# 02402 · Test Quiz 7

## Question 1 of 8

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:

## Question 2 of 8

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:

## Question 3 of 8

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 $?

## Question 4 of 8

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:

## Question 5 of 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 Python-code was run:

Q1 = np.array([1.46,1.66,1.87,2.11,2.37,2.67,2.94,3.31,3.72,4.24]) Q2 = np.array([1.76,1.96,2.24,2.46,2.80,3.11,3.46,3.95,4.42,4.99]) fit = smf.ols(‘Q2 ~ Q1’, data=pd.DataFrame({‘Q1’:Q1, ‘Q2’:Q2})).fit() print(fit.summary(slim=True))

It gave the following output:

OLS Regression Results============================================================================== Dep. Variable: Q2 R-squared: 0.999 Model: OLS Adj. R-squared: 0.999 No. Observations: 10 F-statistic: 1.074e+04 Covariance Type: nonrobust Prob (F-statistic): 8.40e-14 ============================================================================== coef std err t P>|t| [0.025 0.975] —————————————————————————— Intercept 0.0121 0.032 0.383 0.712 -0.061 0.085 Q1 1.1776 0.011 103.622 0.000 1.151 1.204 ==============================================================================

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 $-1/x^2$)

## Question 6 of 8

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:

## Question 7 of 8

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 Python:

Pulse_end = np.array([173, 175, 174, 183, 181, 180, 170, 182, 188, 178, 181, 183, 185]) Pulse_1min = np.array([120, 115, 122, 123, 125, 140, 108, 133, 134, 121, 130, 126, 128]) print(Pulse_end.mean()) 179.4615 print(Pulse_1min.mean()) 125 print(Pulse_end.std(ddof=1)) 5.18998 print(Pulse_1min.std(ddof=1)) 8.406347 print(np.std(Pulse_end-Pulse_1min, ddof=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 Python

```
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)
k = 100000
mysamples = np.random.choice(Pulse_end-Pulse_1min, (k, len(Pulse_end)), replace=True)
mymeans = np.mean(mysamples, axis=1)
np.round(np.quantile(mymeans, [0.001,0.005,0.01,0.025,0.050,0.95,0.975,0.99,0.995,0.999],method='averaged_inverted_cdf'),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?

## Question 8 of 8

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?