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Nata [24]
3 years ago
11

According to data from the United States Elections Project, only 36 percent of eligible voters voted in the 2014 elections. For

random samples of size 40, which of the following best describes the sampling distribution of the (p-hat), the sample proportion of people who voted in the 2014 elections?
Mathematics
1 answer:
Sonbull [250]3 years ago
8 0

Using the Central Limit Theorem, it is found that the sample proportion of people who voted in the 2014 elections is approximately normal, with mean of 0.36 and standard error of 0.0759.

<h3>Central Limit Theorem </h3>
  • The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.  
  • For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this problem:

  • 36% of eligible voters voted, hence p = 0.36.
  • Samples of size 40, hence n = 40.

Then:

\mu = p = 0.36

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.36(0.64)}{40}} = 0.0759

Hence, the distribution is approximately normal, with mean of 0.36 and standard error of 0.0759.

To learn more about the Central Limit Theorem, you can take a look at brainly.com/question/16695444

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Answer:

x = 12

Step-by-step explanation:

3x - 1 + x + 7 = 54

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4x + 6 - 6 = 54 - 6

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Answer:

t=\frac{8.19-9.02}{\frac{0.8}{\sqrt{17}}}=-4.28    

The degrees of freedom are given by

df=n-1=17-1=16  

The p value would be given by this probability:

p_v =P(t_{(16)}  

Since the p value is lower than the significance level provided of 0.01 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 9.02 cm^3

Step-by-step explanation:

Information given

\bar X=8.19 cm^3 represent the sample mean

s=0.8 cm^3 represent the sample deviation

n=17 sample size  

\mu_o =9.02 represent the value to verify

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic

p_v represent the p value

System of hypothesis to check

We want to check if the true mean is less than the normal value of 9.02 cm^3, the system of hypothesis would be:  

Null hypothesis:\mu \geq 9.02  

Alternative hypothesis:\mu < 9.02  

The statistic for this case is given by:

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

Replacing the info given we got:

t=\frac{8.19-9.02}{\frac{0.8}{\sqrt{17}}}=-4.28    

The degrees of freedom are given by

df=n-1=17-1=16  

The p value would be given by this probability:

p_v =P(t_{(16)}  

Since the p value is lower than the significance level provided of 0.01 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 9.02 cm^3

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