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Ksivusya [100]
3 years ago
7

4.8 Twitter users and news, Part I: A poll conducted in 2013 found that 55% of U.S. adult Twitter users get at least some news o

n Twitter (Pew, 2013). The standard error for this estimate was 2.6%, and a normal distribution may be used to model the sample proportion. Construct a 99% confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter, and interpret the confidence interval in context. (please round all percentages to 2 decimal places)
Mathematics
1 answer:
TiliK225 [7]3 years ago
7 0

Answer: Required 99% confidence interval  : (48.30%, 61.7%)

Step-by-step explanation:

Confidence interval for PROPORTION:  p\pm z^* \times S.E., where p = sample proportion , z* = two-tailed critical z-value  and S.E. = standard error.

As per given: p=55%

Critical z-value for 99% confidence = 2.576

S.E. = 2.6%

Then, required confidence interval :

55\% \pm (2.576)\times 2.6\%\\\\\approx55\% \pm6.70\%\\\\=(55\%-6.70\%,\ 55\%+6.70\%)\\\\=(48.30\%,\ 61.7\%)

Required 99% confidence interval  : (48.30%, 61.7%)

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As long as the two equations represent the same straight line on the coordinate plane (they overlap), there will be infinite many intersections and infinite number of solutions.

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2 years ago
Suppose the p-value is 0.1038. what is the correct interpretation of this p-value in context?
Basile [38]

The p value 0.1038 represent the proportion of students who use drugs at this college is equal to 0.157, the probability of getting a statistic as extreme or more extreme as that observed is 0.1038.

According to the statement

we have given that the P value and we have to find the correct interpretation of this p-value in context.

So, For this purpose, we know that the

The P - Value is the In null-hypothesis significance testing, the p-value is the probability of obtaining test results at least as extreme as the result actually observed.

And in the given statement ,

The given p value is 0.1038.

And this p value is a If the proportion of students who use drugs at this college is equal to 0.157, the probability of getting a statistic as extreme or more extreme as that observed is 0.1038.

So, The p value 0.1038 represent the proportion of students who use drugs at this college is equal to 0.157, the probability of getting a statistic as extreme or more extreme as that observed is 0.1038.

Learn more about P value here

brainly.com/question/4621112

Disclaimer: This question was incomplete. Please find the full content below.

Question:

Suppose the p-value is 0.1038. What is the correct interpretation of this p-value in context?

a.) If the proportion of students who use drugs at this college is greater than 0.157, the probability of getting a statistic as extreme or more extreme as that observed is 0.1038.

b.) There is a 0.1038 probability that the proportion of students who use drugs at this college is greater than 0.157.

c.) If the proportion of students who use drugs at this college is equal to 0.157, the probability of getting a statistic as extreme or more extreme as that observed is 0.1038.

d.) There is a 0.1038 probability that the proportion of students who use drugs at this college is equal to 0.157.

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