Test Quiz 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())
	sigma_res = np.sqrt(fit.mse_resid)
print(sigma_res)

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 ==============================================================================

sigma_res=0.03126

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

Proportion explained=1.1776 and standard deviation=0.01136

Proportion explained=0.0121 and standard deviation=0.03153

Proportion explained=0.03153 and standard deviation=0.01136

Proportion explained=0.999 and standard deviation=0.03126

Proportion explained=0.03126 and standard deviation=0.999

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 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())

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 ==============================================================================

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)

Yes, since the 95% confidence interval for the slope contains 1.1832

Yes, since the relevant P-value is 0.712

No, since the relevant P-value is $0$

Yes, since the relevant P-value is $0$

No, since the 95% confidence interval for the intercept does not contain 1.1832

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 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())

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 ==============================================================================

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:

$5.25\pm 1.860\cdot s_e\cdot \sqrt{1+1/10 + \frac{(4.45-2.635)^2}{7.5655}}$

$5.25\pm 1.860\cdot s_e\cdot \sqrt{1/10 + \frac{(4.45-2.635)^2}{10.499}}$

$5.25\pm 3.355\cdot s_e\cdot \sqrt{1+1/10 + \frac{(4.45-2.635)^2}{7.5655}}$

$5.25\pm 1.960\cdot s_e\cdot \sqrt{1/10 + \frac{(4.45-2.635)^2}{10.499}}$

$5.25\pm 3.355\cdot s_e\cdot \sqrt{1/10 + \frac{(4.45-2.635)^2}{10.499}}$

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 Python for the measurements of the systolic blood pressure:

x = np.array([143,134,138,138,135,131,135,139,141,143,142,141,149,140])   k = 10000   mybootsrapsamples = np.random.choice(x, size=(k, len(x)), replace=True)   mybootstrapmeans = np.mean(mybootsrapsamples, axis=1)

And eight different percentiles of the bootstrap distribution was:

print(np.quantile(mybootstrapmeans, [0.005,0.01,0.025,0.05,0.95,0.975,0.99,0.995],
method='averaged_inverted_cdf'))

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)

$\left[136.5715 ; 141.8571 \right]$

$\left[ 135.5294; 142.8992 \right]$

$\left[136.5000;141.9286\right]$

$\left[136.9286;141.5714\right]$

$\left[136.2143; 142.2143\right]$

02402 · Test Quiz 8

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