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Ket [755]
3 years ago
10

Listed below are annual data for various years. The data are weights​ (metric tons) of imported lemons and car crash fatality ra

tes per​ 100,000 population. Construct a​ scatterplot, find the value of the linear correlation coefficient​ r, and find the​P-value using α=0.05. Is there sufficient evidence to conclude that there is a linear correlation between lemon imports and crash fatality​ rates? Do the results suggest that imported lemons cause car​ fatalities?
Lemon_Imports_(x) Crash_Fatality_Rate_(y)
230 15.8
264 15.6
359 15.5
482 15.3
531 14.9

1. What are the null and alternative​ hypotheses?
2. Construct a scatterplot.
3. The linear correlation coefficient r is
4. The test statistic t is
5. The​ P-value is

Because the​ P-value is ____ than the significance level 0.05​, there ____ sufficient evidence to support the claim that there is a linear correlation between lemon imports and crash fatality rates for a significance level of α=0.05.

Do the results suggest that imported lemons cause car​fatalities?

A. The results suggest that an increase in imported lemons causes car fatality rates to remain the same.
B. The results do not suggest any​ cause-effect relationship between the two variables.
C. The results suggest that imported lemons cause car fatalities.
D. The results suggest that an increase in imported lemons causes in an increase in car fatality rates.

Mathematics
1 answer:
anyanavicka [17]3 years ago
4 0

Answer:

Because the​ P-value is _(<u>0.02)  less </u> than the significance level 0.05​, there <u> is </u> sufficient evidence to support the claim that there is a linear correlation between lemon imports and crash fatality rates for a significance level of α=0.05.

C. The results suggest that imported lemons cause car fatalities.

Step-by-step explanation:

Hello!

The study variables are:

X₁: Weight of imported lemons.

X₂: Car crash fatality rate.

The objective is to test if the imported lemons affect the occurrence of car fatalities. To do so you are asked to use a linear correlation test.

I've made a Scatterplot with the given data, it is attached to the answer.

To be able to use the parametric linear correlation you can use the parametric test (Person) or the nonparametric test Spearman. For Person, you need your variables to have a bivariate normal distribution. Since one of the variables is a discrete variable (ratio of car crashes) and the sample is way too small to make an approximation to a normal distribution, the best test to use is Spearman's rank correlation.

This correlation coefficient (rs) takes values from -1 to 1

If rs = -1 this means that there is a negative correlation between the variables

If rs= 1 this means there is a positive correlation between the variables

If rs =0 then there is no correlation between the variables.

The hypothesis is:

H₀: There is no linear association between X₁ and X₂

H₁: There is a linear association between X₁ and X₂

α: 0.05

To calculate the Spearman's correlation coefficient you have to assign ranks to the observed values of each variable, from the smallest to the highest). Then you have to calculate the difference (d)between the ranks and the square of that difference (d²). (see attachment)

The formula for the correlation coefficient is:

rs= 1 - \frac{6* (sum of d^2)}{(n-1)n(n+1)}

rs= 1 - \frac{6* (40)}{4*5*6}

rs= -1

For this value of the correlation coefficient, the p-value is 0.02

Since the p-value (0.02) is less than the significance level (0.05) the decision is to reject the null hypothesis. In other words, there is a linear correlation between the imported lemons and the car crash fatality ration, this means that the modification in the lemon import will affect the car crash fatality ratio.

Note: the correlation coefficient is negative, so you could say that there is a correlation between the variables and this is negative (meaning that when the lemon import increases, the car crash fatality ratio decreases)

I hope it helps!

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Answer:

El perímetro de la región impresa es 72 cm y su área es 288 cm².  

Step-by-step explanation:

1. Tenemos el perímetro de la hoja de papel:

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Como sabemos el margen superior, inferior, izquierdo y derecho podemos encontrar la relación entre el largo y ancho del rectángulo interno (región impresa) con el largo (l) y ancho (a) del rectángulo externo (hoja de papel):      

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a_{2} = a_{1} - (m_{d} + m_{iz}) = a_{1} - (2 cm + 5 cm) = a_{1} - 7 cm   (3)    

El perímetro del rectángulo interno es:

P_{2} = 2l_{2} + 2a_{2}    (4)

Introduciendo la ecuación (2) y (3) en (4):

P_{2} = 2l_{2} + 2a_{2} = 2(l_{1} - 5 cm) + 2(a_{1} - 7 cm) = 2l_{1} + 2a_{1} - 10 cm - 14 cm = 96 cm - 24 cm = 72 cm  

Por lo tanto el perímetro del rectángulo interno (región impresa) es 72 cm.

 

2. Ahora para encontrar el área rectángulo interno debemos encontrar el largo y ancho del mismo, sabiendo que:

l_{2} = 2a_{2}     (5)

Introduciendo (5) en (4):

P_{2} = 2l_{2} + 2a_{2} = 2*2a_{2} + 2a_{2} = 6a_{2}

a_{2} = \frac{P_{2}}{6} = \frac{72 cm}{6} = 12 cm

l_{2} = 2a_{2} = 2*12 cm = 24 cm

Entonces el área es:

A_{2} = l_{2}*a_{2} = 12 cm*24 cm = 288 cm^{2}

Por lo tanto el área del rectágulo interno (región impresa) es 288 cm².      

Espero que te sea de utilidad!  

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