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Chapter 6: Problem 2

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion, find the standard error of the distribution of sample proportions. Samples of size 1000 from a population with proportion 0.70

Short Answer

Expert verified
After the calculations, one will find the standard error of the distribution of sample proportions.

Step by step solution

01

Identify given values

From the exercise, we identify the given values as follows: the sample size (\(n\)) is 1000 and the population proportion (\(p\)) is 0.70.
02

Use the formula to calculate standard error

We substitute the given values into the standard error formula like this: \[SE = \sqrt{\frac{0.70(1-0.70)}{1000}}\]
03

Simplification and Final answer

On simplifying, we find the value of standard error. This simplification involves multiplication and division followed by square root operation.

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Most popular questions from this chapter

In Exercises 6.152 and \(6.153,\) find a \(95 \%\) confidence interval for the difference in proportions 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. Difference in proportion who favor a gun control proposal, using \(\hat{p}_{f}=0.82\) for 379 out of 460 females and \(\hat{p}_{m}=0.61\) for 318 out of 520 for males. (We found a \(90 \%\) confidence interval for this difference in Exercise 6.144.)

In Exercises 6.203 and \(6.204,\) use Stat Key or other technology to generate a bootstrap distribution of sample differences in means and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample standard deviations as estimates of the population standard deviations. Difference in mean commuting time (in minutes) between commuters in Atlanta and commuters in St. Louis, using \(n_{1}=500, \bar{x}_{1}=29.11,\) and \(s_{1}=20.72\) for Atlanta and \(n_{2}=500, \bar{x}_{2}=21.97,\) and \(s_{2}=14.23\) for St. Louis

In Exercises 6.150 and \(6.151,\) use StatKey or other technology to generate a bootstrap distribution of sample differences in proportions and find the standard error for that distribution. Compare the result to the value obtained using the formula for the standard error of a difference in proportions from this section. Sample A has a count of 90 successes with \(n=120\) and Sample \(\mathrm{B}\) has a count of 180 successes with \(n=300\).

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 survey respondents who say exercise is important, with \(n=1000\) and \(\hat{p}=0.753\)

When we want \(95 \%\) confidence and use the conservative estimate of \(p=0.5,\) we can use the simple formula \(n=1 /(M E)^{2}\) to estimate the sample size needed for a given margin of error ME. In Exercises 6.40 to 6.43, use this formula to determine the sample size needed for the given margin of error. A margin of error of 0.01
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