# 02402 · Test Quiz 13

## Question 1 of 13

Two different brands of tablets with the same active compound are compared with respect to their solubility. For each of the two brands 10 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. One measurement failed, so the following values for percent solubility were found:

Brand F | 45 | 47 | 48 | 49 | 49 | 50 | 52 | 52 | 53 | 54 |

Brand G | 48 | 48 | 49 | 49 | 52 | 54 | 54 | 55 | 55 |

A 99% confidence interval for the difference between the two means, which is not based on an assumption about normality of the data, is wanted. The following R code is executed:

```
x=c(45,47,48,49,49,50,52,52,53,54)
y=c(48,48,49,49,52,54,54,55,55)
k = 10000
xsamples = replicate(k, sample (x, replace = TRUE))
ysamples = replicate(k, sample (y, replace = TRUE))
mymeandifs = apply(xsamples, 2, mean)-apply(ysamples, 2, mean)
myquantiles=quantile(mymeandifs, c(0.005,0.01,0.025,0.05,0.25,
0.5,0.75,0.95,0.975,0.99,0.995))
round(myquantiles,2)
```

The result, which then is the rounded (the R-function round is in the last line applied to round to two decimal points) percentiles of the bootstrap distribution of differences of means, is:

```
0.5% 1% 2.5% 5% 25% 50% 75% 95% 97.5% 99% 99.5%
-4.93 -4.64 -4.14 -3.79 -2.54 -1.67 -0.80 0.40 0.83 1.29 1.52
```

The 99% confidence interval for the difference between the two means based on this is?

## Question 2 of 13

A fast-food chain uses a biological degradable material for packaging their burgers. The thermal conductivity of the material is an important feature. The data in the table below comes from an experiment where thermal conductivity is measured as a function of the material density. It is assumes that the relationship can be described by a simple linear model. The following values are measured:

Material density (g/cm$^{3})$ | .175 | .220 | .225 | .226 | .250 | .277 |

Thermal conductivity (W/mK) | .0480 | .0525 | .0540 | .0535 | .0570 | .0610 |

Some computational statistics: \(\bar x = 0.2288\,,\,\bar y = 0.05433\,,\,{S_{xx}} = 0.005767,\,{S_{yy}} = 0.00009583 \,\mbox{and} \,\,{S_{xy}} = 0.0007383\)

The following lines were run in R:

```
x=c(.175,.220,.225,.226,.25,.277)
y=c(.048,.0525,.054,.0535,.057,.061)
summary(lm(y~x))
```

with the following results (however, two of the values have been substituted by “A” and “B”):

```
Call:
lm(formula = y ~ x)
Residuals:
1 2 3 4 5 6
5.590e-04 -7.024e-04 1.575e-04 -4.706e-04 -4.332e-05 4.998e-04
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.025036 A 14.42 0.000134 ***
x 0.128031 B 17.03 6.97e-05 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.0005709 on 4 degrees of freedom
Multiple R-squared: 0.9864, Adjusted R-squared: 0.983
F-statistic: 290 on 1 and 4 DF, p-value: 6.973e-05
```

The percentage explained variation $r^{2}$ and the residual standard deviation $s_{e}$ are?

## Question 3 of 13

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

A fast-food chain uses a biological degradable material for packaging their burgers. The thermal conductivity of the material is an important feature. The data in the table below comes from an experiment where thermal conductivity is measured as a function of the material density. It is assumes that the relationship can be described by a simple linear model. The following values are measured:

Material density (g/cm$^{3})$ | .175 | .220 | .225 | .226 | .250 | .277 |

Thermal conductivity (W/mK) | .0480 | .0525 | .0540 | .0535 | .0570 | .0610 |

Some computational statistics: \(\bar x = 0.2288\,,\,\bar y = 0.05433\,,\,{S_{xx}} = 0.005767,\,{S_{yy}} = 0.00009583 \,\mbox{and} \,\,{S_{xy}} = 0.0007383\)

The following lines were run in R:

```
x=c(.175,.220,.225,.226,.25,.277)
y=c(.048,.0525,.054,.0535,.057,.061)
summary(lm(y~x))
```

with the following results (however, two of the values have been substituted by “A” and “B”):

```
Call:
lm(formula = y ~ x)
Residuals:
1 2 3 4 5 6
5.590e-04 -7.024e-04 1.575e-04 -4.706e-04 -4.332e-05 4.998e-04
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.025036 A 14.42 0.000134 ***
x 0.128031 B 17.03 6.97e-05 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.0005709 on 4 degrees of freedom
Multiple R-squared: 0.9864, Adjusted R-squared: 0.983
F-statistic: 290 on 1 and 4 DF, p-value: 6.973e-05
```

From the theory of the tested materials the slope of the line is expected to be $ \beta = 0.155$. Is this in correspondance with the observed slope, if significance level of 5% is used (both answer and argument must be correct)?

