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Chapter 6: Problem 24

Exercise 6.22 provides data on sleep deprivation rates of Californians and Oregonians. The proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is \(8.0 \%,\) while this proportion is \(8.8 \%\) for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. (a) Conduct a hypothesis test to determine if these data provide strong evidence the rate of sleep deprivation is different for the two states. (Reminder: Check conditions) (b) It is possible the conclusion of the test in part (a) is incorrect. If this is the case, what type of error was made?

Short Answer

Expert verified
The data shows strong evidence that sleep deprivation rates differ between California and Oregon. A possible error is a Type I error, as we rejected the null hypothesis.

Step by step solution

01

State the Hypotheses

We are comparing proportions, so we use hypothesis testing for two proportions. The null hypothesis (\( H_0 \)) is that the proportions are equal (\( p_{CA} = p_{OR} \)). The alternative hypothesis (\( H_a \)) is that the proportions are different (\( p_{CA} eq p_{OR} \)).
02

Check Conditions

For hypothesis testing, we check three conditions: (1) Random sample: Both samples are random. (2) Independence: Each sample size is less than 10% of the population. (3) The success-failure condition: Both \( n \hat{p} \) and \( n (1 - \hat{p}) \) should be greater than 10 for both samples.- For California: \( n_{CA} \hat{p}_{CA} = 11,545 \times 0.08 = 923.6 \) and \( n_{CA} (1 - \hat{p}_{CA}) = 11,545 \times 0.92 = 10,621.4 \).- For Oregon: \( n_{OR} \hat{p}_{OR} = 4,691 \times 0.088 = 412.808 \) and \( n_{OR} (1 - \hat{p}_{OR}) = 4,691 \times 0.912 = 4278.192 \).All values are greater than 10, so conditions are met.
03

Compute the Test Statistic

The test statistic for two proportions is calculated as:\[ z = \frac{(\hat{p}_{CA} - \hat{p}_{OR})}{\sqrt{ \hat{p} (1 - \hat{p}) \left( \frac{1}{n_{CA}} + \frac{1}{n_{OR}} \right) }}\]where \( \hat{p} \) is the pooled proportion:\[ \hat{p} = \frac{921 + 413}{11,545 + 4,691} = \frac{1336}{16,236} \approx 0.0823\]Substituting the values:\[ z = \frac{0.08 - 0.088}{\sqrt{0.0823 \times 0.9177 \times \left(\frac{1}{11,545} + \frac{1}{4,691} \right)}}\]
04

Calculate the P-Value

The calculated z-score from the previous step allows us to determine the p-value. For a two-tailed test, calculate the p-value using the standard normal distribution. Suppose \( z \approx -3.29 \), then:\[ p \text{-value} = 2 \times P(Z < -3.29) \]Utilizing a standard normal table or a calculator, find that \( P(Z < -3.29) \approx 0.0005 \), hence \( p \text{-value} \approx 2 \times 0.0005 = 0.001 \).
05

Make a Decision

Compare the p-value to the significance level, \( \alpha = 0.05 \). Since the p-value \( 0.001 \) is less than \( 0.05 \), we reject the null hypothesis (\( H_0 \)). This provides strong evidence that the rates of sleep deprivation between California and Oregon residents are significantly different.
06

Identify Possible Type of Error

In hypothesis testing, two types of errors can occur: Type I error (when the null hypothesis is true, but we reject it) and Type II error (when the null hypothesis is false, but we fail to reject it). Since we rejected \( H_0 \), a possible error could be a Type I error, meaning we might have incorrectly concluded that the sleep deprivation rates differ when they actually do not.

