X-Git-Url: https://code.communitydata.science/stats_class_2020.git/blobdiff_plain/efa6913590499d105277c2907cde2a96aa7bd51f..9d912e87ec9ee97b3ffb0e3b7dc8b2361ccca9f2:/os_exercises/ch5_exercises_solutions.html?ds=inline diff --git a/os_exercises/ch5_exercises_solutions.html b/os_exercises/ch5_exercises_solutions.html index ef20494..a30a272 100644 --- a/os_exercises/ch5_exercises_solutions.html +++ b/os_exercises/ch5_exercises_solutions.html @@ -1569,15 +1569,15 @@ SE = \sqrt{\frac{p(1-p)}{n}}\\ \end{array}\]

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5.8 Twitter users and news, Part I

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5.8 Twitter users & news I

The general formula for a confidence interval is \(point~estimate~±~z^*\times~SE\). Where \(z^*\) corresponds to the z-score for the desired value of \(\alpha\).

To estimate the interval from the data described in the question, identify the three different values. The point estimate is 45%, \(z^* = 2.58\) for a 99% confidence level (that’s the number of standard deviations around the mean that ensure that 99% of a Z-score distribution is included), and \(SE = 2.4\%\). With this we can plug and chug:

\[52\% ± 2.58 \times 2.4\%\] And that yields: \[95\% CI = (45.8\%, 58.2\%)\]

Which means that from this data we are 99% confident that between 45.8% and 58.2% U.S. adult Twitter users get some news through the site.

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5.10 Twitter users and news, Part II

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5.10 Twitter users & news II

  1. False. See the answer to exercise 5.8 above. With \(\alpha = 0.01\), we can consult the 99% confidence interval. It includes 50% but also goes lower. A null hypothesis of \(p=0.50\) would not be rejected at this level.

  2. False. The standard error of the sample proportion does not contain any information about the proportion of the population included in the sample. It estimates the variability of the sample proportion.