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According to a Washington Post- \(A B C\) News poll, 331 of 502 randomly selected U.S. adults interviewed said they would not be bothered if the National Security Agency collected records of personal telephone calls they had made. Is there sufficient evidence to conclude that a majority of U.S. adults feel this way? Test the appropriate hypotheses using a .01 significance level.

Short Answer

Expert verified
The conclusion about whether the majority of U.S adults feel that way or not depends on the calculated Z-score and the comparison of its p-value with the given significance level of 0.01.

Step by step solution

01

State the Hypotheses

The null hypothesis (H0) assumes that less than or equal to 50% of U.S adults feel this way. The alternative hypothesis (H1) assumes that more than 50% of U.S adults feel this way. This yields to: H0: p ≤ 0.5 H1: p > 0.5
02

Calculate the Test Statistic

The test statistic for hypothesis testing for a population proportion is calculated using the Z-score formula: Z = (p̂ - P0) / sqrt ((P0(1 - P0)) / n)where p̂ is the sample proportion (331 / 502 = 0.659), P0 is the assumed population proportion under the null hypothesis (0.5), and n is the sample size (502). This results in: Z = (0.659 - 0.5) / sqrt ((0.5 * 0.5) / 502)
03

Find the Z-Score Using Tables or Calculation

From the normal distribution table or calculation, find the Z-score that corresponds to our calculated test statistic.
04

Reject or Fail to Reject the Null Hypothesis

If the Z-score corresponds to a p-value less than the significance level (0.01), then reject H0, otherwise fail to reject H0. This will provide the evidence to conclude if more than 50% of U.S adults feel this way.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis, denoted as \( H_0 \), represents a statement of no effect or no difference. It often includes a condition of equality, such as \( \leq \) (less than or equal to), \( \geq \) (greater than or equal to), or \( = \) (equal to). In the context of the given exercise, the null hypothesis assumes that 50% or fewer of U.S. adults would not be bothered by the National Security Agency collecting personal phone call records. Mathematically, it is stated as:
  • \( H_0: p \leq 0.5 \)
Here, \( p \) represents the population proportion. The null hypothesis serves as the premise that is tested for evidence against it. It is important to remember that failing to reject the null hypothesis does not prove it to be true; it only suggests there is not enough evidence against it.
It is a crucial component in statistical hypothesis testing, providing a benchmark against which to compare observed data.
Alternative Hypothesis
Contrasting the null hypothesis, the alternative hypothesis, denoted as \( H_1 \) or \( H_a \), reflects what one might believe to be true or is trying to prove with the statistical test. It often suggests that there is a significant effect or a difference. In our scenario, the alternative hypothesis assumes that more than 50% of U.S. adults would not be bothered by the NSA gathering their phone records. This is articulated as:
  • \( H_1: p > 0.5 \)
This alternative hypothesis is set up because we want to find evidence to support that a majority (more than half) feels this way. The decision to reject the null hypothesis—in favor of the alternative—is based on whether the evidence, in the form of sample data, supports this or not. It's all about finding legitimate backing for the observed data against the baseline given by the null hypothesis.
Z-score Calculation
The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. In the context of hypothesis testing for population proportions, it helps us determine how far the sample proportion is from the assumed population proportion under the null hypothesis, in terms of standard errors.The formula used is:\[Z = \frac{\hat{p} - P_0}{\sqrt{\frac{P_0(1-P_0)}{n}}}\]
  • \( \hat{p} \) is the sample proportion, which in our case is calculated as \( \frac{331}{502} = 0.659 \).
  • \( P_0 \) is the hypothesized population proportion under the null hypothesis, equal to 0.5 here.
  • \( n \) is the size of the sample, which is 502 in this situation.
By substituting these values into the formula, we can determine the Z-score, which informs us how many standard deviations away our sample proportion is from the population proportion prescribed by our null hypothesis. A larger absolute value of the Z-score indicates stronger evidence against the null hypothesis.
Population Proportion
Population proportion is a key concept in statistics. It refers to the fraction of the total population that possesses a certain attribute. In this exercise, the population proportion (represented by \( p \)) signifies the percentage of all U.S. adults who are not disturbed by the NSA collecting personal phone records.
The sample proportion (\( \hat{p} \)) is an estimate of the population proportion, derived from the sample data—in this scenario, it was calculated to be 0.659 (or 65.9%).
Understanding population proportion is vital, as it sets the parameter, \( P_0 \), in hypothesis tests regarding proportions. When carrying out such a test, this is the value under the null hypothesis against which the sample is tested. If our sample provides compelling evidence, we can reject the null hypothesis and argue that the actual population proportion differs from \( P_0 \). This forms the basis for drawing conclusions about an entire population based on a representative sample.