## Question 4 of 13

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

A fast-food chain uses a biological degradable material for packaging their burgers. The thermal conductivity of the material is an important feature. The data in the table below comes from an experiment where thermal conductivity is measured as a function of the material density. It is assumes that the relationship can be described by a simple linear model. The following values are measured:

Material density (g/cm$^{3})$ | .175 | .220 | .225 | .226 | .250 | .277 |

Thermal conductivity (W/mK) | .0480 | .0525 | .0540 | .0535 | .0570 | .0610 |

Some computational statistics: \(\bar x = 0.2288\,,\,\bar y = 0.05433\,,\,{S_{xx}} = 0.005767,\,{S_{yy}} = 0.00009583 \,\mbox{and} \,\,{S_{xy}} = 0.0007383\)

The following lines were run in R:

```
x=c(.175,.220,.225,.226,.25,.277)
y=c(.048,.0525,.054,.0535,.057,.061)
summary(lm(y~x))
```

with the following results (note, two of the values have been substituted by “A” and “B”):

```
Call:
lm(formula = y ~ x)
Residuals:
1 2 3 4 5 6
5.590e-04 -7.024e-04 1.575e-04 -4.706e-04 -4.332e-05 4.998e-04
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.025036 A 14.42 0.000134 ***
x 0.128031 B 17.03 6.97e-05 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.0005709 on 4 degrees of freedom
Multiple R-squared: 0.9864, Adjusted R-squared: 0.983
F-statistic: 290 on 1 and 4 DF, p-value: 6.973e-05
```

A 95% confidence interval for the thermal conductivity, if the density is 0.200 (g/cm$^{3}$), becomes:

## Question 5 of 13

In a period of 4 month, the number of male and female participants was registered in a smoking cessation course. It was registered how many of those who after the course had been smoking and how many who had stopped. The following numbers was recorded during the 4 months:

Gender | Still smoking | Stopped smoking |
---|---|---|

Women | 91 | 352 |

Men | 32 | 212 |

Previously completed courses had shown that the probability that a randomly selected participant stopped smoking was 80%.

In a similar smoking cessation course 20 smokers attended. What is the probability that at least 18 out of the 20 participants will stop smoking after participating in this course? (Below $B(x;n,p)$ denotes the probability distribution function for the binomial distribution)

## Question 6 of 13

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

In a period of 4 month, the number of male and female participants was registered in a smoking cessation course. It was registered how many of those who after the course had been smoking and how many who had stopped. The following numbers was recorded during the 4 months:

Gender | Still smoking | Stopped smoking |
---|---|---|

Women | 91 | 352 |

Men | 32 | 212 |

If one randomly asks some participants after a smoking cessation course whether they had stopped smoking, how many participants should at least be asked, in order to get a probability above 50% that at least 1 of those asked STILL was smoking?

## Question 7 of 13

The systolic blood pressure (SBP) was measured on 8 parkinsonian patients and 21 healthy subjects. The purpose of the study was to examine whether there are differences in the SBP between the 2 groups. The following values were calculated for the parkinson group: $\overline{y}_1 = 132.86$ and $s_1 = 15.34$, and for the healthy subjects: $\overline{y}_2 = 127.44$ and $s_2 = 18.23$. It is assumed that the two populations follow the normal distribution.

What is the test statistic of a hypothesis test at a significance level of $\alpha = 0.05$?

