Test Quiz 11

Five different brands of tablets with the same active compound are compared with respect to their solubility. For each brand four tablets were investigated. For each tablet, percent solubility is measured after the tablet have been kept in 1000 ml de-ionized water for a while. The following data are found:

A	B	C	D	E
39	36	48	60	30
43	37	46	65	32
38	37	43	62	39
35	35	41	67	31

In the following table for a usual oneway analysis of variance, some of the results are shown. Further, the means for the brands are given. (It can be assumed that the data follows a normal distribution with the same variance in each group)

Source	DF	SS	MS	F
Brand	$x$	$w$	585.925	61.2465
Error	$y$	143.50	9.567
Total	$z$	2487.20

The means of the five brands are:

Brand	Number	Mean
A	4	38.75
B	4	36.25
C	4	44.50
D	4	63.50
E	4	33.00

The values for $x$, $y$, $z$ and $w$ are:

$x = 15$ , $y = 4$ , $z = 19$ and $w = 2343.7$

$x = 4$ , $y = 15 $, $z = 19$ and $w = 585.925$

$x = 10$ , $y = 5$ , $z = 19$ and $w = 2343.7$

$x = 4$ , $y = 15$ , $z = 19 $ and $w = 2343.7 $

$x = 4$ , $y = 16 $, $z = 20$ and $w = 585.925$

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

Five different brands of tablets with the same active compound are compared with respect to their solubility. For each brand four tablets were investigated. For each tablet, percent solubility is measured after the tablet have been kept in 1000 ml de-ionized water for a while. The following data are found:

A	B	C	D	E
39	36	48	60	30
43	37	46	65	32
38	37	43	62	39
35	35	41	67	31

In the following table for a usual oneway analysis of variance, some of the results are shown. Further, the means for the brands are given. (It can be assumed that the data follows a normal distribution with the same variance in each group)

Source	DF	SS	MS	F
Brand	$x$	$w$	585.925	61.2465
Error	$y$	143.50	9.567
Total	$z$	2487.20

The means of the five brands are:

Brand	Number	Mean
A	4	38.75
B	4	36.25
C	4	44.50
D	4	63.50
E	4	33.00

The result of the hypothesis test of no difference in mean solubility for the five brands is: (Both conclusion and argument must be correct)

At least one mean value is different from one of the others since the P-value is $ \approx $0

All means are different from each other since the P-value is $ \approx $1

No means are different from each other since the P-value is $ \approx $0

At least one mean value is different from one of the others since the P-value is $ \approx $1

All means are different from each other since the P-value is $ \approx $0

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

Five different brands of tablets with the same active compound are compared with respect to their solubility. For each brand four tablets were investigated. For each tablet, percent solubility is measured after the tablet have been kept in 1000 ml de-ionized water for a while. The following data are found:

A	B	C	D	E
39	36	48	60	30
43	37	46	65	32
38	37	43	62	39
35	35	41	67	31

In the following table for a usual oneway analysis of variance, some of the results are shown. Further, the means for the brands are given. (It can be assumed that the data follows a normal distribution with the same variance in each group)

Source	DF	SS	MS	F
Brand	$x$	$w$	585.925	61.2465
Error	$y$	143.50	9.567
Total	$z$	2487.20

The means of the five brands are:

Brand	Number	Mean
A	4	38.75
B	4	36.25
C	4	44.50
D	4	63.50
E	4	33.00

The 99% confidence interval for the single pre-planned comparison $\mu_A-\mu_B$ is:

$2.5\pm 3.09$

$2.5\pm 1.55$

$2.5\pm 4.29$

$2.5\pm 3.83$

$2.5\pm 6.44$

It is known that the quotient between surface ares and volume has an influence on how fast a tablet is dissolved. To investigate this the following data was collected:

Surface area/Volume ($m{m^2}/m{m^3})$ (x)	0.60	0.75	0.90	1.05	1.50
% dissolved (y)	33.00	36.25	38.75	44.50	63.50

In R the following has been run:

x=c(0.6,.75,0.9,1.05,1.5)
y=c(33,36.25,38.75,44.50,63.5)
summary(lm(y~x))

with the following result:

Call:
lm(formula = y ~ x)

Residuals:
  	1       2       3       4       5 
 	2.2358  0.3042 -2.3774 -1.8090  1.6462 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)   
(Intercept)   10.038      3.448   2.911  0.06193 . 
x             34.544      3.419  10.104  0.00206 **
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 

Residual standard error: 2.361 on 3 degrees of freedom
Multiple R-squared: 0.9715,     Adjusted R-squared: 0.9619 
F-statistic: 102.1 on 1 and 3 DF,  p-value: 0.002065

