Test Quiz 7

The polar moment of inertia of a tube is calculated by the formula:

${I_P} = \frac{\pi }{ {32}} \cdot \left( { {D^4} - {d^4}} \right)$ where D is the outer diameter of the tube and d is the inner diameter of the tube

We assume that $\left( { {\mu _D};{\sigma _D}} \right) = (45;0.3\,)\,mm\quad \mbox{and}\quad \left( { {\mu _d};{\sigma _d}} \right) = (32;0.4{\kern 1pt} )\,mm$ and that the two diameters are independent. An approximate value for the variance of the polar moment of inertia is:

$\begin{array}{l} \sigma_{ {I_P}}^2 \approx {\left( {\frac{\pi }{ {32}} \cdot 4 \cdot \left({4{5^3} - 3{2^3}} \right)} \right)^2} \cdot 0.{5^2}\ \end{array}$

$\sigma_{ {I_P}}^2 \approx {\left( {\frac{\pi }{ {32}} \cdot 45} \right)^2}\cdot 0.{3^2} + {\left( {\frac{\pi }{ {32}} \cdot 32}\right)^2}\cdot 0.{4^2}$

$\sigma_{ {I_P}}^2 \approx {\left( {\frac{\pi }{ {32}} \cdot \left( {4{5^4} - 3{2^4}} \right)} \right)^2} \cdot 0.{7^2}$

$\sigma_{ {I_P}}^2 \approx {\left( {\frac{\pi }{ {32}} \cdot 4 \cdot \left({4{5^3} - 3{2^3}} \right)} \right)^2} \cdot 0.{7^2}$

$\sigma_{ {I_P}}^2 \approx {\left( {\frac{\pi }{ {32}} \cdot 4 \cdot 4{5^3}}\right)^2} \cdot 0.{3^2} + {\left( {\frac{\pi }{ {32}} \cdot 4 \cdot 3{2^3}} \right)^2}\cdot 0.{4^2}$

In a scientific study the following 20 measurements were obtained:

4.3	3.1	1.9	4.9	4.9	3.8	6.4	3.1	1.8	1.3
3.7	2.8	6.8	2.1	9.1	6.2	1.9	4.1	1.7	6.5

with average ${\overline x}=4.02$ mm and standard deviation $s=2.121$ mm.

In continuation of the investigation of $ \sigma $ a 99% confidence interval for the standard deviation $ \sigma $ is wanted. This becomes:

$\frac{\sqrt{19}\cdot 2.121}{\sqrt{32.852}}<\sigma< \frac{\sqrt{19}\cdot 2.121}{\sqrt{8.907}}$

$\frac{\sqrt{19}\cdot 2.121}{\sqrt{8.907}}<\sigma< \frac{\sqrt{19}\cdot 2.121}{\sqrt{32.852}}$

$\frac{\sqrt{20}\cdot 2.121^2}{\sqrt{8.907}}<\sigma< \frac{\sqrt{20}\cdot 2.121^2}{\sqrt{32.852}}$

$\frac{\sqrt{19}\cdot 2.121}{\sqrt{38.582}}<\sigma< \frac{\sqrt{19}\cdot 2.121}{\sqrt{6.844}}$

$\frac{19\cdot 2.121^2}{\sqrt{38.582}}<\sigma< \frac{19\cdot 2.121}{\sqrt{6.844}}$

A trick dice is designed so it has the following probabilities for the 6 possible outcomes:

1’er	2’er	3’er	4’er	5’er	6’er
0.1	0.1	0.1	0.2	0.2	0.3

Let $ X $ be the random outcome of a single throw of the dice, that is, the number of dots on the face-up of the dice. Denote the mean of $ X $, $ \mu $ and the variance of $ X $, $ \sigma^2 $.

Let $ Y $ be the total number of dots in 50 throws with this dice. What is the mean and variance of $ Y $?

$E(Y)=50\mu$ and $Var(Y)=2500\sigma^2$

$E(Y)=\mu$ and $Var(Y)=\sigma^2$

$E(Y)=50$ and $Var(Y)=\sigma^2$

$E(Y)=\mu$ and $Var(Y)=\sigma^2/50$

$E(Y)=50\mu$ and $Var(Y)=50\sigma^2$

The following data are measurements of albumin in blood samples from a group of people consisting of 7 males and 8 females. The standard deviation and the mean of the measurements are given separately for the two groups:

									Mean	Standard deviation
Males	38	36	37	41	37	43	41		$\overline{x}_1=39.0$	$s_1=2.6458$
Females	45	39	41	44	47	46	44	42	$\overline{x}_2=43.5$	$s_2=2.6726$

