Exercise Quiz 13

By illegal importation of dogs from Eastern Europe, often dogs are received that are not vaccinated against a specific disease. A new product is on the market for the treatment of non-vaccinated dogs, if they are attacked by the disease.

So far the disease has caused a great mortality among imported dogs, but the producer of the new product claims that at least 80% of infected dogs will survive if treated with the product. In an animal clinique 30 dogs were treated with the new product, and 19 of them were cured.

Does this result disprove the manufacturer’s claim, if a significance level of $ \alpha = 5\% $ is used (both the conclusion and the argument must be in order)?

Yes, since the $p$-value $= P(X\leq 19)=0.0256$, where $X$ is a $B(x;30,0.8)$-distribution

No, since the $p$-value $= P(X\leq 19)=0.0256$, where $X$ is a $B(x;30,0.8)$-distribution

No, since the $p$-value $= P(X\leq 19)=0.9506$, where $X$ is a $B(x;30,0.5)$-distribution

Yes, since the $p$-value $= P(X\geq 20)=0.0494$, where $X$ is a $B(x;30,0.5)$-distribution

No, since the $p$-value $= P(X\geq 20)=0.9744$, where $X$ is a $B(x;30,0.8)$-distribution

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

By illegal importation of dogs from Eastern Europe, often dogs are received that are not vaccinated against a specific disease. A new product is on the market for the treatment of non-vaccinated dogs, if they are attacked by the disease.

So far the disease has caused a great mortality among imported dogs, but the producer of the new product claims that at least 80% of infected dogs will survive if treated with the product. In an animal clinique 30 dogs were treated with the new product, and 19 of them were cured.

Veterinarians are required to report cases of illness. In one period, there were on average 10 reported cases of illness per workday. If you use the usual probability distribution for the number of events per time unit with an average of 10 per day, the probability of getting 16 reported cases or more on any given day is?

0.118

0.951

0.049

0.882

0.027

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 0.35 mm. 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 ($y_{\text{max}}$)	2.62	2.71	2.18	2.25	2.72	2.34	2.63	1.86	2.84	2.93
Min. measurement in mm ($y_{\text{min}}$)	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: $\bar{y}_{\text{max}}=2.508,\;\bar{y}_{\text{min}}=2.103,\;s_{y_{\text{max}}}=0.3373,\;s_{y_{\text{min}}}=0.2834,\;s_{D}=0.09664$

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

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

we get the following $t$-statistic:

$1.96$

$\frac{\sqrt{9}(2.508-2.103)}{\sqrt{(0.3372^2+0.2834^2)/2}}=3.90 $

$\frac{\sqrt{10}(2.508-2.103-0.35)}{\sqrt{(0.3372^2+0.2834^2)/2}}=0.56 $

$\frac{2.508-2.103}{(0.3372+0.2834)/(2\cdot\sqrt{10})}= 4.12$

$\frac{2.508-2.103-0.35}{0.09664/\sqrt{10}}= 1.80$

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

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 0.35 mm. 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 ($y_{\text{max}}$)	2.62	2.71	2.18	2.25	2.72	2.34	2.63	1.86	2.84	2.93
Min. measurement in mm ($y_{\text{min}}$)	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: $\bar{y}_{\text{max}}=2.508,\;\bar{y}_{\text{min}}=2.103,\;s_{y_{\text{max}}}=0.3373,\;s_{y_{\text{min}}}=0.2834,\;s_{D}=0.09664$

For one of the foils produced, the requirement is that the mean thickness must be $ \mu = 2.55$mm with a standard deviation of $\sigma = 0.10$mm. How many thickness measurements must be made, if it is required that a 95% confidence interval for the foil thickness has a width of 0.1mm maximum:

85, since $(1.65\cdot 2.55/0.05)=84.15$

2498, since $(1.96\cdot 2.55/0.10)^2=2498$

16, since $(1.96\cdot 0.1/0.05)^2=15.37$

1, since $(1.65\cdot 0.05/0.1)^2=0.68$

6, since $(2.262\cdot 0.1/0.1)^2=5.120036$

As part of the food quality control in a supermarket 5 fresh chickens with different origin than Denmark is sampled from the cold counter for control. It is assumed that there is a total of 15 chickens in the counter and that three of these are infected with salmonella. The random variable $X$ describes the number of chickens infected with salmonella among the 5 sampled.

