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A poll conducted in 2013 found that \(52 \%\) of U.S. adult Twitter users get at least some news on Twitter, and the standard error for this estimate was \(2.4 \% .\) Identify each of the following statements as true or false. Provide an explanation to justify each of your answers. (a) The data provide statistically significant evidence that more than half of U.S. adult Twitter users get some news through Twitter. Use a significance level of \(\alpha=0.01\). (b) Since the standard error is \(2.4 \%\), we can conclude that \(97.6 \%\) of all U.S. adult Twitter users were included in the study. (c) If we want to reduce the standard error of the estimate, we should collect less data. (d) If we construct a \(90 \%\) confidence interval for the percentage of U.S. adults Twitter users who get some news through Twitter, this confidence interval will be wider than a corresponding \(99 \%\) confidence interval.

Short Answer

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(a) False, (b) False, (c) False, (d) False.

Step by step solution

01

Analyze Statement (a)

To determine if more than half of U.S. adult Twitter users obtain news on Twitter at a significance level of \( \alpha = 0.01 \), we can perform a hypothesis test. - Null Hypothesis \( H_0 \): \( p = 0.5 \)- Alternative Hypothesis \( H_a \): \( p > 0.5 \)The test statistic for a proportion can be calculated using:\[ z = \frac{\hat{p} - p_0}{SE}\],where \( \hat{p} = 0.52 \), \( p_0 = 0.5 \), and \( SE = 0.024 \).Substituting these values gives:\[ z = \frac{0.52 - 0.5}{0.024} = \frac{0.02}{0.024} \approx 0.833.\]The critical z-value for \( \alpha = 0.01 \) (one-tailed test) is approximately 2.33.Since the calculated z-value (0.833) is less than 2.33, we fail to reject the null hypothesis. Therefore, **this statement is false**.
02

Analyze Statement (b)

The statement suggests concluding that 97.6% of Twitter users were surveyed based on the standard error of 2.4%. In reality, the standard error is a measure of variability that depends on the sample size, not the percentage of population surveyed. Therefore, this conclusion is incorrect. Thus, **this statement is false**.
03

Analyze Statement (c)

The standard error decreases with an increase in sample size because it is calculated using the formula:\[ SE = \frac{\sigma}{\sqrt{n}}\],where \( \sigma \) is the standard deviation and \( n \) is the sample size. Therefore, collecting less data would actually increase the standard error. Hence, **this statement is false**.
04

Analyze Statement (d)

A 90% confidence interval has a smaller critical value compared to a 99% confidence interval, as the Z-values for 90% and 99% intervals are different (1.645 vs 2.576 approximately). Thus, a 90% confidence interval will be narrower than a 99% confidence interval. Hence, **this statement is false**.

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Key Concepts

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

Confidence Intervals
Confidence intervals are a way to estimate a population parameter based on a sample statistic. They provide a range of values within which the true parameter is likely to fall, with a certain level of confidence. The level of confidence is typically expressed as a percentage, such as 90%, 95%, or 99%. For example, a 90% confidence interval suggests that if we were to take 100 different samples and construct a confidence interval from each of them, approximately 90 of the intervals would contain the true population parameter.

The width of the confidence interval is influenced by the sample size, standard deviation, and the confidence level itself. - A larger sample size tends to give a more precise (narrower) interval. - A higher confidence level results in a wider interval because we require more certainty. - A smaller variance or standard deviation in the sample also narrows the interval. In the original exercise, the assertion about the width of the confidence interval relates to the differences between 90% and 99% intervals, where a 90% confidence interval is narrower due to a smaller critical value compared to a 99% interval.
Standard Error
The standard error (SE) measures the variability or dispersion of a sample statistic from the actual population parameter. It is essentially the standard deviation of the sampling distribution, and it gets smaller as the sample size increases.

