Test Quiz 6

The life time of batteries purchased from two different producers is investigated. The lifetime of a number of batteries from each manufacturer is recorded and the following data is available:

Producer A: ${n_A} = 15\quad \left( { { {\bar x}_A},{s_A}} \right) = \left( {122.4;30.5} \right)$

Producer B: ${n_B} = 10\quad \left( { { {\bar x}_B},{s_B}} \right) = \left( {145.9;22.3} \right)$

Data is assumed to follow a normal distribution in each group. What is the test statistic for and the conclusion about the hypothesis $ {\mu_A} = {\mu}_B $: (Both $t$-value and the conclusion must be valid)

$t_{ {obs}} = - 2.22$, H$_{0}$ is rejected on a 5\% significance level

The t-statistic becomes (the default, and so-called, whelch 2-sample version): $t_{ {obs}}=\frac{122.4-145.9}{\sqrt{30.5^2/15+22.3^2/10}}=$ In R: > (122.4-145.9)/(sqrt(30.5^2/15+22.3^2/10)) [1] -2.223067

So $t_{ {obs}}=-2.09$ and we need the degrees of freedom, her found in R:

		ms <- c(122.4, 145.9)
		vs <- c(30.5^2, 22.3^2)
		ns <- c(15, 10)
		tobs <- (ms[1]-ms[2])/sqrt(vs[1]/ns[1]+vs[2]/ns[2])
		nu <- ((vs[1]/ns[1]+vs[2]/ns[2])^2)/((vs[1]/ns[1])^2/(ns[1]-1)+(vs[2]/ns[2])^2/(ns[2]-1))
		nu
		tobs
		> nu
		[1] 22.72472
		> tobs
		[1] -2.223067

since $t_{0.975}(22.73)=2.07$ (in R: \texttt{qt(0.975,22.73)}), the answer is 1: (The observed $t_{\textup{obs}}$ is more extreme than the critical $t$)

		$t =  - 2.22$, H$_{0}$ is rejected on a 5% significance level

$t_{ {obs}} = - 2.09$, H$_{0}$ is rejected on a 5% significance level

$t_{ {obs}} = - 2.09$, H$_{0}$ is rejected on a 1% significance level

$t_{ {obs}} = - 2.22$, H$_{0}$ is accepted on a 1% significance level

$t_{ {obs}} = - 2.22$, H$_{0}$ is accepted on a 5% significance level

Data in the table below comes from a study where 14 patient’s systolic blood pressure was measured (in mmHg) before a certain medical drug treatment and 1 month after starting the treatment.

Patient no	1	2	3	4	5	6	7	8	9	10	11	12	13	14
Blood pressure before treatment	155	164	177	167	156	198	155	187	185	177	150	145	189	176
Blood pressure after treatment	165	154	172	156	150	185	145	182	165	166	145	150	169	172

A 95% confidence interval for the change in blood pressure becomes:

$170.07 - 162.57 \pm 1.960 \cdot {\textstyle{ {8.21} \over {\sqrt {14} }}} = \left[ {3.20;11.80} \right]$

$170.07 - 162.57 \pm 2.056 \cdot {\textstyle{ {8.21} \over {\sqrt {14} }}} = \left[ {2.99;12.01} \right]$

$170.07 - 162.57 \pm 2.160 \cdot {\textstyle{ {8.21} \over {\sqrt {14} }}} = \left[ {2.76;12.24} \right]$

This is a paired 2-sample situation, which means that the 14 set of differences are the data to consider. Hence the relevant critical t-value becomes $t_{0.025}(13)=2.160$, and the only possible correct answer is 3:

\[170.07 - 162.57 \pm 2.160 \cdot{\textstyle{ {8.21} \over {\sqrt {14} }}} = \left[ {2.76;12.24} \right]\]

By the way it could also be found in R as:

		x1=c(155,164,177,167,156,198,155,187,185,177,150,145,189,176)
		x2=c(165,154,172,156,150,185,145,182,165,166,145,150,169,172)
		
		mean(x1)
		mean(x2)
		ds=x1-x2
		sd(ds)
		mean(ds)
		t.test(ds)

$170.07 - 162.57 \pm 2.056 \cdot 14.78 \cdot \sqrt { {\textstyle{1 \over {14}}} + {\textstyle{1 \over {14}}}} = \left[ { - 3.99;18.99} \right]$

$170.07 - 162.57 \pm 1.960 \cdot 14.78 \cdot \sqrt { {\textstyle{1 \over {14}}} + {\textstyle{1 \over {14}}}} = \left[ { - 3.45;18.45} \right]$

