statistics 2015/2016 fkekk assignment

8/16/2019 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



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.



Polygon

Figure 2: Polygon for the height of student.



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



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



(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?



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)



= √

= 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.



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



= ∑

=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



= = √ 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.



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:



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



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



= ∑ = 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



= = 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)



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

statistics 2015/2016 fkekk assignment

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