## Question 8 of 13

The organizers of Copenhagen marathon want to test whether there is a correlation between how many marathon race the participants previously have completed, and the time they completed the 42.195 km at Copenhagen Marathon in May 2009. In the table below the participants are divided into 3 different groups depending on how many marathons they have completed and the race times are divided into 5 different groups, where n.c.\ stands for not completed:

Race time in hours | [0;3) | [3;4) | [4;5) | [5;6) | n.c. | Total |
---|---|---|---|---|---|---|

0 marathon | 51 | 1281 | 811 | 125 | 194 | 2462 |

$\leq$ 10 marathon | 82 | 1523 | 1077 | 108 | 134 | 2924 |

$>$ 10 marathon | 92 | 1812 | 1298 | 122 | 120 | 3444 |

Total | 225 | 4616 | 3186 | 355 | 448 | 8830 |

The expected frequencies are calculated and given in the table below:

Race time in hours | [0;3) | [3;4) | [4;5) | [5;6) | n.c. | Total |
---|---|---|---|---|---|---|

0 marathon | 63 | 1287 | 888 | 99 | 125 | 2462 |

$\leq$ 10 marathon | 74 | 1529 | 1055 | 118 | 148 | 2924 |

$>$ 10 marathon | 88 | 1800 | 1243 | 138 | 175 | 3444 |

Total | 225 | 4616 | 3186 | 355 | 448 | 8830 |

The test statistic can be calculated as: $\chi^2 = \frac {(51-63)^2}{63}+ \frac {(1281-1287)^2}{1287}+ \dots + \frac {(120-175)^2}{175} = 79.25$

The conclusion for the above test at a significance level of $\alpha = 0.01$ is?

## Question 9 of 13

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

The organizers of Copenhagen marathon want to test whether there is a correlation between how many marathon race the participants previously have completed, and the time they completed the 42.195 km at Copenhagen Marathon in May 2009. In the table below the participants are divided into 3 different groups depending on how many marathons they have completed and the race times are divided into 5 different groups, where n.c.\ stands for not completed:

Race time in hours | [0;3) | [3;4) | [4;5) | [5;6) | n.c. | Total |
---|---|---|---|---|---|---|

0 marathon | 51 | 1281 | 811 | 125 | 194 | 2462 |

$\leq$ 10 marathon | 82 | 1523 | 1077 | 108 | 134 | 2924 |

$>$ 10 marathon | 92 | 1812 | 1298 | 122 | 120 | 3444 |

Total | 225 | 4616 | 3186 | 355 | 448 | 8830 |

The expected frequencies are calculated and given in the table below:

Race time in hours | [0;3) | [3;4) | [4;5) | [5;6) | n.c. | Total |
---|---|---|---|---|---|---|

0 marathon | 63 | 1287 | 888 | 99 | 125 | 2462 |

$\leq$ 10 marathon | 74 | 1529 | 1055 | 118 | 148 | 2924 |

$>$ 10 marathon | 88 | 1800 | 1243 | 138 | 175 | 3444 |

Total | 225 | 4616 | 3186 | 355 | 448 | 8830 |

The test statistic can be calculated as: $\chi^2 = \frac {(51-63)^2}{63}+ \frac {(1281-1287)^2}{1287}+ \dots + \frac {(120-175)^2}{175} = 79.25$

If one had chosen only to divide the race times in the following 3 groups: [0;3), [3;5), and [5;$\infty$) (incl.\ the n.c.\ group), then the critical value at a test with $\alpha= 0.05$, would be:

## Question 10 of 13

The content of the heavy metal Cadmium in canned tuna were measured in tuna from 3 different manufacturers of canned tuna. From each manufacturer 5 cans of tuna were randomly selected. The following quantities of Cadmium in $\mu$g$/$kg were measured in the 15 different cans of tuna:

Manufacturer 1 | 57 | 52 | 62 | 49 | 43 |

Manufacturer 2 | 55 | 74 | 62 | 42 | 52 |

Manufacturer 3 | 61 | 54 | 55 | 53 | 51 |

The aim is to test whether there is a difference in the amount of Cadmium in canned tuna from the 3 manufacturers. The table below shows the result of an analysis of variance of content of Cadmium in the 15 cans:

Source | DF | SS | MS | F |
---|---|---|---|---|

Manufacturer | 48.4 | 24.2 | 0.35 | |

Error | 838.0 | 69.8 | ||

Total | 886.4 |

The column with degrees of freedom which is not filled in is (mentioned in the order manufacturer, error, total):

## Question 11 of 13

A fast-food chain uses a biodegradable material for packaging of their burgers. The heat conductivity of the material is an essential characteristic. The data in the table below comes from a study where heat conductivity is measured as a function of the density of the material. It is assumed that the relationship can be described by a simpl linear regression model. The following values are measured:

The density of the product (g/cm$^3$) | .175 | .220 | .225 | .226 | .250 | .277 |

Heat conductivity (W/mK) | .0480 | .0525 | .0540 | .0535 | .0570 | .0610 |

The following R-code was run:

```
x=c(.175,.220,.225,.226,.250,.277)
y=c(.0480,.0525,.0540,.0535,.0570,.0610)
summary(lm(y~x))
```

It gave the following output:

```
all:
lm(formula = y ~ x)
Residuals:
1 2 3 4 5 6
5.590e-04 -7.024e-04 1.575e-04 -4.706e-04 -4.332e-05 4.998e-04
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.025036 0.001736 14.42 0.000134 ***
x 0.128031 0.007518 17.03 6.97e-05 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.0005709 on 4 degrees of freedom
Multiple R-squared: 0.9864, Adjusted R-squared: 0.983
F-statistic: 290 on 1 and 4 DF, p-value: 6.973e-05
```

What is the correlationen between the density and the heat conductivity?

## Question 12 of 13

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

A fast-food chain uses a biodegradable material for packaging of their burgers. The heat conductivity of the material is an essential characteristic. The data in the table below comes from a study where heat conductivity is measured as a function of the density of the material. It is assumed that the relationship can be described by a simpl linear regression model. The following values are measured:

The density of the product (g/cm$^3$) | .175 | .220 | .225 | .226 | .250 | .277 |

Heat conductivity (W/mK) | .0480 | .0525 | .0540 | .0535 | .0570 | .0610 |

The following R-code was run:

```
x=c(.175,.220,.225,.226,.250,.277)
y=c(.0480,.0525,.0540,.0535,.0570,.0610)
summary(lm(y~x))
```

It gave the following output:

```
all:
lm(formula = y ~ x)
Residuals:
1 2 3 4 5 6
5.590e-04 -7.024e-04 1.575e-04 -4.706e-04 -4.332e-05 4.998e-04
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.025036 0.001736 14.42 0.000134 ***
x 0.128031 0.007518 17.03 6.97e-05 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.0005709 on 4 degrees of freedom
Multiple R-squared: 0.9864, Adjusted R-squared: 0.983
F-statistic: 290 on 1 and 4 DF, p-value: 6.973e-05
```

Which statement below is the only correct one?

## Question 13 of 13

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

A fast-food chain uses a biodegradable material for packaging of their burgers. The heat conductivity of the material is an essential characteristic. The data in the table below comes from a study where heat conductivity is measured as a function of the density of the material. It is assumed that the relationship can be described by a simpl linear regression model. The following values are measured:

The density of the product (g/cm$^3$) | .175 | .220 | .225 | .226 | .250 | .277 |

Heat conductivity (W/mK) | .0480 | .0525 | .0540 | .0535 | .0570 | .0610 |

The following R-code was run:

```
x=c(.175,.220,.225,.226,.250,.277)
y=c(.0480,.0525,.0540,.0535,.0570,.0610)
summary(lm(y~x))
```

It gave the following output:

```
all:
lm(formula = y ~ x)
Residuals:
1 2 3 4 5 6
5.590e-04 -7.024e-04 1.575e-04 -4.706e-04 -4.332e-05 4.998e-04
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.025036 0.001736 14.42 0.000134 ***
x 0.128031 0.007518 17.03 6.97e-05 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.0005709 on 4 degrees of freedom
Multiple R-squared: 0.9864, Adjusted R-squared: 0.983
F-statistic: 290 on 1 and 4 DF, p-value: 6.973e-05
```

The density was measured for five other samples of material, and the following measurements was obtained $ (g/cm^3)$: \(0.179, 0.181, 0.201, 0.280, 0.282\) The hypothesis that the mean value of these is the same as the mean value of the density of the original 6 samples is to be tested. For that purpose, the following R-code was run:

```
x1=c(0.175,0.220,0.225,0.226,0.250,0.277)
x2=c(0.179,0.181,0.201,0.280,0.282)
k=10000
x1samples=replicate(k,sample(x1,replace=TRUE))
x2samples=replicate(k,sample(x2,replace=TRUE))
mymeandifs=apply(x1samples,2,mean)-apply(x2samples,2,mean)
hist(mymeandifs)
```

and the histogram for the 10000 bootstrap outcomes of mean differences became:

What is to be concluded? (as well conclusion as argument must be valid)