In a sports study one wants to investigate whether there is a difference in energy consumption for various types of training. We have (for a single person) measured the energy consumed in 10 jogs of 30 minutes and 10 bike rides of 30 minutes. Measurements, expressed in kcal, is given in the table below:

Jogs	Bike rides
314	294
340	317
331	317
333	310
329	327
322	300
332	293
330	321
338	307
325	304

The following R code was run:

	x1=c(314,340,331,333,329,322,332,330,338,325)
	x2=c(294,317,317,310,327,300,293,321,307,304)
	var(x1)
	var(x2)
	t.test(x1,x2)
	t.test(x1,x2,pair=T,mu=20)

with the following results:

> var(x1)
[1] 57.82222
> var(x2)
[1] 132
> t.test(x1,x2)

Welch Two Sample t-test

data:  x1 and x2
t = 4.6823, df = 15.615, p-value = 0.0002658
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 	11.14535 29.65465
sample estimates:
mean of x mean of y 
	329.4     309.0 

> t.test(x1,x2,pair=T,mu=20)

	Paired t-test

data:  x1 and x2
t = 0.1209, df = 9, p-value = 0.9064
alternative hypothesis: true difference in means is not equal to 20
95 percent confidence interval:
 	12.91852 27.88148
sample estimates:
mean of the differences 
                   20.4

Consider only data for the bike rides. Denote the true mean energy consumption by bike rides $ \mu $. A 99% confidence interval for $ \mu $ is:

$0\pm 309\cdot 132/9$

$309\pm 1.96\cdot 132/10$

$309\pm 3.250\cdot \sqrt{132/10}$

$309\pm 2.326\cdot \sqrt{132/9}$

$309\pm 2.821 \cdot \sqrt{132/10}$

Let $ X_ {1} $ describe the weight of a random cola bottle filled with 0.25 l. Let $ X_ {2} $ describe the weight of a random empty cola bottle. It is assumed that these variables are normally distributed and independent with mean and variance as given here:

\[{X_1}:\quad \;({\mu _{ {X_1}}};\sigma _{ {X_1}}^2) = (626.9 g;4.{2^2} {g^2})\;\quad \mbox{and} \quad {X_2}:\quad ({\mu _{ {X_2}}};\sigma _{ {X_2}}^2) = (364.7 g;3.{6^2} {g^2})\;\]

The colas are sold in boxes of 24 bottles. It is assumed that the box weight is 1500 g (and potential variation of this number is ignored). The probability that the weight of a random box of filled colas is under 16500 g becomes:

0.027

0.013

0.325

0.973

0.675

The probability density functions of three normal distributions are shown below:

normals

It is given that the standard deviation of the distribution in the middle is $\sigma = 0.3$. If you have to mention the three distributions in order after the size of the standard deviation, where you mention the distribution with the lowest standard deviation first and the distribution with the largest standard deviation at last, the correct order is?

a, b, c

a, c, b

b, c, a,

c, a, b

b, a, c

In order for a taxi company to make money, the number of customers for a given taxi driver should at least be 4 per. hour. In a period of 3 hours, 9 costumers were observed at a random selected taxi driver. It is desired to investigate whether there is reason to believe that the taxi company lose money on this taxi driver in the long run?

The $p$-value for such a test is given by:

The $p$-value is equal to $P(X \leq 9)=0.242$, where $X \sim \mathit{Po}(12)$ (Poisson distributed)

The $p$-value is equal to $1-P(X \ne 8)=0.845$, where $X \sim \mathit{Po}(12)$

The $p$-value is equal to $P(X \leq 3)=0.433$, where $X \sim \mathit{Po}(4)$

The $p$-value is equal to $1-P(X \ne 2)=0.762$, where $X \sim \mathit{Po}(4)$

This is not possible to answer from the above information

The probability density function of a normal distribution is shown below:

It is given that the standard deviation of the distribution is $\sigma = 0.3$.

The 2.5% and 97.5% percentile for the distribution in the middle is approximately?

$\pm 0.6$

$\pm 1.6$

$\pm 1.96$

$\pm 1$

$\pm 0.3$

02323 · Test Quiz 6

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Jogs	Bike rides
314	294
340	317
331	317
333	310
329	327
322	300
332	293
330	321
338	307
325	304

Jogs	Bike rides
314	294
340	317
331	317
333	310
329	327
322	300
332	293
330	321
338	307
325	304

Jogs	Bike rides
314	294
340	317
331	317
333	310
329	327
322	300
332	293
330	321
338	307
325	304