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masya89 [10]
3 years ago
11

"The majority of U.S. car owners still buy American cars."

Mathematics
1 answer:
Vikki [24]3 years ago
3 0

Question:

The majority of U.S. car owners still buy American cars." Ideally, what would the sample be in the study that generated this stat?

answer choices

a) Everyone in the United States who owns a car

b) Everyone in the United States who owns an American car

c) A randomly selected group of car owners in the U.S.

d) A randomly selected group of Americans who own American cars

Answer:

A randomly selected group of car owners in the U.S.

Step-by-step explanation:

In this study, we are told the majority of U.S car owners still buy American cars. The sample in the study that generated this stat would be "a randomly selected group of car owners in the U.S".

⬤ Option A is incorrect, because collecting a sample from every car owner in U.S is impossible as the sample size would be rather tool large.

⬤ Option B is also incorrect because just like in option A, the sample size would be too large. Also this does not  take into consideration those who don't use American cars

⬤ Option D is incorrect, this study deals with all car owners in the U.S, not Americans who own American cars. Therefore selecting a sample from Americans who own American cars would be wrong.

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What’s the answer for -3(r-4)_>0
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Answer:

r<4

Step-by-step explanation:

-3(r-4)>0 step 1: distribute the -3

-3r+12>0  explain: -3 times r =-3r, -3 times -4 = 12

next, isolate the -3r by putting 12 on the other side

-3r+12>0    subtract 12 from both sides

-3r>-12    now, divide both sides by -3    (you have to flip the > because you are dividing by a negative) -12 divided by -3 is 4

so you should end up with r<4

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Answer:

3t=15

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In a recent Super Bowl, a TV network predicted that 50 % of the audience would express an interest in seeing one of its forthcom
coldgirl [10]

Answer:

z= -0.968

We can conclude that we fail to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that  they would watch one of the television shows not differs from 0.5 or 50% .  

Step-by-step explanation:

1) Data given and notation n  

n=106 represent the random sample taken

X=48 represent the people who says that  they would watch one of the television shows.

\hat p=\frac{48}{106}=0.453 estimated proportion of people who says that  they would watch one of the television shows.

p_o=0.5 is the value that we want to test

\alpha represent the significance level  

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that 50% of people who says that  they would watch one of the television shows.:  

Null hypothesis:p=0.5  

Alternative hypothesis:p \neq 0.5  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.453 -0.5}{\sqrt{\frac{0.5(1-0.5)}{106}}}=-0.968  

4) Statistical decision  

P value method or p value approach . "This method consists on determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level is not provided, but we can assume \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(z  

So based on the p value obtained and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we fail to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that  they would watch one of the television shows not differs from 0.5 or 50% .  

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