator: N * Mean Value per Cluster Est. Expected value: 29 775.8
  3-b)Estimate of the sampling variance of the estimator222.7
  3-c)Estimate of the error of the estimator14.9
8.6III1-a)Estimate of the total value of the population19 915.4 
  1-b)Estimate of the error of your estimator32.4 
  2-a)Estimate of the total landing of all the vessels21 205 
  2-b)Estimate of the sampling variance of your estimator and its errorSampling Var
Sampling Error
38 482 456
6 203
  2-c)Estimate of an approximate 95% confidence interval of the total landing3 981–38 428 
8.7I1-a)Estimate of the total weight of fish landed248 785 
 (two1-b)Estimate of the error of the estimation48 003 
 stages)2-a)Estimate of the proportion of boxes of fish in the total landings0.228 
  2-b)Estimate of the error of the estimation0.053 
8.7II1Estimate of the total weight of fish landed192 090 
  2Estimate of the error of the estimation50 285 

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