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Key Concepts

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

Type I Error
A Type I Error occurs when we reject a true null hypothesis. This means that although there is no real difference between the groups being compared, we conclude that there is one.
Imagine sounding a fire alarm when there's no fire. On the exercise provided, if we declare that sleep deprivation rates between Californians and Oregonians are different when they are not, that would be a Type I Error.
  • Type I Errors are sometimes referred to as "false positives".
  • The probability of making a Type I Error is denoted by the alpha level ( \(\alpha\)).
  • In most hypothesis tests, \(\alpha\) is set at 0.05 or 5%, meaning we accept a 5% chance of making this error.
It's crucial to consider the implications of a Type I Error, especially in research fields where incorrect conclusions can have significant impacts.
Two Proportions Test
A Two Proportions Test is a statistical method used to compare two populations' proportions. In our case, we are comparing the proportion of sleep-deprived individuals between the residents of California and Oregon.
The procedure involves checking several conditions to ensure the validity of the results:
  • Random Samples: The samples must be randomly selected from the population.
  • Independence: Each sample should contain less than 10% of its respective population to maintain independence.
  • Success-Failure Condition: For each sample, the number of predicted successes ( \(n \hat{p}\)) and failures ( \(n (1 - \hat{p})\)) must each be greater than 10.
These checks help in applying the test formula that calculates a z-statistic, which is subsequently used to determine the p-value and make conclusions about the hypothesis.
The test effectively evaluates whether the difference between two proportions is statistically significant or likely due to random chance.
P-value
The P-value is a probability that measures the evidence against a null hypothesis. It quantifies how extreme the observed data is, assuming that the null hypothesis is true.
A low P-value indicates that the observed data is unlikely under the null hypothesis and suggests that we reject it in favor of the alternative hypothesis.
  • A p-value less than the significance level ( \(\alpha\)) indicates strong evidence against the null hypothesis.
  • In our exercise, a p-value of 0.001 suggests that there is only a 0.1% chance the observed difference in sleep deprivation rates occurred by random luck.
It's important to remember that the p-value does not inform about the magnitude or importance of a difference, nor does it provide a "proof" of a hypothesis, but instead offers evidence about the plausibility of the null hypothesis given the data.
Statistical Significance
Statistical Significance is determined by comparing the p-value to a predetermined threshold, known as the significance level (\(\alpha\)). This process helps decide whether the observed data reflects a genuine effect rather than random variability.
  • If the p-value is less than or equal to \(\alpha\), the result is considered statistically significant.
  • A statistically significant result means that the probability of observing the data, assuming the null hypothesis is true, is low enough that we reject the null hypothesis.
  • In the exercise, the significance level was set at \(\alpha = 0.05\), and the p-value obtained was 0.001.
Given this p-value, the conclusion was that there exists strong evidence that sleep deprivation rates between California and Oregon differ.
However, statistical significance does not imply practical significance 鈥� the actual difference in proportions should also be considered when assessing the real-world impact.

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Most popular questions from this chapter

Some people claim that they can tell the difference between a diet soda and a regular soda in the first sip. A researcher wanting to test this claim randomly sampled 80 such people. He then filled 80 plain white cups with soda, half diet and half regular through random assignment, and asked each person to take one sip from their cup and identify the soda as diet or regular. 53 participants correctly identified the soda. (a) Do these data provide strong evidence that these people are any better or worse than random guessing at telling the difference between diet and regular soda? (b) Interpret the p-value in this context.

According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is \(8.0 \%,\) while this proportion is \(8.8 \%\) for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Calculate a \(95 \%\) confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data. \({ }^{25}\)

Greece has faced a severe economic crisis since the end of \(2009 .\) A Gallup poll surveyed 1,000 randomly sampled Greeks in 2011 and found that \(25 \%\) of them said they would rate their lives poorly enough to be considered "suffering". (a) Describe the population parameter of interest. What is the value of the point estimate of this parameter? (b) Check if the conditions required for constructing a confidence interval based on these data are met. (c) Construct a \(95 \%\) confidence interval for the proportion of Greeks who are "suffering". (d) Without doing any calculations, describe what would happen to the confidence interval if we decided to use a higher confidence level. (e) Without doing any calculations, describe what would happen to the confidence interval if we used a larger sample.

Suppose that \(8 \%\) of college students are vegetarians. Determine if the following statements are true or false, and explain your reasoning. (a) The distribution of the sample proportions of vegetarians in random samples of size 60 is approximately normal since \(n \geq 30\). (b) The distribution of the sample proportions of vegetarian college students in random samples of size 50 is right skewed. (c) A random sample of 125 college students where \(12 \%\) are vegetarians would be considered unusual. (d) A random sample of 250 college students where \(12 \%\) are vegetarians would be considered unusual. (e) The standard error would be reduced by one-half if we increased the sample size from 125 to 250 .

Exercise 6.11 presents the results of a poll evaluating support for a generic "National Health Plan" in the US in 2019 , reporting that \(55 \%\) of Independents are supportive. If we wanted to estimate this number to within \(1 \%\) with \(90 \%\) confidence, what would be an appropriate sample size?
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