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tangare [24]
3 years ago
6

A random sample of voters is polled and 34 plan to vote for Candidate A, 56 plan to vote for Candidate B, and 12 are undecided.

What inference can be made from this random sample? Candidate A will most likely win the election. Candidate B will most likely win the election. There is not enough information to make an inference. Candidate A and Candidate B will have the same amount of votes.
Mathematics
2 answers:
lina2011 [118]3 years ago
7 0
<h2>Answer:</h2>

The inference that can be made from this random sample is:

Candidate B will most likely win the election.

<h2>Step-by-step explanation:</h2>

We are given that:

A random sample of voters is polled and 34 plan to vote for Candidate A, 56 plan to vote for Candidate B, and 12 are undecided.

Since from the simulation we could see that  more person voted for  the Candidate B and less for the candidate A.

Even if all the undecided person will vote for Candidate A then also the total number of votes for candidate A will be less.

( Since 12+34=46<56)

Hence, there are most chances that Candidate B will win.

ella [17]3 years ago
6 0
Candidate B is more likely to be elected because out of the sample, it had more votes than candidate A. It is safe to assume this because the number of undecided votes is quite small and not likely to go all to one candidate.
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he one‑sample t statistic from a sample of n = 23 observations for the two‑sided test of H 0 : μ = 15 versus H α : μ &gt; 15 has
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Answer:

t = 2.24

The first step is calculate the degrees of freedom, on this case:  

df=n-1=23-1=22  

Since is a one side right tailed test the p value would be:  

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And for this case we can conclude that:

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And we will reject the null hypothesis at \alpha=0.025 since p_v < \alpha

Step-by-step explanation:

Data given and notation  

\bar X represent the mean height for the sample  

s represent the sample standard deviation

n=23 sample size  

\mu_o =15 represent the value that we want to 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 higher than 15, the system of hypothesis would be:  

Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

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, and the statistic is given by:  

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

t-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

For this case the statistic is given:

t = 2.24

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=23-1=22  

Since is a one side right tailed test the p value would be:  

p_v =P(t_{(22)}>2.24)=0.01776  

And for this case we can conclude that:

0.01 < p_v < 0.025

And we will reject the null hypothesis at \alpha=0.025 since p_v < \alpha

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Answer:

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