statistics 2015/2016 fkekk assignment
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QUESTION 1
From real statistic notes, survey conducted on height of student by Internet
a) Find the mean and standard deviation of your raw data
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b) Conduct a frequency table for your data set. State clearly the range and intervals used
Intervals and Range
c)
Based on the frequency table above, construct a histogram and a polygonHistogram
Figure 1: Histogram for the height of student.
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Polygon
Figure 2: Polygon for the height of student.
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d) If your raw data containing decimal numbers, round off those to value to the nearest
integer. Find the probability that you will get an:
i. Odd number
ii. Even number
iii. Prime number
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QUESTION 2
A certain machine makes electrical resistors that is approximately normally distributed. A
random sample of 25 such resistors is selected as below:
(a) Calculate for the sample mean and sample error for the data.
40.82 40.55 36.79 44.48 37.60 39.87 41.59 42.03 38.80 37.99
40.66 39.28 40.51 41.77 39.78 38.74 38.65 41.51 43.16 40.03
42.30 40.16 39.02 40.16 42.64
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(b) Based on past experience the population mean resistance is known to be 40 ohms with a
population standard deviation of 2 ohms.
(i) What is the probability that the sample mean will be at least 40.72 ohms?
(ii) There is an 85 percent chance that the sample average will fall between two values
symmetrically distributed around the population mean. What are those two values?
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QUESTION 3
A machine produces metal rods used in an automobile suspension system. A random sample of
30 rods is selected and the diameter is measured. The resulting data and the diameter (in mm) are
follows:
a)
Find point estimation for the true mean diameter of all rods.Mean, µ =AVERAGE(B:B) = 8.3393
Standard Deviation =STDEV(B:B) = 0.1759
Variance = 1/(D:D-1)*(F:F-(G:G/D:D)) = 0.0309 =K:K /SQRT(D:D) = 0.0056
n x x^2 n Σx Σx² (Σx)² mean (μ) standard confide
1 8.24 67.8976 30 250.18 2087.232 62590.03 8.339333 0.175949 0
2 8.21 67.4041
3 8.23 67.7329
4 8.25 68.0625
5 8.26 68.2276
6 8.23 67.7329
7 8.2 67.24
8 8.26 68.2276
9 8.19 67.0761
10 8.23 67.7329
11 8.2 67.24
12 8.28 68.5584
13 8.24 67.8976
14 8.25 68.062515 8.24 67.8976
16 8.35 69.7225
17 8.42 70.8964
18 8.2 67.24
19 8.55 73.1025
a) Find a 98.3% confidence interval on mean rod diameter.
Confidence level = 98.3%
Confidence coefficient = 0.983
Significance level, α = 0.017 2 =0.0085
=−2.39 (Refer normal distribution table)
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= √
= 0.1759√ 30
=0.0321
µ = ± =8.3393± (2.39)(0.0321)
=8.3393±0.0767 = (8.416 , 8.2626)
b) Comment on your result in (b)
We are 98.3% confident that the interval on mean rod diameter from 8.4161 to 8.2626
contains the corresponding true proportion.
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QUESTION 4
4. An engineer measured the Brinell hardness of 36 pieces of ductile iron that were sub
critically annealed. The resulting data are as follows:
The engineer claimed that the mean Brinell hardness of all such ductile iron pieces is less than
170.
(a) Do you agree with the claim of the engineer? Test at 6.5% significance level.
=6144
=1052438
=37748736
, = − − (∑ )
= 13 6 − 1 [1052438− 3774873636 ]
= 135 1052438−1048576
= 135 3862
=110.3429
, = √
= √ 110.3429
=10.5044
= √
= 10.5044√ 36
=1.7507
170 167 174 179 179 156 183 175 155 168 180 182
156 163 156 187 156 187 174 159 165 161 159 175
183 179 174 179 170 179 187 167 159 172 152 177
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= ∑
=6144
36
=170.6667
= µ ≤ ; = µ >
= − µ
= 170.6667−1701.7507
=0.3808
= . , = .
Since, ∗ falls in non-rejection region, so we accept at = . . Hence, we agree with the
claim of the engineer. The mean Brinell hardness of all such ductile iron pieces is less than 170.
(a) Repeat the test by using p-value approach. =6144
=1052438
=37748736
= − − (∑ )
= 13 6 − 1 [1052438− 3774873636 ]
= 135 1052438−1048576 = 135 3862
=110.3429
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= = √ 110.3429
=10.5044
= √
= 10.5044√ 36
=1.7507
= ∑
= 614436
=170.6667
= µ ≤ ; = µ >
∗ = − µ
= 170.6667−1701.7507 =0.3808
− = (>.) = (<0.3808) =0.3520
= . , − > . = .
