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Give information about the proportion of a sample that agrees with a certain statement. Use StatKey or other technology to estimate the standard error from a bootstrap distribution generated from the sample. Then use the standard error to give a \(95 \%\) confidence interval for the proportion of the population to agree with the statement. StatKey tip: Use "CI for Single Proportion" and then "Edit Data" to enter the sample information. In a random sample of 400 people, 112 agree and 288 disagree.

Short Answer

Expert verified
The 95% confidence interval for the proportion of the population agreeing with the statement is estimated using the standard error calculated from the bootstrap distribution generated from the sample data of '112 agree and 288 disagree' out of 400 people sampled.

Step by step solution

01

Input Data Into StatKey

Input the sample data into StatKey or any statistical tool of choice. In this case, input the number of people who agree as '112' and the total number of people sampled as '400'. This information is required to calculate the proportion of people who agree with the statement.
02

Generate Bootstrap Distribution

Generate a bootstrap distribution of proportions using the sample data. This will create a distribution of sample proportions from the available data.
03

Calculate Standard Error

Calculate the standard error from the bootstrap distribution. This error is a measure of the statistical accuracy of an estimator or a statistical measurement.
04

Estimate Confidence Interval

Use the standard error to estimate the 95% confidence interval for the proportion of the population that agrees with the statement. A 95% confidence interval is the range of values that you can be 95% confident contains the true population proportion.

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Most popular questions from this chapter

\(\mathbf{3 . 1 2 3}\) What Proportion Have Pesticides Detected? In addition to the quantitative variable pesticide concentration, the researchers also report whether or not the pesticide was detected in the urine (at standard detection levels). Before the participants started eating organic, 111 of the 240 measurements (combining all pesticides and people) yielded a positive pesticide detection. While eating organic, only 24 of the 240 measurements resulted in a positive pesticide detection. (a) Calculate the sample difference in proportions: proportion of measurements resulting in pesticide detection while eating non- organic minus proportion of measurements resulting in pesticide detection while eating organic. (b) Figure 3.33 gives a bootstrap distribution for the difference in proportions, based on \(1000 \mathrm{sim}-\) ulated bootstrap samples. Approximate a \(98 \%\) confidence interval. (c) Interpret this interval in context.

Exercises 3.71 to 3.73 consider the question (using fish) of whether uncommitted members of a group make it more democratic. It has been argued that individuals with weak preferences are particularly vulnerable to a vocal opinionated minority. However, recent studies, including computer simulations, observational studies with humans, and experiments with fish, all suggest that adding uncommitted members to a group might make for more democratic decisions by taking control away from an opinionated minority. \({ }^{36}\) In the experiment with fish, golden shiners (small freshwater fish who have a very strong tendency to stick together in schools) were trained to swim toward either yellow or blue marks to receive a treat. Those swimming toward the yellow mark were trained more to develop stronger preferences and became the fish version of individuals with strong opinions. When a minority of five opinionated fish (wanting to aim for the yellow mark) were mixed with a majority of six less opinionated fish (wanting to aim for the blue mark), the group swam toward the minority yellow mark almost all the time. When some untrained fish with no prior preferences were added, however, the majority opinion prevailed most of the time. \({ }^{37}\) Exercises 3.71 to 3.73 elaborate on this study. How Often Does the Fish Majority Win? In a school of fish with a minority of strongly opinionated fish wanting to aim for the yellow mark and a majority of less passionate fish wanting to aim for the blue mark, as described under Fish Democracies above, a \(95 \%\) confidence interval for the proportion of times the majority wins (they go to the blue mark) is 0.09 to \(0.26 .\) Interpret this confidence interval. Is it plausible that fish in this situation are equally likely to go for either of the two options?

College Graduates In Example 3.1 on page 197, we see that \(27.5 \%\) of US adults are college graduates. (a) Use StatKey or other technology to generate a sampling distribution for the sample proportion of college graduates using a sample size of \(n=50 .\) Generate at least 1000 sample proportions. Give the shape and center of the sampling distribution and give the standard error. (b) Repeat part (a) using a sample size of \(n=500\).

Proportion of registered voters in a county who voted in the last election, using data from the county voting records.

Exercises 3.71 to 3.73 consider the question (using fish) of whether uncommitted members of a group make it more democratic. It has been argued that individuals with weak preferences are particularly vulnerable to a vocal opinionated minority. However, recent studies, including computer simulations, observational studies with humans, and experiments with fish, all suggest that adding uncommitted members to a group might make for more democratic decisions by taking control away from an opinionated minority. \({ }^{36}\) In the experiment with fish, golden shiners (small freshwater fish who have a very strong tendency to stick together in schools) were trained to swim toward either yellow or blue marks to receive a treat. Those swimming toward the yellow mark were trained more to develop stronger preferences and became the fish version of individuals with strong opinions. When a minority of five opinionated fish (wanting to aim for the yellow mark) were mixed with a majority of six less opinionated fish (wanting to aim for the blue mark), the group swam toward the minority yellow mark almost all the time. When some untrained fish with no prior preferences were added, however, the majority opinion prevailed most of the time. \({ }^{37}\) Exercises 3.71 to 3.73 elaborate on this study. Training Fish to Pick a Color Fish can be trained quite easily. With just seven days of training, golden shiner fish learn to pick a color (yellow or blue) to receive a treat, and the fish will swim to that color immediately. On the first day of training, however, it takes them some time. In the study described under Fish Democracies above, the mean time for the fish in the study to reach the yellow mark is \(\bar{x}=51\) seconds with a standard error for this statistic of 2.4 . Find and interpret a \(95 \%\) confidence interval for the mean time it takes a golden shiner fish to reach the yellow mark. Is it plausible that the average time it takes fish to find the mark is 60 seconds? Is it plausible that it is 55 seconds?

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