We assume that the albumin content in the blood can be modelled by a normal distribution in each group. There seems to be some systematic difference in the mean albumin content between the males and the females. Give a 95% confidence interval for this difference:

$4.5\pm 2.164\cdot\sqrt{(1/15)(2.6458^2 + 2.6726^2)/2}$

$4.5\pm 2.164\cdot\sqrt{(1/6+1/7)(6\cdot 2.6458^2 + 7\cdot 2.6726^2)/13}$

$4.5\pm 1.96\cdot\sqrt{(1/7+1/8)(2.6458^2 + 2.6726^2)/2}$

$4.5\pm 2.13\cdot\sqrt{(1/7+1/8)(2.6458^2 + 2.6726^2)/2}$

$4.5\pm 2.164\cdot\sqrt{2.6458^2/7 + 2.6726^2/8}$

For oil nozzles, it is assumed that the nozzle capacity at a given atomizing pressure can be written as: $Q_2=Q_1\cdot \sqrt{\frac{P_2}{P_1}},$ where $ Q_1 $ is the reference capacity at the pressure $ P_1 $ and $ Q_2 $ is the capacity at the pressure $ P_2 $. To verify the formula, a study was performed where the capacity at 14 bar is investigated as a function of the capacity at 10 bar (reference pressure). The following values are found:

Capacity $Q_1$ (kg/h)	1.46	1.66	1.87	2.11	2.37	2.67	2.94	3.31	3.72	4.24
at the pressure 10 bar
Capacity $Q_2$ (kg/h)	1.76	1.96	2.24	2.46	2.80	3.11	3.46	3.95	4.42	4.99
at the pressure 14 bar

The following computations can be used ($x=Q_1$ and $y=Q_2$) $\bar{x}=2.635, \bar{y}=3.115, S_{xx}=7.5655, S_{yy}=10.499, S_{xy}=8.9090$ The following R-code was run:

Q1=c(1.46,1.66,1.87,2.11,2.37,2.67,2.94,3.31,3.72,4.24)
Q2=c(1.76,1.96,2.24,2.46,2.80,3.11,3.46,3.95,4.42,4.99)
summary(lm(Q2~Q1))

It gave the following output:

Call:
lm(formula = Q2 ~ Q1)

Residuals:
      Min        1Q    Median        3Q       Max 
-0.046215 -0.014807 -0.004898  0.026954  0.040131 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  0.01207    0.03153   0.383    0.712    
Q1           1.17758    0.01136 103.622  8.4e-14 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 

Residual standard error: 0.03126 on 8 degrees of freedom
Multiple R-squared: 0.9993,     Adjusted R-squared: 0.9992 
F-statistic: 1.074e+04 on 1 and 8 DF,  p-value: 8.403e-14

Assume that $ P_1 = 10$ bar has a measurement error of $ \sigma_1 = 0.01$ and $ P_2 = 14$ bar has a measurement error of $ \sigma_2 = 0.02$. What is then the approximately $\sigma_ {P_2/P_1} $ the measuring error of $P_2/P_1 = 1.4$? (As help you can potentially use that the derivative of the function $1/x$ is $-1/x^2$)

$\sqrt{0.02^2/100 + 0.01^2\cdot 14^2/10000}=0.00244$

$\sqrt{(0.02^2 + 0.01^2)/2}=0.0158$

$(0.02 + 0.01)/2=0.015$

$\sqrt{(14\cdot 0.02^2 + 10\cdot 0.01^2)/24}=0.01658$

$\sqrt{0.02^2/100 + 0.01^2/100}=0.00224$

For a device for measuring blood pressure at home the accuracy was investigated. Therefore repeated measurements of blood pressure of a person, with a time interval of 5 min and under as identical circumstances as possible.

Measurement no	1	2	3	4	5	6	7	8	9	10	11	12	13	14
Systolic pressure (mmHg)	143	134	138	138	135	131	135	139	141	143	142	141	149	140

Data is assumed normally distributed, and parameter estimates for the two blood pressure measurements are:

\[({\bar x_S};{s_S}) = (139.21;4.58)\quad \quad \quad \quad \quad \quad ({\bar x_D};{s_D}) = (91.43;3.61)\quad \quad\]

A 95% confidence interval for the variance of the diastolic blood pressure becomes:

$\left[ { {\textstyle{ {14 \cdot 3.6{1^2}} \over {26.12}}};{\textstyle{ {14 \cdot 3.6{1^2}} \over {5.63}}}}\right]$

$\left[ { {\textstyle{ {13 \cdot 3.6{1^2}} \over {22.36}}};{\textstyle{ {13 \cdot 3.6{1^2}} \over {5.89}}}} \right]$