Yes, we agree with the claim of engineer. Since − > , so we accept at =
.. The mean Brinell hardness of all such ductile iron pieces is less than 170.
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QUESTION 5
An article in the Journal of Sound and Vibration (Vol. 151, 1991, pp. 383-394) described a study
investigating the relationship between noise exposure and hypertension. The following data are
representative of those reported in the article.
y x1 60
0 63
1 65
2 70
5 70
1 70
4 80
6 90
2 80
3 80
5 85
4 89
6 90
8 90
4 90
5 90
7 94
9 100
7 100
6 100
(a) Draw a scatter plot of y (blood pressure rise in millimeters of mercury) versus x (sound
pressure level in decibels). Is it reasonable to assume that y and x is linearly related?
[2 marks]
Answer:
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Figure 5.1: Scatter of y versus x
Comment:From the scatter plot of y versus x shown in Figure 5.1, it appears that the variables have a
positive linear correlation. As we reading from left to right, the blood pressure is rising when
the sound pressure level tends to increase. Hence, it can be assume that y and x is linearly
related.
(b) Find the correlation between these two variables. Interpret your result.
[2 marks]
Answer:
x y xy x² y²
60 1 60 3600 1
63 0 0 3969 0
65 1 65 4225 1
70 2 140 4900 4
70 5 350 4900 25
70 1 70 4900 1
80 4 320 6400 16
90 6 540 8100 3680 2 160 6400 4
80 3 240 6400 9
85 5 425 7225 25
89 4 356 7921 16
90 6 540 8100 36
90 8 720 8100 64
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90 4 360 8100 16
90 5 450 8100 25
94 7 658 8836 49
100 9 900 10000 81
100 7 700 10000 49
100 6 600 10000 36∑x = 1656 ∑y = 86 ∑xy = 7654 ∑x² =140176 ∑ y² =494
With these sums and = 20, the linear correlation coefficient, r is
=
= ∑− (∑)(∑)
∑ − (∑)
∑ − (∑)
= 7654− (1656)(86)20 140176− (1656)20 494− (86)20
= 533.2√ 3059.2√ 124.2
≈0.8650 (4 decimal places)
Applying the formula in Excel to ensure the calculated linear correlation coefficient is correct:
CORREL(B2:B21,C2:C21) 0.86502
Comment:
The result ≈0.8650 suggests a strong positive linear correlation. Therefore, when the sound
pressure level increases, the blood pressure tends to rise up.
(c)
Find the simple linear regression model using least squares method.[3 marks]
Answer:
= ∑ = 165620 =82.8
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= ∑ = 8620 =4.3
= = 533.23059.2 =0.1743
= −=4.3− (0.1743)(82.8) =−10.1315 = +
=−10.1315+0.1743
Therefore, the estimated regression model is: =−10.1315+0.1743
By using Excel to compute a and b:
a:
b:
(d) Find the predicted mean rise in blood pressure level associated with a sound pressure level of
85 decibels.
[1 mark]
Answer:
For = 85 : =−10.1315+0.1743(85)
=4.6840
Thus, when the sound pressure level is 85 decibels, the predicted rise in blood pressure level
is 4.6840 millimeters of mercury.
(e) Compute a 99% confidence interval (CI) for the slope, B.
[2 marks]
Answer:
= − − 2 = 124.2−(0.1743)(533.2)2 0 − 2 =1.3179
INTERCEPT(C2:C21,B2:B21) -10.132
SLOPE( C2:C21,B2:B21) 0.17429
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= = 1.3179√ 3059.2 =0.0238
= 0 . 0 1
=0.005
. for degrees of freedom 18 = 2.878
Hence, a 99% confidence interval for B is:
± =0.1743±(2.878)(0.0238)
= (0.1058 , 0.2428)
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FACULTY OF COMPUTER ENGINEERING AND ELECTRONICS ENGINEERING
BACHELOR IN INDUSTRIAL ELECTRONICS ENGINEERING
(BENE)
BENH 2112
STATISTICS ASSIGNMENT
NO NAME MATRIX NO
1 SUREN A/L GNANASEGARAN BO21410064
2 AHMAD SYAHMI BIN ABDUL HAMID B021410028
3 WONG JIA LI B021310081
4 NURFARAHIN RAIHANAH BT ABD RAZAK B021310250
5 MUNIRAH BT MUKHTAR B021310133
LECTURER NAME : MISS FARAH SHAHNAZ BINTI FEROZ
SUBMISSION DATE : 2ND JUNE 2016