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N76 [4]
3 years ago
7

Suppose the following regression equation was generated from the sample data of 50 cities relating number of cigarette packs sol

d per 1000 residents in one week to tax in dollars on one pack of cigarettes and if smoking is allowed in bars:
PACKS i= 57221.431732 − 1423.696906TAXi + 155.441784BARSi + ei.
BARS i= 1 if city i allows smoking in bars and BARSi = 0 if city i does not allow smoking in bars. This equation has an R2 value of 0.351292, and the coefficient of BARSi has a P-value of 0.086529. Which of the following conclusions is valid?
A. According to the regression equation, regardless of whether or not smoking is allowed in bars, the number of cigarette packs sold per 1000 people decreases by approximately 1424 for each additional dollar of cigarette tax.
B. There is evidence at the 0.05 level of significance to support the claim that cities with a smoking ban have lower cigarette sales than those without a smoking ban.
C. According to the regression equation, cities that allow smoking in bars have lower cigarette sales than cities that do not allow smoking in bars.
D. According to the regression equation, cities that allow smoking in bars sell approximately 155 fewer packs of cigarettes per 1000 people than cities that do not allow smoking in bars.
Mathematics
1 answer:
frez [133]3 years ago
4 0

Answer:

A) According to the regression equation, regardless of whether or not smoking is allowed in bars, the number of cigarette packs sold per 1000 people decreases by approximately 1424 for each additional dollar of cigarette tax.

Step-by-step explanation:

Given the regression equation:

PACKS i= 57221.431732 − 1423.696906TAXi + 155.441784BARSi + eᵢ.

BARS i= 1 if city i allows smoking in bars

BARSi = 0 if city i does not allow smoking in bars

R2 = 0.351292

P-value = 0.086529

Conlusion:

Simnce p value, 0.0865 is greater than level of significance, 0.05, BARS is not significant. Thus, allowing smoking in bars increase cigarette sales, since the coefficient of BARS is positive.

Correct answer is option A.

According to the regression equation, regardless of whether or not smoking is allowed in bars, the number of cigarette packs sold per 1000 people decreases by approximately 1424 for each additional dollar of cigarette tax.

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Step-by-step explanation:

<u><em>Step(i):-</em></u>

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      \frac{5+2\sqrt{3} }{7+4\sqrt{3} }

Rationalize

=    \frac{5+2\sqrt{3} }{7+4\sqrt{3} } X \frac{7-4\sqrt{3} }{7-4\sqrt{3} }

=    \frac{(5+2\sqrt{3})(7-4\sqrt{3} ) }{7^{2} -(4\sqrt{3})^{2}  }

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<u><em>Step(ii):-</em></u>

<em>Multiply</em>

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According to one investigator’s model, the data are like 400 draws made at random from a large box. The null hypothesis says tha
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Answer:

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p_v =2*P(Z>2.2)=0.0278  

If we compare the p value and the significance level assumed \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the mean is significantly diffrent fro 50 at 1% of signficance.  

Step-by-step explanation:

1) Data given and notation  

\bar X=52.75 represent the mean height for the sample  

s=25 represent the sample standard deviation

n=400 sample size  

\mu_o =50 represent the value that we want to test

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 50 or not, the system of hypothesis would be:  

Null hypothesis:\mu = 50  

Alternative hypothesis:\mu \neq 50  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value but we can assume it as z distribution, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

z=\frac{52.75-50}{\frac{25}{\sqrt{400}}}=2.2    

P-value

Since is a two sided test the p value would be:  

p_v =2*P(Z>2.2)=0.0278  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the mean is significantly diffrent fro 50 at 1% of signficance.  

3 0
4 years ago
Helped nephew with his homework and would like to know if it’s correct
AleksAgata [21]

the answers that you've provided are, in fact, correct.

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3 years ago
Read 2 more answers
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