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According to a large national survey conducted by the Pew Research Center ("What Americans Think About NSA Surveillance, National Security and Privacy," May \(2,2015,\) wwW.pewresearch.org, retrieved December 1,2016 ), \(54 \%\) of adult Americans disapprove of the National Security Agency collecting records of phone and Internet data. Suppose that this estimate was based on a random sample of 1000 adult Americans. a. Is there convincing evidence that a majority of adult Americans feel this way? Test the relevant hypotheses using a 0.05 significance level. b. The actual sample size was much larger than 1000 . If you had used the actual sample size when doing the calculations for the test in Part (a), would the \(P\) -value have been larger than, the same as, or smaller than the \(P\) -value you obtained in Part (a)? Provide a justification for your answer.

Short Answer

Expert verified
In summary, based on a 0.05 significance level, we found convincing evidence that a majority of adult Americans disapprove of the NSA collecting phone and Internet data as our p-value of 0.02494 was less than the significance level. If we had used the actual (larger) sample size in the calculations, the p-value would have been smaller than the p-value obtained in Part (a) due to the smaller standard deviation in the sampling distribution of sample proportions, making the difference in proportions appear more extreme.

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis (\(H_0\)) states that there is no majority who disapprove the NSA collecting phone and Internet data, meaning the proportion is equal to 0.5. The alternative hypothesis (\(H_a\)) is that a majority disapproves, meaning the proportion is greater than 0.5. \(H_0: p = 0.5\) \(H_a: p > 0.5\)
02

Calculate the Test Statistic and p-value

To conduct a one-sample z-test for proportions, we have the sample proportion, \(\hat{p}\), which is the proportion of the 1000 adult Americans who disapprove (54% or 0.54). To calculate the test statistic, we can use the formula: \(z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\), where \(p_0\) is the null hypothesis's proportion (0.5) and n is the sample size (1000). \(z = \frac{0.54 - 0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}} = 1.96\) Now calculate the p-value using the z-score: The p-value can be found by looking up the z-score (1.96) in a standard normal distribution table or using a calculator. The p-value for z = 1.96 is approximately 0.02494.
03

Compare the p-value to the significance level and draw conclusions

Since the p-value (0.02494) is less than the significance level (0.05), we reject the null hypothesis. Therefore, there is convincing evidence that a majority of adult Americans disapprove of the NSA collecting phone and Internet data. #b. Comparing P-value with the actual sample size#
04

Analyze the effect of a larger sample size on the test statistic and p-value

When the sample size increases, the standard deviation of the sampling distribution of sample proportions decreases. This results in a narrower distribution, and thus, the same difference in proportions (actual and null hypothesis) would result in a more extreme z-score which would lead to a smaller p-value.
05

Conclusion on the effect of a larger sample size on the p-value

If we had used the actual sample size when doing the calculations for the test in Part (a), the p-value would have been smaller than the p-value obtained in Part (a). The justification for this conclusion is that a larger sample size leads to smaller standard deviation in the sampling distribution of sample proportions, making the same difference in proportions appear even more extreme, thus resulting in a smaller p-value.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental procedure in statistics used to determine whether there is enough evidence in a sample of data to infer that a certain condition is true for the entire population. In the context of the exercise provided, we wanted to know if a majority of adult Americans disapprove of the NSA's data collection practices.

To begin, we establish two contradictory statements: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). The null hypothesis usually represents a default position or a claim of 'no effect'. In contrast, the alternative hypothesis represents what we want to prove or suspect may be true. After calculating the test statistic and comparing the p-value to a pre-determined significance level, we can decide whether to reject the null hypothesis in favor of the alternative hypothesis, or fail to reject it.
One-Sample Z-Test for Proportions
A one-sample z-test for proportions is used when we wish to compare the sample proportion to a known population proportion. When dealing with large samples (generally n > 30) and known population standard deviations, the z-test is the ideal method.

The formula for the test statistic in a one-sample z-test is \(z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\) where \(\hat{p}\) is the sample proportion, \(p_0\) is the hypothesized population proportion, and \(n\) is the sample size. For our exercise, the sample proportion (\(\hat{p}\)) was 54%, and we compared it against the null hypothesis proportion (\(p_0\)) of 50%. The resultant z-score helps determine how far our sample proportion lies from the null hypothesis' expected proportion.
P-Value Significance
The p-value is a crucial concept in hypothesis testing. It helps us understand the strength of the evidence against the null hypothesis. A p-value is the probability of observing a test statistic as extreme as, or more extreme than, the value computed from the sample data, given that the null hypothesis is true.

If the p-value is less than or equal to the significance level (\(\alpha\)), often set at 0.05, it suggests that the observed data is unlikely under the null hypothesis and thus provides evidence to reject the null hypothesis. In the Pew Research Center example, with a p-value of approximately 0.02494, we reject the null hypothesis because this p-value is less than the 0.05 threshold, indicating significant evidence against the null hypothesis.
Sampling Distribution
A sampling distribution is a probability distribution of a statistic obtained from a large number of samples drawn from a specific population. This distribution tells us how the sample statistic would behave if we were to take many samples from the population.

In our case, we are dealing with the sampling distribution of the sample proportions. The central limit theorem states that, given a sufficiently large sample size, the sampling distribution of the sample mean will be normally distributed regardless of the population’s distribution. As the sample size increases, the standard deviation of the sampling distribution (also known as the standard error) decreases, leading to a narrower distribution. This effectively makes any observed deviation from the null hypothesis appear more significant, potentially resulting in a smaller p-value, as noted in steps 4 and 5 of our original exercise solution.

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Most popular questions from this chapter

Refer to the instructions given prior to Exercise \(10.57 .\) The paper referenced in the previous exercise also reported that when each of the 1178 students who participated in the study was asked if he or she played video games at least once a day, 271 responded "yes." The researchers were interested in using this information to decide if there is convincing evidence that more than \(20 \%\) of students age 8 to 18 play video games at least once a day.

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Which of the following specify legitimate pairs of null and alternative hypotheses? a. \(H_{0}: p=0.25 \quad H_{i}: p>0.25\) b. \(H_{0}: p<0.40 \quad H_{i}: p>0.40\) c. \(H_{0}: p=0.40 \quad H: p<0.65\) d. \(H_{0}: p \neq 0.50 \quad H_{i}: p=0.50\) e. \(H_{\mathrm{n}}: p=0.50 \quad H_{i}: p>0.50\) f. \(H_{0}: \hat{p}=0.25 \quad H_{e^{\prime}} \hat{p}>0.25\)

Explain why failing to reject the null hypothesis in a hypothesis test does not mean there is convincing evidence that the null hypothesis is true.

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