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91Ó°ÊÓ

A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels yearround to reduce air pollution. If more than 315 voters respond positively, we will conclude that at least \(60 \%\) of the voters favor the use of these fuels. (a) Find the probability of type I error if exactly \(60 \%\) of the voters favor the use of these fuels. (b) What is the type II error probability \(\beta\) if \(75 \%\) of the voters favor this action?

Short Answer

Expert verified
Type I error probability is calculated using the z-score for p = 0.6, and Type II error \( \beta \) is found using the z-score for p = 0.75.

Step by step solution

01

Define Type I and Type II Errors

A Type I error occurs when we reject the null hypothesis when it is true. For this scenario, that's concluding that at least 60% favor the fuels when only 60% do. A Type II error occurs when we fail to reject the null hypothesis while the alternative is true. In this context, it's not concluding that at least 60% favor the fuels when, in fact, 75% do.
02

State the Null and Alternate Hypotheses

The null hypothesis ( H_0 ) is that 60% ( p = 0.6 ) of the voters favor using the fuels. The alternate hypothesis ( H_1 ) is that more than 60% favor using the fuels ( p > 0.6 ).
03

Determine Critical Value for Type I Error

Calculate the critical value of the sample proportion needed to reject the null hypothesis. This is where more than 315 out of 500 responses are positive. Therefore, the critical sample proportion ( p_c ) is 316/500 = 0.632.
04

Type I Error Calculation

If exactly 60% of voters favor the fuels, use the normal approximation to calculate the probability of observing at least 0.632.The standard deviation, \( \sigma \), is calculated by \( \sigma = \sqrt{\frac{0.6 \times 0.4}{500}}\).Calculate the z-score: \( z = \frac{0.632 - 0.6}{\sigma} \).Find the corresponding p-value for this z-score from the standard normal table.
05

Evaluate Type II Error Conditions

Set the new population proportion at 0.75 as per the new scenario. We need to calculate the probability that we fail to observe at least 0.632 when the true proportion is 0.75.
06

Calculate Type II Error Probability

When 75% favor the use, use the normal approximation to calculate the probability of observing less than 0.632.Find \( \sigma \) as \( = \sqrt{\frac{0.75 \times 0.25}{500}} \).Find \( z \) using \( \frac{0.632 - 0.75}{\sigma} \).Look up this z-score's probability from the standard normal distribution table, which represents \( \beta \) (Type II error probability).

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Key Concepts

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

Type I Error
In hypothesis testing, a Type I error occurs when the null hypothesis is rejected even though it is true. It is like a false alarm. In our exercise, the null hypothesis states that 60% of voters favor using the fuels. A Type I error would mean concluding that more than 60% of voters support the fuels when actually only 60% do. This error can be thought of as jumping the gun because we mistakenly believe there is an effect or a difference when there isn’t one. The probability of making a Type I error is represented by the Greek letter alpha (\( \alpha \)). This probability is usually set in advance by the researcher and is known as the significance level of the test. Typically, \( \alpha \) is set at 0.05 or 5%, which means there is a 5% risk of making a Type I error.
Type II Error
A Type II error happens when we fail to reject a false null hypothesis. Essentially, it's missing the real effect or difference that exists. In this exercise, it occurs when the test fails to show that more than 60% of voters favor the use of fuels, despite 75% actually supporting it. This kind of error is sometimes referred to as a "miss" in statistics. The probability of making a Type II error is denoted by the Greek letter beta (\( \beta \)). It reflects the chance that a real effect or true positive goes undetected in the study. Type II errors can often be reduced by increasing the sample size or by selecting a more sensitive test. A smaller \( \beta \) value indicates a higher power of the test, which measures the test's ability to detect an effect when there is one.
Critical Value
The critical value in hypothesis testing is a threshold that determines whether the test statistic is extreme enough to reject the null hypothesis. It's a key component in deciding the outcome of a hypothesis test. In our voter exercise, we set a critical sample proportion of 0.632 as the threshold. If the observed sample proportion exceeds this critical value, we will reject the null hypothesis.Critical values depend on the significance level (\( \alpha \)) and the distribution of the test statistic. For instance, a 5% significance level in a standard normal distribution will have a critical value around 1.96 for two-tailed tests. The critical value helps manage the risk of Type I error by establishing how extreme the data must be to conclude that something significant is happening.
Z-score
A z-score is a measure that describes a value's position relative to the mean of a group of values, expressed in terms of standard deviations. In terms of hypothesis testing, z-scores help determine whether to reject the null hypothesis. For our exercise, we calculate the z-score to see how extreme the observed sample proportion (0.632) is under the assumption that the null hypothesis is true (60% support).The formula for calculating a z-score when testing a proportion is:\[ z = \frac{{\text{{Observed Proportion}} - \text{{Expected Proportion}}}}{{\sigma}} \]where \( \sigma \) is the standard deviation of the sample proportion, calculated using the formula:\[ \sigma = \sqrt{\frac{{p(1 - p)}}{n}} \]The z-score allows us to convert our sample result into a standard normal distribution; thus, making it easier to compare how likely or unlikely our sample result is.
Normal Distribution
The normal distribution, or Gaussian distribution, is a continuous probability distribution that is symmetrical around its mean, showing that data near the mean are more frequent in occurrence than data far from it. It appears in many fields of study, especially in the context of hypothesis testing, because many statistics tend to be normally distributed given a large enough sample size due to the Central Limit Theorem. In hypothesis testing, we often assume that our test statistic follows a normal distribution. This assumption allows us to use z-scores and critical values effectively. The properties of a normal distribution include:
  • The mean, median, and mode of the distribution are all equal.
  • The curve is bell-shaped and symmetrical about the mean.
  • Approximately 68% of the data falls within one standard deviation of the mean. About 95% falls within two, and 99.7% falls within three.
Understanding normal distribution is crucial in interpreting z-scores and making decisions about hypotheses based on critical values.

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Most popular questions from this chapter

A consumer products company is formulating a new shampoo and is interested in foam height (in millimeters). Foam height is approximately normally distributed and has a standard deviation of 20 millimeters. The company wishes to test \(H_{0}: \mu=175\) millimeters versus \(H_{1}: \mu>175\) millimeters, using the results of \(n=10\) samples. (a) Find the type I error probability \(\alpha\) if the critical region is \(\bar{x}>185\) (b) What is the probability of type II error if the true mean foam height is 185 millimeters? (c) Find \(\beta\) for the true mean of 195 millimeters.

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A 1992 article in the Journal of the American Medical Association ("A Critical Appraisal of 98.6 Degrees \(\mathrm{F}\), the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich") reported body temperature, gender, and heart rate for a number of subjects. The body temperatures for 25 female subjects follow: 97.8,97.2,97.4,97.6 97.8,97.9,98.0,98.0,98.0,98.1,98.2,98.3,98.3,98.4,98.4 \(98.4,98.5,98.6,98.6,98.7,98.8,98.8,98.9,98.9,\) and \(99.0 .\) (a) Test the hypothesis \(H_{0}: \mu=98.6\) versus \(H_{1}: \mu \neq 98.6,\) using \(\alpha=0.05 .\) Find the \(P\) -value. (b) Check the assumption that female body temperature is normally distributed. (c) Compute the power of the test if the true mean female body temperature is as low as \(98.0 .\) (d) What sample size would be required to detect a true mean female body temperature as low as 98.2 if you wanted the power of the test to be at least \(0.9 ?\) (e) Explain how the question in part (a) could be answered by constructing a two-sided confidence interval on the mean female body temperature.

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