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In Exercises 6.7 and 6.8 , compute the standard error for sample proportions from a population with the given proportion using three different sample sizes. What effect does increasing the sample size have on the standard error? Using this information about the effect on the standard error, discuss the effect of increasing the sample size on the accuracy of using a sample proportion to estimate a population proportion. A population with proportion \(p=0.75\) for sample sizes of \(n=40, n=300,\) and \(n=1000 .\)

In Exercises 6.9 and 6.10 , indicate whether the Central Limit Theorem applies so that the sample proportions follow a normal distribution. In each case below, is the sample size large enough so that the sample proportions follow a normal distribution? (a) \(n=500\) and \(p=0.1\) (b) \(n=25\) and \(p=0.5\) (c) \(n=30\) and \(p=0.2\) (d) \(n=100\) and \(p=0.92\)

In Exercises 6.11 to 6.14, use the normal distribution to find a confidence interval for a proportion \(p\) given the relevant sample results. Give the best point estimate for \(p,\) the margin of error, and the confidence interval. Assume the results come from a random sample. A \(95 \%\) confidence interval for \(p\) given that \(\hat{p}=\) 0.38 and \(n=500\)

A survey \(^{4}\) of 1060 randomly selected US teens ages 13 to 17 found that 605 of them say they have made a new friend online. (a) Find and interpret a \(90 \%\) confidence interval for the proportion, \(p\), of all US teens who have made a new friend online. (b) Give the best estimate for \(p\) and give the margin of error for the estimate. (c) Use the interval to determine whether we can be \(90 \%\) confident that more than half of US teens have made a new friend online.

In a survey of 1000 US adults, twenty percent say they never exercise. This is the highest level seen in five years. \(^{5}\) Find and interpret a \(99 \%\) confidence interval for the proportion of US adults who say they never exercise. What is the margin of error, with \(99 \%\) confidence?

In a survey of 2255 randomly selected US adults (age 18 or older), 1787 of them use the Internet regularly. Of the Internet users, 1054 use a social networking site. \({ }^{7}\) Find and interpret a \(95 \%\) confidence interval for each of the following proportions: (a) Proportion of US adults who use the Internet regularly. (b) Proportion of US adult Internet users who use a social networking site. (c) Proportion of all US adults who use a social networking site. Use the confidence interval to estimate whether it is plausible that \(50 \%\) of all US adults use a social networking site.

In Exercises 6.28 to 6.31, use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of peanuts in mixed nuts, with \(n=100\) and \(\hat{p}=0.52\)

In Exercises 6.32 and 6.33, find a \(95 \%\) confidence interval for the proportion two ways: using StatKey or other technology and percentiles from a bootstrap distribution, and using the normal distribution and the formula for standard error. Compare the results. Proportion of home team wins in soccer, using \(\hat{p}=0.583\) with \(n=120\)

We examine the effect of different inputs on determining the sample size needed to obtain a specific margin of error when finding a confidence interval for a proportion. Find the sample size needed to give a margin of error to estimate a proportion within \(\pm 3 \%\) with \(99 \%\) confidence. With \(95 \%\) confidence. With \(90 \%\) confidence. (Assume no prior knowledge about the population proportion \(p\).) Comment on the relationship between the sample size and the confidence level desired.

A survey of 1000 adults in the US conducted in March 2011 asked "Do you favor or oppose 'sin taxes' on soda and junk food?" The proportion in favor of taxing these foods was \(32 \% .10\) (a) Find a \(95 \%\) confidence interval for the proportion of US adults favoring taxes on soda and junk food. (b) What is the margin of error? (c) If we want a margin of error of only \(1 \%\) (with \(95 \%\) confidence \()\), what sample size is needed?