Mat1P98 The Lengths Of Human Assessment Answers
1. Which variables are continuous/numerical? Which are ordinal Which are nominal
2. Calculate following summary statistics: mean, median, max and standard deviation for each of the continuous variables, and count for each categorical variable. Is there any evidence of extreme values Briefly discuss.
3. Plot histograms for each of the continuous variables and create summary statistics. Based on the histogram and summary statistics answer the following and provide brief explanations:
a. Which variables have the largest variability
b. Which variables seems skewed
c. Are there any values that seem extreme
4. Which, if any, of the variables have missing values
a. What are the methods of handling missing values
b. Demonstrate the output (summary statistics and transformation plot) for each method in (4-a).
c. Apply the 3 methods of missing value handling discussed in the lectures. Which method of handling missing values is most suitable for this data set Discuss briefly referring to the data set.
Answer:
The requisite z value is 0.8750.
- b) The area 0.1908 can be interpreted as the probability of the length of human pregnancy exceeding 280 days.
- c) The z value for top 10% is 1.28
Hence, the minimum pregnancy time = Z*standard deviation + Mean = 1.28*16 + 266 = 286.48 days
- d) The z value for bottom 5% is -1.645
Hence, the maximum pregnancy time = Z*standard deviation + Mean = -1.645*16 + 266 = 239.68 days
Question 2
It is known that µ = 138.5 seconds and σ = 29 seconds
- a) We use the following formula
Z = (X-µ)/ σ
Here X= 100
Z = (100-138.5)/29 = -1.3276
P(Z<-1.3276) = 0.092
- b) We use the following formula
Z = (X-µ)/ σ
Here X= 160
Z = (160-138.5)/29 = 0.7414
P(Z>0.7414) = 1- P(Z≤0.7414) = 1- 0.7708 = 0.2292
- c) Here X1 = 120 seconds
Z1 = (120-138.5)/29 = -0.6379
P(Z<Z1) = 0.2618
Also, X2 = 180 seconds
Z2 = (180-138.5)/29 = 1.43
P(Z<Z1) = 0.9238
Hence, P (Z1<Z<Z2) = 0.9238-0.2618 = 0.6620
- d) Here X = 300 seconds
Z = (180-138.5)/29 = 1.43
P(Z<1.43) = 0.9238
Hence, P(Z>5.569) = 1-0.9238 = 0.0762
Thus, it can be concluded that it would not be unusual for a car to spend in excess of 3 minutes in the drive thru as the underlying probability exceeds 0.05.
Question 3
- a) Population mean (µ) = (151+122+120+131)/4 = 131 minutes
- b) The possible samples with 2 samples are as follows.
1) (151,122)
2) (151,120)
3) (151, 131)
4) (122, 120)
5) (122, 131)
6) (120, 131)
- c) The respective means of the various sample are presented below in a tabular form.
|
Sample |
Constituent |
Sample Mean (Minutes) |
|
1 |
(151,122) |
136.5 |
|
2 |
(151,120) |
135.5 |
|
3 |
(151,131) |
141 |
|
4 |
(122,120) |
121 |
|
5 |
(122,131) |
126.5 |
|
6 |
(120,131) |
125.5 |
Each of the samples has a probability of occurrence of (1/6)
Hence, sample mean based on the above samples = (1/6)*(136.5+135.5+141+121+126.5+125.5) = 130.92 minutes
- d) In accordance with the central limit theorem, the best estimate of the sample mean is the population mean which seems to be satisfied in the given case considering the fact that the sample mean is almost equal to the population mean.
Question 4
The relevant excel screenshot with the answer is shown below.
The formula view for the computations is shown below.
Standard deviation for sample = Population standard deviation/Sample size0.5
Question 5
- a) Sample mean = 182.9 lb
The sample standard deviation as per central limit theorem can be indicated as shown below.
The requisite computation of relevant z value is based on the following formula.
Z = (X-µ)/ σ
Z = (140-182.9)/5.77 = -7.44
P (Z>-7.44) = 1-P(Z<-7.44) = 1-0.0001 = 0.9999
- b) Sample mean = 182.9 lb
The sample standard deviation as per central limit theorem can be indicated as shown below.
The requisite computation of relevant z value is based on the following formula.
Z = (X-µ)/ σ
Z = (174-182.9)/10.91 = -0.82
P (Z>-0.82) = 1-P(Z<-0.82) = 1-0.2061 = 0.7939
Based on the above computations, it is apparent that the new rating cannot be inferred as safe considering a significant probability of overloading.
Question 6
Sample proportion (p) = Favourable Cases/Total sample size = 37/125 = 0.296
Standard error = √(p*(1-p)/n) = √(0.296*(1-0.296)/125) = 0.0408
Z value at 95% confidence = 1.96
Margin of error = 1.96*0.0408 = 0.08
Hence, lower limit of 95% confidence interval = 0.296-0.08 = 0.216
Further, upper limit of 95% confidence interval = 0.296 + 0.08 = 0.376
It is apparent that Harris has reported a value of 35% or 0.35 which tends to lie within the 95% confidence interval obtained above and therefore interactive findings of Harris are validated.
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