Mathematically, it is calculated as:\[ SE = \frac{\sigma}{\sqrt{n}} \]Where \( \sigma \) is the standard deviation of the population, and \( n \) is the sample size. - A larger sample size (\( n \)) reduces the standard error, resulting in a more accurate estimate of the population parameter. - Conversely, a smaller sample size increases the standard error, making the estimation less reliable.In the context of the original exercise, the standard error of 2.4% allows us to understand the precision of the 52% estimate of Twitter users. The statement related to 97.6% is misleading because the SE does not directly inform us about the proportion of the population surveyed.
Sample Size
Sample size is a critical component in statistics as it directly influences the reliability and precision of statistical estimates. A larger sample size tends to increase the accuracy of estimates by providing more data that can reflect the population more accurately. - When the sample size increases, the standard error generally decreases, which leads to a narrower confidence interval, leading to more precise estimates of the population parameter. - A small sample size might not represent the population well and can lead to larger errors in estimation. From the exercise, it was noted that tampering with the sample size would directly affect the standard error. Collecting less data (reducing sample size) results in a larger standard error, contrary to the claim made in the original question.
Z-Test for Proportions
A Z-test for proportions is a statistical method used to determine if there is a significant difference between the observed sample proportion and a known proportion in the population. - It is used when the data follows a normal distribution, and the sample size is large enough.- The Z-test helps us decide whether to reject the null hypothesis by comparing a calculated Z-statistic against a critical Z-value.In the mathematical formula for Z:\[ z = \frac{\hat{p} - p_0}{SE} \]- \( \hat{p} \) is the sample proportion,- \( p_0 \) is the hypothesized population proportion,- And \( SE \) is the standard error.In the original exercise, a Z-test was used to determine if more than half of U.S. adult Twitter users get news from Twitter with a significance level of \( \alpha = 0.01 \). The calculated Z-statistic was less than the critical Z-value, leading to the conclusion of failing to reject the null hypothesis.

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Most popular questions from this chapter

In each part below, there is a value of interest and two scenarios (I and II). For each part, report if the value of interest is larger under scenario I, scenario II, or whether the value is equal under the scenarios. (a) The standard error of \(\hat{p}\) when (I) \(n=125\) or (II) \(n=500\). (b) The margin of error of a confidence interval when the confidence level is (I) \(90 \%\) or (II) \(80 \%\). (c) The p-value for a Z-statistic of 2.5 calculated based on a (I) sample with \(n=500\) or based on a (II) sample with \(n=1000\). (d) The probability of making a Type 2 Error when the alternative hypothesis is true and the significance level is (I) 0.05 or (II) 0.10 .

The General Social Survey asked the question: "After an average work day, about how many hours do you have to relax or pursue activities that you enjoy?" to a random sample of 1,155 Americans. \(^{26}\) A \(95 \%\) confidence interval for the mean number of hours spent relaxing or pursuing activities they enjoy was (1.38,1.92) (a) Interpret this interval in context of the data. (b) Suppose another set of researchers reported a confidence interval with a larger margin of error based on the same sample of 1,155 Americans. How does their confidence level compare to the confidence level of the interval stated above? (c) Suppose next year a new survey asking the same question is conducted, and this time the sample size is 2,500 . Assuming that the population characteristics, with respect to how much time people spend relaxing after work, have not changed much within a year. How will the margin of error of the \(95 \%\) confidence interval constructed based on data from the new survey compare to the margin of error of the interval stated above?

A poll conducted in 2013 found that \(52 \%\) of U.S. adult Twitter users get at least some news on Twitter. \(^{13}\). The standard error for this estimate was \(2.4 \%\), and a normal distribution may be used to model the sample proportion. Construct a \(99 \%\) confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter, and interpret the confidence interval in context.

As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. (a) What population is under consideration in the data set? (b) What parameter is being estimated? (c) What is the point estimate for the parameter? (d) What is the name of the statistic can we use to measure the uncertainty of the point estimate? (e) Compute the value from part (d) for this context. (f) The historical rate of defects is \(10 \%\). Should the engineer be surprised by the observed rate of defects during the current week? (g) Suppose the true population value was found to be \(10 \%\). If we use this proportion to recompute the value in part (e) using \(p=0.1\) instead of \(\hat{p},\) does the resulting value change much?

A food safety inspector is called upon to investigate a restaurant with a few customer reports of poor sanitation practices. The food safety inspector uses a hypothesis testing framework to evaluate whether regulations are not being met. If he decides the restaurant is in gross violation, its license to serve food will be revoked. (a) Write the hypotheses in words. (b) What is a Type 1 Error in this context? (c) What is a Type 2 Error in this context? (d) Which error is more problematic for the restaurant owner? Why? (e) Which error is more problematic for the diners? Why? (f) As a diner, would you prefer that the food safety inspector requires strong evidence or very strong evidence of health concerns before revoking a restaurant's license? Explain your reasoning.

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