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Most popular questions from this chapter

Suppose that you are an inspector for the Fish and Game Department and that you are given the task of determining whether to prohibit fishing along part of the Oregon coast. You will close an area to fishing if it is determined that fish in that region have an unacceptably high mercury content. a. Assuming that a mercury concentration of \(5 \mathrm{ppm}\) is considered the maximum safe concentration, which of the following pairs of hypotheses would you test: $$ H_{0}: \mu=5 \text { versus } H_{a}: \mu>5 $$ or $$ H_{0}: \mu=5 \text { versus } H_{a}: \mu<5 $$ Give the reasons for your choice. b. Would you prefer a significance level of \(.1\) or \(.01\) for your test? Explain.

A manufacturer of hand-held calculators receives large shipments of printed circuits from a supplier. It is too costly and time-consuming to inspect all incoming circuits, so when each shipment arrives, a sample is selected for inspection. Information from the sample is then used to test \(H_{0}: \pi=.05\) versus \(H_{a}: \pi>.05\), where \(\pi\) is the true proportion of defective circuits in the shipment. If the null hypothesis is not rejected, the shipment is accepted, and the circuits are used in the production of calculators. If the null hypothesis is rejected, the entire shipment is returned to the supplier because of inferior quality. (A shipment is defined to be of inferior quality if it contains more than \(5 \%\) defective circuits.) a. In this context, define Type I and Type II errors. b. From the calculator manufacturer's point of view, which type of error is considered more serious? c. From the printed circuit supplier's point of view, which type of error is considered more serious?

Many older homes have electrical systems that use fuses rather than circuit breakers. A manufacturer of 40 -amp fuses wants to make sure that the mean amperage at which its fuses burn out is in fact \(40 .\) If the mean amperage is lower than 40 , customers will complain because the fuses require replacement too often. If the mean amperage is higher than 40 , the manufacturer might be liable for damage to an electrical system as a result of fuse malfunction. To verify the mean amperage of the fuses, a sample of fuses is selected and tested. If a hypothesis test is performed using the resulting data, what null and alternative hypotheses would be of interest to the manufacturer?

The amount of shaft wear after a fixed mileage was determined for each of 7 randomly selected internal combustion engines, resulting in a mean of \(0.0372\) in. and a standard deviation of \(0.0125 \mathrm{in}\). a. Assuming that the distribution of shaft wear is normal, test at level \(.05\) the hypotheses \(H_{0}: \mu=.035\) versus \(H_{a}:\) \(\mu>.035\) b. Using \(\sigma=0.0125, \alpha=.05\), and Appendix Table 5 . what is the approximate value of \(\beta\), the probability of a Type II error, when \(\mu=.04\) ? c. What is the approximate power of the test when \(\mu=\) \(.04\) and \(\alpha=.05\) ?

A random sample of \(n=44\) individuals with a B.S. degree in accounting who started with a Big Eight accounting firm and subsequently changed jobs resulted in a sample mean time to change of \(35.02\) months and a sample standard deviation of \(18.94\) months ("The Debate over Post-Baccalaureate Education: One University's Experience," Issues in Accounting Education \([1992]: 18-36)\). Can it be concluded that the true average time to change exceeds 2 years? Test the appropriate hypotheses using a significance level of \(.01\).

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