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91Ó°ÊÓ

In a nationwide poll of 1000 randomly sampled adults conducted in June \(2011,83 \%\) said they think children spend too much time on their computers and other electronic devices (but \(37 \%\) say time spent on a computer is better than time spent in front of a \(\mathrm{TV}) .{ }^{9}\) Find and interpret a \(95 \%\) confidence interval for the proportion of adults who believe children spend too much time on electronic devices. What is the margin of error for this result? Is it plausible that the proportion of all adults who feel this way is less than \(80 \%\) ? Is it plausible that the proportion is greater than \(85 \% ?\)

Short Answer

Expert verified
Once you calculate the margin of error, confidence interval, and compare it to the proportion of adults provided, you can determine whether it is plausible that the proportion of all adults who think children use electronics too much is less than 80% or greater than 85%.

Step by step solution

01

Calculate the sample proportion

First, get the sample proportion by dividing the number of adults who think children spend too much time on their electronic devices, 0.83, by the total number of adults, 1000. So the sample proportion \( p = 0.83. \)
02

Calculate the standard error

Standard error (SE) is a measure of the variability in the sampling distribution. Use the formula \( SE = \sqrt{ \frac{p(1-p)}{n} } \) where p is the sample proportion and n is the sample size. Plug in the values from Step 1 into the equation: \( SE = \sqrt{\frac{0.83(1-0.83)}{1000}}\).
03

Find the 95% confidence interval

A 95% confidence interval can be found using the formula \( CI = p \pm Z * SE \) where Z is the Z score. For a 95% confidence interval, the Z score is 1.96. Using the sample proportion and standard error calculated previously, plug into the formula: \( CI = 0.83 \pm 1.96 * SE \). Use the lower and upper limits of this interval as the confidence interval.
04

Calculate the margin of error

The margin of error (MOE) is also known as the range of values. The MOE can be calculated by \( MOE = Z * SE \). Using the Z score of 1.96 and the standard error calculated previously, the calculation here would be \( MOE = 1.96 * SE \).
05

Interpret the results

The confidence interval is the range where one can expect the true population proportion to lie with 95% confidence. Based on this, decide whether it's plausible if the proportion of all adults who feel children spend too much time on electronic devices is less than 80% or more than 85%. The two values should be compared with the confidence interval. If they're outside the interval, they might not be plausible.

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Key Concepts

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

Sample Proportion
When conducting surveys, researchers are often interested in estimating certain characteristics of a population. One such characteristic is the **sample proportion**. In the context of this survey, the sample proportion represents the estimated percentage of adults who believe that children spend too much time on electronic devices.
To find the sample proportion, we divide the number of favorable outcomes (those who agree) by the total sample size. In our exercise, the sample proportion for those who think kids spend too much time on devices is 0.83 or 83%. That means 83% of the sample population expressed this opinion.
  • Numerator: Number of affirmative responses, in this case corresponding to 83% of the sample.
  • Denominator: Total number of survey participants, which is 1000.
Understanding sample proportion is crucial as it forms the basis for further statistical analysis like estimating the true proportion in the whole population.
Standard Error
The **standard error** is a statistical measure that provides insight into the reliability of the sample proportion. Essentially, it reflects how much the sample proportion would vary from sample to sample if we were to repeat the survey multiple times, each with the same size.To calculate the standard error, you can use the formula:\[ SE = \sqrt{ \frac{p(1-p)}{n} } \]Where:
  • \(p\) is the sample proportion (0.83 in this survey).
  • \(n\) is the total number of respondents (1000 adults).
The standard error serves as a measure of the sampling distribution’s spread. A smaller standard error indicates that the sample proportion is a more accurate reflection of the true population proportion.
Margin of Error
The **margin of error** provides a range around the sample proportion, within which we can assert with a certain level of confidence that the true population proportion lies.To compute the margin of error, multiply the standard error by the critical value associated with the desired confidence level. For a 95% confidence interval, this critical value, also known as the Z-score, is 1.96.The formula is:\[ MOE = Z * SE \]The margin of error allows us to create a confidence interval by adding and subtracting it from our sample proportion. These intervals are critical as they offer both a lower and an upper bound, indicating the range within which the true population value likely falls.
Statistical Interpretation
**Statistical interpretation** enables us to make informed conclusions and decisions based on data analysis. A 95% confidence interval means that if we were to take 100 different samples and compute a confidence interval for each sample, approximately 95 of those intervals would contain the true population proportion. In our example, if the confidence interval results in values between, say, 80% to 86%, this implies:
  • The true proportion of adults who think kids spend too much time on devices is very likely to fall between these bounds.
  • The idea that the proportion is less than 80% is not justifiable within this interval.
  • Similarly, it’s unlikely that it’s more than 85%.
Thus, statistical interpretation takes the abstract numbers and translates them into practical suggestions.
Survey Statistics
**Survey statistics** deals with the practical application of statistical methods to interpret survey data. When utilizing surveys, like the one in this exercise, you're gathering quantitative data to estimate parameters like proportions. This exercise exemplifies how survey statistics help make inferences about larger populations from smaller samples. The reliability of survey statistics heavily depends on:
  • Quality of the sample (random and representative).
  • Accuracy of measurement methods (e.g., precise formulation of survey questions).
  • Appropriate data analysis techniques.
Survey statistics not only aids in understanding public opinion but also drives decision-making processes, influencing policies and strategies based on empirical data.

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Most popular questions from this chapter

A \(90 \%\) confidence interval for \(\mu\) using the sample results \(\bar{x}=3.1, s=0.4,\) and \(n=100\)

A margin of error within \(\pm 2 \%\) with \(95 \%\) confidence. An initial small sample has \(\hat{p}=0.78\).

(a) Find the relevant sample proportions in each group and the pooled proportion. (b) Complete the hypothesis test using the normal distribution and show all details. Table 6.10 gives flight arrival numbers from a random sample of flights for two airlines. Test whether there is a difference between the two airlines in the percent of flights that arrive late.

On page 1 1 in Section \(1.1,\) we describe studies to investigate whether there is evidence of pheromones (subconscious chemical signals) in female tears that affect sexual arousal in men. In one of the studies, \(^{71} 50\) men had a pad attached to the upper lip that contained either female tears or a salt solution dripped down the same female's face. Each subject participated twice, on consecutive days, once with tears and once with saline, randomized for order, and doubleblind. Testosterone levels were measured before sniffing and after sniffing on both days. While normal testosterone levels vary significantly between different men, average levels for the group were the same before sniffing on both days and after sniffing the salt solution (about \(155 \mathrm{pg} / \mathrm{mL}\) ) but were reduced after sniffing the tears (about \(133 \mathrm{pg} / \mathrm{mL}\) ). The mean difference in testosterone levels after sniffing the tears was 21.7 with standard deviation \(46.5 .\) (a) Why did the investigators choose a matchedpairs design for this experiment? (b) Test to see if testosterone levels are significantly reduced after sniffing tears? (c) Can we conclude that sniffing female tears reduces testosterone levels (which is a significant indicator of sexual arousal in men)?

Standard Error from a Formula and Simulation In Exercises 6.15 to \(6.18,\) find the mean and standard error of the sample proportions two ways: (a) Use StatKey or other technology to simulate at least 1000 sample proportions. Give the mean and standard error and comment on whether the distribution appears to be normal. (b) Use the formulas in the Central Limit Theorem to compute the mean and standard error. Are the results similar to those found in part (a)? Sample proportions of sample size \(n=10\) from a population with \(p=0.2\)

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