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Chapter 6: Problem 116

In a recent study, 342006 randomly selected \({ }^{34}\) Hampton, K., Goulet, L., Rainie, L., and Purcell, K., "Social Networking Sites and Our Lives," Pew Research Center, pewresearch.org, June 16,2011. US adults (age 18 or older) were asked to give the number of people in the last six months "with whom you discussed matters that are important to you." The average number of close confidants was 2.2 with a standard deviation of 1.4 (a) Find the margin of error for this estimate if we want \(99 \%\) confidence. (b) Find and interpret a \(99 \%\) confidence interval for average number of close confidants.

Short Answer

Expert verified
To answer part (a), calculate the margin of error using the values provided in step 2. The value will result in a very small decimal that represents the uncertainty percentage in the population average. For part (b), use the value from step 3 to calculate the confidence interval which is the range in which we can expect the real population mean to lie 99% of the time.

Step by step solution

01

Calculate the Z-score

The first step involves finding out the Z-score value for a 99% confidence interval. Utilize a Z-table or calculator to find this. For a 99% confidence level, the Z-score is 2.576.
02

Calculate the Margin of Error

The margin of error is calculated using the following formula: \( m = z \cdot \(\frac{s}{\sqrt{n}}\) \) where, m is the margin of error, z is the Z-score, s is the standard deviation and n is the sample size. Plug in the provided values: \( m = 2.576 \cdot \(\frac{1.4}{\sqrt{342006}}\) \)
03

Calculate the Confidence Interval

The Confidence Interval is calculated as: \( CI = \(\overline{x} 卤 m \) where, CI = Confidence Interval, \( \overline{x} \) = Sample mean, m = Margin of error. Plug in the calculated margin of error and the given mean: \( CI = 2.2 卤 m \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Margin of Error
When you're conducting a study or survey, you always want your results to be as accurate as possible. However, there's usually a small amount of error due to sampling. This is where the **Margin of Error** (MOE) comes into play. It quantifies the range our estimate could vary in the population.

The formula for calculating the margin of error is: \[m = z \cdot \frac{s}{\sqrt{n}}\] - **m** is the margin of error. - **z** represents the Z-score corresponding to the desired confidence level. - **s** is the standard deviation, which shows how much variation or dispersion exists from the average. - **n** is the sample size.

In simpler terms, the margin of error tells us how much we can expect our sample results to differ from the actual population values, within a certain level of probability.
Z-score Calculation
The **Z-score Calculation** plays a crucial role in determining how many standard deviations an element is from the mean. For confidence intervals, Z-scores help us understand how certain we can be about our sampling results.

To find the Z-score for a specific confidence level: - Use a standard Z-table (found in most statistics textbooks) or a statistical calculator. - Identify the Z-score that matches your desired confidence level. For a 99% confidence level, the common Z-score is 2.576.

This Z-score tells us that 99% of sample means will fall within this many standard deviations from the mean under normal distribution. This understanding is significant because it indicates how sure we can be about whether the sample mean truly represents the population mean.
Standard Deviation
**Standard Deviation** is a fundamental concept in statistics that measures the amount of variation or dispersion in a set of values. It's a bit like a fingerprint of your data, showing how spread out numbers are around the mean.

Here's why standard deviation is critical: - A low standard deviation means that the data points are close to the mean, showing less variability. - A high standard deviation indicates that the data points are spread out over a large range of values.
Using standard deviation in our margin of error calculation helps provide a clearer picture of how much confidence we can place in our results. It influences how wide our confidence interval will be, ultimately impacting our conclusions about the survey or study.

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Most popular questions from this chapter

Describes scores on the Critical Reading portion of the Scholastic Aptitude Test (SAT) for college-bound students in the class of 2010. Critical Reading scores are approximately normally distributed with mean \(\mu=501\) and standard deviation \(\sigma=112\) (a) For each sample size below, use a normal distribution to find the percentage of sample means that will be greater than or equal to \(525 .\) Assume the samples are random samples. i. \(n=1\) ii. \(n=10\) iii. \(n=100\) iv. \(n=1000\) (b) Considering your answers from part (a), discuss the effect of the sample size on the likelihood of a sample mean being as far from the population mean as \(\bar{x}=525\) is from \(\mu=501\).

Difference in mean commuting distance (in miles) between commuters in Atlanta and commuters in St. Louis, using \(n_{1}=500, \bar{x}_{1}=18.16,\) and \(s_{1}=13.80\) for Atlanta and \(n_{2}=500, \bar{x}_{2}=14.16,\) and \(s_{2}=10.75\) for St. Louis.

According to the 2006 Australia Census, \(^{43} 25.5 \%\) of Australian women over the age of 25 had a college degree, while the percentage for Australian men was \(21.4 \% .\) Suppose we select random samples of 200 women and 200 men from this population and look at the differences in proportions with college degrees, \(\hat{p}_{f}-\hat{p}_{m}\), in those samples. (a) Describe the distribution (center, spread,shape) for the difference in sample proportions. Include a rough sketch of the distribution with values labeled on the horizontal axis. (b) What is the chance that the proportion with college degrees in the men's sample is actually more than the proportion in the women's sample? (Hint: Think about what must be true about \(\hat{p}_{f}-\hat{p}_{m}\) when this happens.)

Use a t-distribution. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the area in a t-distribution less than -1.4 if the samples have sizes \(n_{1}=30\) and \(n_{2}=40\).

NBA Free Throws In Exercise 6.10, we learn that the percent of free throws made in the \(\mathrm{NBA}\) (National Basketball Association) has been about \(75 \%\) for the last 50 years. If we take random samples of free throws in the NBA and compute the proportion of free throws made, what percent of samples of size \(n=200\) will have a sample proportion greater than \(80 \%\) ? Use the fact that the sample proportions are normally distributed and compute the mean and standard deviation of the distribution.
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