$\left[ { {\textstyle{ {13 \cdot 3.6{1^2}} \over {24.74}}};{\textstyle{ {13 \cdot 3.6{1^2}} \over {5.01}}}} \right]$

$\left[ { {\textstyle{ {14 \cdot 3.61} \over {26.12}}};{\textstyle{ {14 \cdot 3.61} \over {5.63}}}} \right]$

$\left[ { {\textstyle{ {13 \cdot 3.61} \over {24.74}}};{\textstyle{ {13 \cdot 3.61} \over {5.01}}}} \right]$

13 runners had their pulse measured at the end of a workout and 1 minute after again and we got the following pulse measurements:

Runner	1	2	3	4	5	6	7	8	9	10	11	12	13
Pulse end	173	175	174	183	181	180	170	182	188	178	181	183	185
Pulse 1min	120	115	122	123	125	140	108	133	134	121	130	126	128

The following was run in R:

	> Pulse_end=c(173,175,174,183,181,180,170,182,188,178,181,183,185)
	> Pulse_1min=c(120,115,122,123,125,140,108,133,134,121,130,126,128)
	> mean(Pulse_end)
	[1] 179.4615
	> mean(Pulse_1min)
	[1] 125
	> sd(Pulse_end)
	[1] 5.18998
	> sd(Pulse_1min)
	[1] 8.406347
	> sd(Pulse_end-Pulse_1min)
	[1] 5.767949

A 95% confidence interval for the mean pulse drop is wanted WITHOUT use of any normal distribution assumptions. Hence the following was run in R

k = 100000
mysamples = replicate(k, sample(Pulse_end-Pulse_1min, replace = TRUE)) 
mymeans = apply(mysamples, 2, mean) 
round(quantile(mymeans,c(0.001,0.005,0.01,0.025,0.050,0.95,0.975,0.99,0.995,0.999)),2)

here “round” rounds the given quantiles at 2 decimal points. The following quantiles were the results:

0.1%  0.5%    1%  2.5%    5%   95% 97.5%   99% 99.5% 99.9% 
49.00 50.15 50.54 51.23 51.77 56.85 57.23 57.69 57.92 58.54

What will the 95% confidence interval be based on this?

$[50.15;57.92]$

$[51.23;57.23]$

$[49.00;58.54]$

$[51.77;56.85]$

None of the above

The so-called BMI (Body Mass Index) is a measure of the weight-height-relation, and is defined as the weight ($W$) in kg divided by the squared height ($H$) in meters: $BMI=\frac{W}{H^2}.$

Assume that the BMI-distribution in a population is a lognormal distribution with significance level $\alpha=3.1$ and $\beta=0.15$ (hence that log(BMI) is normal distributed with mean $3.1$ and standard deviation $0.15$).

If a person should determine his/her own $\log(BMI)$-value, the person should then compute: $\log(BMI)=\log(W)-2\log(H)$ Assume now that the standard deviation of a height measurement is $\sigma_H=0.005m$ and the standard deviation of a weight measurement is $\sigma_W=1.5kg$.

In addition the partial derivatives of $\log(BMI)$ is given here:

\[\frac{\partial \log(BMI)}{\partial W}=\frac{1}{W}\;\mbox{ and }\; \frac{\partial \log(BMI)}{\partial H}=-\frac{2}{H}\]

If a person measures his/her height at 1.67m and his/her weight at 64.3 kg, and hence a $\log(BMI)$-value at $\log(64.3)-2\cdot \log(1.67)=3.14$, what is then the approximate standard deviation of this $\log(BMI)$-measurement?

$1.5/64.3+0.005/1.67=0.026$

$\sqrt{1.5^2+0.005^2}=1.50$

$\sqrt{1.5^2/64.3^2+4\cdot 0.005^2/1.67^2}=0.024$

$1.96\cdot (1.5/64.3+0.005/1.67)=0.052$

$\log(\sqrt{64.3})-\log(1.67^2)=1.06$

02402 · Test Quiz 7

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4.3	3.1	1.9	4.9	4.9	3.8	6.4	3.1	1.8	1.3
3.7	2.8	6.8	2.1	9.1	6.2	1.9	4.1	1.7	6.5

4.3	3.1	1.9	4.9	4.9	3.8	6.4	3.1	1.8	1.3
3.7	2.8	6.8	2.1	9.1	6.2	1.9	4.1	1.7	6.5

4.3	3.1	1.9	4.9	4.9	3.8	6.4	3.1	1.8	1.3
3.7	2.8	6.8	2.1	9.1	6.2	1.9	4.1	1.7	6.5