What is the mean and standard deviation of $X$?

$\mu_X=3$ and $\sigma_X=4.47$

$\mu_X=1$ and $\sigma_X=0.894$

$\mu_X=1$ and $\sigma_X=0.200$

$\mu_X=1$ and $\sigma_X=0.800$

$\mu_X=1$ and $\sigma_X=0.756$

As part of the food quality control in a supermarket 5 fresh chickens with different origin than Denmark is sampled from the cold counter for control. It is assumed that there is a total of 15 chickens in the counter and that three of these are infected with salmonella. The random variable $X$ describes the number of chickens infected with salmonella among the 5 sampled.

Chickens are received in batches of 1500. A batch is controlled by sampling 15 chickens for control and the batch is accepted if all the controlled chickens are free of salmonella. Assume that the chickens with different origin than Denmark contains 10% chickens infected with salmonella. Assume that the Danish produced chicken contains 1% chickens infected with salmonella.

The probability of accepting a batch, if chickens have another origin than Denmark respectively if it is Danish-produced chickens are:

Another origin: between $0$ og $1\%$. Denmark: around $50\%$

Another origin: between $20$ og $21\%$. Denmark: around $86\%$

The probability of accepting a batch is $P(X=15)$ where is the random variable consisting of the number of chickens free of salmonella in each case. Formally, the correct distribution would be a hypergeometric for each case. However, due to the high number of chickens in a batch ($N=1500$) compared to the sample size $(n=15)$ the probabilities can nicely be approximated by binomial probabilities:

For another origin $P(X=15)=0.90^{15}=0.2059$ In R this could be found in any of three following ways: (note that the last call gives the probability using the hypergeometric distribution)

> dbinom(15, 15 , 0.9) 
[1] 0.2058911
> dbinom(0, 15 , 0.1) 
[1] 0.2058911
> dhyper(15, 1350,150 , 15) 
[1] 0.2042851

For Denmark: $P(X=15)=0.99^{15}=0.8601$

In R this could be found in any of three following ways:

> dbinom(15, 15 , 0.99) 
[1] 0.8600584
> dbinom(0, 15 , 0.01) 
[1] 0.8600584
> dhyper(15, 1485,15 , 15) 
[1] 0.8594465

and thus the correct answer is found.

Another origin: around $10\%$. Denmark: around $99\%$

Another origin: around $0\%$. Denmark: around $0\%$

Another origin: around $90\%$. Denmark: around $99\%$

The length of a cut off aluminum profile is supposed to be in the range: $179.8 \text{ mm} \leq L \leq 180.2 \text{ mm}$ It is assumed that the length L of the profiles can be described by a normal random variable with parameters: $\mu_L=180 \text{ mm}\mbox{ and } \sigma_L=0.08 \text{ mm}$

The percentage of the profiles, which have a length which is outside the desired range, will be:

About 50%

About 0.6%

About 2%

About 1.2%

About 0.2%

The length of a cut off aluminum profile is supposed to be in the range: $179.8\text{ mm} \leq L \leq 180.2\text{ mm}$ It is assumed that the length L of the profiles can be described by a normal random variable with parameters: $\mu_L=180\text{ mm}\mbox{ and } \sigma_L=0.08\text{ mm}$

On the profile a tube is to be installed for which it is given that the length of the tube R can be described by a normal random variable with mean $ \mu_R = 180$mm. The requirement for the profile and the tube to fit together is: $-0.25\text{ mm} < L - R < 0.25\text{ mm}$

If this requirement is to be met in 99\% of the cases, the requirement that must be made to the standard deviation $\sigma_R$ on the length of the tube is: ($\mu_L = 180$ mm and $\sigma_L = 0.08$ mm is still assumed)

$\sigma_R=0.055$ mm

$\sigma_R=0.097$ mm

$\sigma_R=2.576$ mm

$\sigma_R=0.250$ mm

$\sigma_R=0.500$ mm

02402 · Exercise Quiz 13

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