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Gelneren [198K]
2 years ago
8

A national survey conducted in 2011 among a simple random sample of 1,507 adults shows that 56% of Americans think the Civil War

is still relevant to American politics and political life.(a) Conduct a hypothesis test to determine if these data provide strong evidence that the majority of the Americans think the Civil War is still relevant.(b) Interpret the p-value in this context.(c) Calculate a 90% confidence interval for the proportion of Americans who think the Civil War is still relevant. Interpret the interval in this context, and comment on whether or not the confidence interval agrees with the conclusion of the hypothesis test.
Mathematics
1 answer:
Drupady [299]2 years ago
7 0

Answer:

a)z=4.658

p_v =P(z>4.65)=1-P(z  

b) Using the significance level assumed \alpha=0.05 we see that p_v so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life is  higher than 50%.

c) The 90% confidence interval would be given (0.527;0.593).

We are confident that about 54% to 59% of all Americans think the Civil War is relevant.

Step-by-step explanation:

I )Part a

1) Data given and notation

n=1507 represent the random sample taken  

X represent the Americans who thinks that the Civil War is still relevant to American politics and political life

\hat p estimated proportion of Americans who thinks that the Civil War is still relevant to American politics and political life in the sample

p_o=0.5 is the value that we want to test since the problem says majority    

\alpha=0.05 represent the significance level (no given, but is assumed)  

z would represent the statistic (variable of interest)  

p_v represent the p value (variable of interest)  

p= population proportion of Americans who thinks that the Civil War is still relevant to American politics and political life

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life exceeds 50%(Majority). :  

Null Hypothesis: p \leq 0.5

Alternative Hypothesis: p >0.5

We assume that the proportion follows a normal distribution.  

This is a one tail upper test for the proportion of  union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion \hat p is significantly (different,higher or less) from a hypothesized value p_o".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

np_o =1507*0.5=753.5>10

n(1-p_o)=1507*(1-0.5)=753.5>10

3) Calculate the statistic  

The statistic is calculated with the following formula:

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

On this case the value of p_o=0.5 is the value that we are testing and n = 1507.

z=\frac{0.56 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1507}}}=4.658

The p value for the test would be:

p_v =P(z>4.65)=1-P(z  

II) Part b

It's important to refresh the p value method or p value approach . "This method is about 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.  

Based on the alternative hypothesis the p value would be given by:

p_v =P(z>4.65)=1-P(z  

Using the significance level assumed \alpha=0.05 we see that p_v so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life is  higher than 50%.

III) Part c

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.56 - 2.58 \sqrt{\frac{0.56(1-0.56)}{1507}}=0.527

0.56 + 2.58 \sqrt{\frac{0.56(1-0.56)}{1507}}=0.593

And the 90% confidence interval would be given (0.527;0.593).

We are confident that about 54% to 59% of all Americans think the Civil War is relevant.

And this result agrees with the result of part b, since the interval not contains the value of 0.5 we can conclude that the proportion of Americans who thinks that the Civil War is still relevant to American politics and political life it's higher than 0.5 at 90% of confidence.

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Answer:

a) 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.

b) 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.

Step-by-step explanation:

To solve this question, it is important to know the Normal probability distribution and the Central Limit Theorem

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

In this problem, we have that:

\mu = 6.3, \sigma = 2.1, n = 49, s = \frac{2.1}{\sqrt{49}} = 0.3

A) What is the chance HLI will find a sample mean between 5.5 and 7.1 hours?

This is the pvalue of Z when X = 7.1 subtracted by the pvalue of Z when X = 5.5.

By the Central Limit Theorem, the formula for Z is:

Z = \frac{X - \mu}{s}

X = 7.1

Z = \frac{7.1 - 6.3}{0.3}

Z = 2.67

Z = 2.67 has a pvalue of 0.9962

X = 5.5

Z = \frac{5.5 - 6.3}{0.3}

Z = -2.67

Z = -2.67 has a pvalue of 0.0038

So there is a 0.9962 - 0.0038 = 0.9924 = 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.

B) Calculate the probability that the sample mean will be between 5.9 and 6.7 hours.

This is the pvalue of Z when X = 6.7 subtracted by the pvalue of Z when X = 5.9

X = 6.7

Z = \frac{6.7 - 6.3}{0.3}

Z = 1.33

Z = 1.33 has a pvalue of 0.9082

X = 5.9

Z = \frac{5.9 - 6.3}{0.3}

Z = -1.33

Z = -1.33 has a pvalue of 0.0918.

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