A 95% confidence interval for the slope in the model behind the analysis is:

$34.544 \pm 1.96\cdot 2.911 $

$10.038 \pm 3.448\cdot 2.911 $

$34.544 \pm 3.182\cdot 3.419$

$10.038 \pm 1.96 \cdot 3.411 $

$10.104 \pm 2.353\cdot 2.911 $

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

It is known that the quotient between surface ares and volume has an influence on how fast a tablet is dissolved. To investigate this the following data was collected:

Surface area/Volume ($m{m^2}/m{m^3})$ (x)	0.60	0.75	0.90	1.05	1.50
% dissolved (y)	33.00	36.25	38.75	44.50	63.50

In R the following has been run:

x=c(0.6,.75,0.9,1.05,1.5)
y=c(33,36.25,38.75,44.50,63.5)
summary(lm(y~x))

with the following result:

Call:
lm(formula = y ~ x)

Residuals:
  	1       2       3       4       5 
 	2.2358  0.3042 -2.3774 -1.8090  1.6462 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)   
(Intercept)   10.038      3.448   2.911  0.06193 . 
x             34.544      3.419  10.104  0.00206 **
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 

Residual standard error: 2.361 on 3 degrees of freedom
Multiple R-squared: 0.9715,     Adjusted R-squared: 0.9619 
F-statistic: 102.1 on 1 and 3 DF,  p-value: 0.002065

For a random tablet of some other brand, where the surface area/volume = 1.25 $m{m^2}/m{m^3}$ it is found that \% dissovled= 47. Is this in correspondance with the given model?

Yes, since the prediction interval $44.582 \pm 3.182 \cdot 2.361 \cdot \sqrt {1 + {\textstyle{1 \over 5}} + {\textstyle{ {(1.00 - 0.96){}^2} \over {0.477}}}} $ contains the value

No, since the confidence interval $53.218 \pm 3.182 \cdot 2.361 \cdot \sqrt { {\textstyle{1 \over 5}} + {\textstyle{ {(1.25 - 0.96){}^2} \over {0.477}}}} $ does not contain the value

Yes, since the prediction interval $53.218 \pm 3.182 \cdot 2.361 \cdot \sqrt {1 + {\textstyle{1 \over 5}} + {\textstyle{ {(1.25 - 0.96){}^2} \over {0.477}}}} $ contains the value

Yes, since the prediction interval $53.218 \pm 1.960 \cdot 2.361 \cdot \sqrt {1 + {\textstyle{1 \over 5}} + {\textstyle{ {(1.25 - 0.96){}^2} \over {0.477}}}} $ contains the value

Yes, since the confidence interval $44.582 \pm 1.960 \cdot 2.361 \cdot \sqrt { {\textstyle{1 \over 5}} + {\textstyle{ {(1.00 - 0.96){}^2} \over {0.477}}}} $ does not contain the value

In the production of a certain foil (film), the foil is controlled by measuring the thickness of the foil in a number of points distributed over the width of the foil. The production is considered stable if the mean of the difference between the maximum and minimum measurements does not exceed 0.35mm. At a given day, the following data are found for 10 foils:

Foil	1	2	3	4	5	6	7	8	9	10
Max. measurement in mm ($y_{max}$)	2.62	2.71	2.18	2.25	2.72	2.34	2.63	1.86	2.84	2.93
Min. measurement in mm ($y_{min}$)	2.14	2.39	1.86	1.92	2.33	2.00	2.25	1.50	2.27	2.37
Max-Min($D$)	0.48	0.32	0.32	0.33	0.39	0.34	0.38	0.36	0.57	0.56

The following statistics may potentially be used: $\bar{y}_{max}=2.508,\;\bar{y}_{min}=2.103,\;s_{y_{max}}=0.3373,\;s_{y_{min}}=0.2834,\;s_{D}=0.09664$

If the following hypothesis test on the mean difference is carried out:

\[\begin{array}{l} {H_0}:{\mu_D} = 0.35\\ {H_1}:{\mu_D} \not= 0.35 \end{array}\]

The critical value for the t-test of this hypothesis: on level $\alpha=0.01$ is: (expressed by R-functions)

qt(0.995,9) $(=3.250)$

The critical value for the test is $t_{\alpha}(9)=t_{0.995}(9)$ (it is a “matched pair’’ sample situation - that is the “paired’’ t-test is what we are using now - so counting the number of differences, $n=10$) and hence the correct answer is:

[1:]  qt(0.995,9) $(=3.250)$

qnorm(0.995) $(=2.576)$

qt(0.995,18) $(=2.878)$

qt(0.005,18) $(=-2.878)$

qnorm(0.025) $(=-1.96)$

02323 · Test Quiz 11

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