Mathematics, 04.03.2020 00:10 aidenbender06

A poll conducted in 2013 found that 52% of U. S. adult Twitter user get at least some news on Twitter, and the standard error for this estimate was 2.4%. Identify each of the following statements as true or false. Provide an explanation justify each of your answers.

a) the data provide statistically significant evidence that more than half U. S. adult twitter users get some news throught Twitter. Use a significance level of alpha = 0.01

b) Since the standard error is 2.4%, we can conclude that 97.6% of all U. S. adult Twitter users were included in the study.

c) If we want to reduce the standard error of estimate, we should collect less data.

d) If we construct a 90% confidence interval for the perventage of U. S. adults Twitter suers who get some news through Twitter, this confidene interval will be wider than a corresponding 99% confidence interval.

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Answer from: brandon1748

a) FALSE we obtain a NOT significant results after conduct the hypothesis tes.

b) FALSE, the standard error is not associated to a certain % of people included in the study

c) FALSE. if we want to reduce the standard error we need to increase the sample size or data

d) FALSE, always if we have a higher confidence level the confidence interval associated to this level would be wider than for a lower confidence interval

Step-by-step explanation:

Data given and notation n

n represent the random sample taken

$\hat p=0.52$ estimated proportion of U.S. adult Twitter user get at least some news on Twitter

p_o=0.5 is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that more than half U.S. adult twitter users get some news throught Twitter:

Null hypothesis: $p\leq 0.5$

Alternative hypothesis: p 0.5

When we conduct a proportion test we need to use the z statistic, 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 .

Calculate the statistic

The standard error is given:

$SE = \sqrt{\frac{p_o (1-p_o)}{n}}=0.024$

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

$z=\frac{0.52 -0.5}{0.024}=0.833$

Statistical decision

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.

The significance level assumed is $\alpha=0.01$ . The next step would be calculate the p value for this test.

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

p_v =P(z0.833)=0.2024

So the p value obtained was a very high value and using the significance level given $\alpha=0.01$ we have $p_v\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Now let's idendity the statements

a) FALSE we obtain a NOT significant results after conduct the hypothesis tes.

b) FALSE, the standard error is not associated to a certain % of people included in the study

c) FALSE. if we want to reduce the standard error we need to increase the sample size or data

d) FALSE, always if we have a higher confidence level the confidence interval associated to this level would be wider than for a lower confidence interval

Answer from: Quest

i need the options

step-by-step explanation:

Answer from: Quest

answer: it's a

step-by-step explanation: