Respuesta :
The 99% confidence interval for the population proportion is given by
[tex]sample \ proportion \pm 2.575\times standard \ error[/tex]
Given that 52% = 0.52 is the standard proportion and the standard error is 2.4% = 0.024.
Therefore, the 99% confidence interval for the fraction of u.s. adult twitter users who get some news on twitter is given by
[tex]0.52\pm2.575\times0.024=0.52\pm0.0618=(0.4582, \ 0.5818)[/tex]
Therefore, the 99% confidence interval for the fraction of u.s. adult twitter users who get some news on twitter is 45.82% and 58.18%.
The interpretation of the confidence interval is as follows: "We are 99% confident that the fraction of U.S. adult twiter users who get some news on twitter is between 45.82% and 58.18%.
[tex]sample \ proportion \pm 2.575\times standard \ error[/tex]
Given that 52% = 0.52 is the standard proportion and the standard error is 2.4% = 0.024.
Therefore, the 99% confidence interval for the fraction of u.s. adult twitter users who get some news on twitter is given by
[tex]0.52\pm2.575\times0.024=0.52\pm0.0618=(0.4582, \ 0.5818)[/tex]
Therefore, the 99% confidence interval for the fraction of u.s. adult twitter users who get some news on twitter is 45.82% and 58.18%.
The interpretation of the confidence interval is as follows: "We are 99% confident that the fraction of U.S. adult twiter users who get some news on twitter is between 45.82% and 58.18%.
For 99% confidence level, the interval will be (0.4582, 0.5818).
It can be interpreted as "45.82% to 58.18% of US adult will get information from twitter with a confidence of 99%".
Given information:
52% or 0.52 fraction of US adult twitter users get at least some news on twitter. It is the sample population in fractions.
The standard error is 2.4% or 0.024 fraction.
Normal distribution may be used to model the sample proportion.
Now, for 99% confidence level, the interval for the fraction of US adult who get some news on twitter will be,
[tex]I=\texttt{sample population}\pm 2.575\times \texttt{standard error}\\I=0.52\pm 2.575\times0.024\\I=0.52\pm 0.0618\\I=(0.4582, 0.5818)[/tex]
Therefore, for 99% confidence level, the interval will be (0.4582, 0.5818).
It can be interpreted as "45.82% to 58.18% of US adult will get information from twitter with a confidence of 99%".
For more details, refer the link:
https://brainly.com/question/15519485?referrer=searchResults