# 02402 · Test Quiz 8

## Question 1 of 4

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
```

What is the proportion of $y$-variability explained and the estimated (residual) standard deviation ($s_e$)?

## Question 2 of 4

If you did the previous exercise, the following is a repetition:

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
```

Is data in correspondence with the hypothesis corresponding to the assumption above that the slope should be $\sqrt{14/10}=1.1832$? (As well answer as argument must be valid)

## Question 3 of 4

If you did the previous exercise, the following is a repetition:

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
```

For a random nozzle with capacity 4.45 kg/h at 10 bar, the capacity is measured at 14 bar. Within which limits is this measurement expected to be with 99% confidence:

## Question 4 of 4

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. The following data were measured:

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 |

Diastolic pressure (mmHg) | 98 | 94 | 96 | 89 | 88 | 95 | 85 | 88 | 89 | 92 | 89 | 92 | 93 | 92 |

The following bootstrap procedure was performed in R for the measurements of the systolic blood pressure:

```
x=c(143,134,138,138,135,131,135,139,141,143,142,141,149,140)
k=10000
mybootstrapsamples=replicate(k,sample(x,replace=TRUE))
mybootstrapmeans=apply(mysamples,2,mean)
```

And eight different percentiles of the bootstrap distribution was:

```
>quantile(mybootstrapmeans,c(0.005,0.01,0.025,0.05,0.95,0.975,0.99,0.995))
0.5% 1% 2.5% 5% 95% 97.5% 99% 99.5%
136.2143 136.5000 136.9286 137.2857 141.2143 141.5714 141.9286 142.2143
```

A 99% confidence interval for the mean value of the systolic blood pressure becomes: (based on the displayed bootstrap distribution)