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Chapter 6: Problem 24

CA vs. OR, Part II. 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
No evidence of difference; potential Type II error if incorrect.

Step by step solution

01

Identify Hypotheses

Define the null and alternative hypotheses. We are testing if there is a difference in proportions of sleep deprivation between California and Oregon residents. Set up the hypotheses as follows:- Null hypothesis, \(H_0\): There is no difference in the sleep deprivation rates between California and Oregon, i.e., \( p_1 = p_2 \).- Alternative hypothesis, \(H_a\): There is a difference in the sleep deprivation rates between the two states, i.e., \( p_1 eq p_2 \).
02

Check Conditions

Ensure sample sizes and proportions are appropriate for using a z-test for two proportions:1. Large sample size: Check that both \( n_1 \times \hat{p}_1 \) and \( n_1 \times (1 - \hat{p}_1) \), and \( n_2 \times \hat{p}_2 \) and \( n_2 \times (1 - \hat{p}_2) \) are all greater than 5. - For California, \( n_1 = 11,545 \), \( \hat{p}_1 = 0.08 \) gives \( 11,545 \times 0.08 = 923.6 \) and \( 11,545 \times 0.92 = 10,621.4 \), both of which are greater than 5. - For Oregon, \( n_2 = 4,691 \), \( \hat{p}_2 = 0.088 \) gives \( 4,691 \times 0.088 = 412.9 \) and \( 4,691 \times 0.912 = 4,278.1 \), both of which are greater than 5. Conditions are satisfied.
03

Calculate Test Statistic

Use the formula for the z-test statistic for two proportions:\[z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1 - \hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}\]First, compute the pooled proportion:\[\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{(0.08 \times 11,545) + (0.088 \times 4,691)}{11,545 + 4,691} = \frac{923.6 + 412.9}{16,236} = \frac{1336.5}{16,236} \approx 0.0823\]Plug the values into the z formula:\[z = \frac{0.08 - 0.088}{\sqrt{0.0823 \times 0.9177 \times \left(\frac{1}{11,545} + \frac{1}{4,691}\right)}} \approx \frac{-0.008}{0.0044} \approx -1.818\]
04

Find p-Value

Using a standard normal distribution, find the p-value associated with \(z = -1.818\). As we are conducting a two-tailed test, the p-value is twice the area to the left of \(z = -1.818\). Using a Z-table or calculator:\[ \text{p-value} \approx 2 \times 0.0346 = 0.0692\]
05

Make a Decision

Compare the p-value to a significance level (typically \(\alpha = 0.05\)). If the p-value is less than \(\alpha\), reject the null hypothesis. Here, \( 0.0692 > 0.05\), so we do not reject the null hypothesis.
06

Conclusion

There is insufficient evidence to conclude that there is a difference in the rate of sleep deprivation between California and Oregon residents.
07

Identify Potential Error

If the conclusion was incorrect, we might have made a Type II error, which means failing to reject the null hypothesis (saying there is no difference) when there actually is a difference.

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Key Concepts

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

p-value
The p-value is a probability that helps us determine the significance of our test results. In hypothesis testing, it quantifies the evidence against the null hypothesis. Put simply, the p-value tells you how likely it is to observe the data you've collected, or something more extreme, assuming the null hypothesis is true.

When you conduct a test and obtain a p-value, you make a decision based on its size relative to your chosen significance level (commonly set at 5"). For instance:
  • A p-value less than 0.05 suggests strong evidence against the null hypothesis, prompting you to reject it.
  • A p-value greater than 0.05 means the evidence is not strong enough to reject the null hypothesis.
In the sleep deprivation exercise, the p-value calculated was approximately 0.0692, which is higher than 0.05. So, we did not reject the null hypothesis, indicating insufficient evidence to claim a difference in sleep deprivation rates between California and Oregon.
z-test for two proportions
The z-test for two proportions is a statistical method used to compare the proportions from two independent groups. It helps determine if there is a significant difference between them.

To conduct a z-test for two proportions, you must ensure that certain conditions are met. These include having large enough sample sizes so that the sampling distribution can be approximated by a normal distribution.

For this test, follow these steps:
  • Calculate the sample proportions for each group.
  • Check that both populations are approximately normally distributed.
  • Calculate the pooled proportion.
  • Use these values to compute the z-statistic.
In the exercise, the z-statistic was approximately -1.818, calculated by comparing the sleep deprivation rates of Californians and Oregonians. The negative sign indicates that the California rate was lower than the Oregon rate, but this difference wasn't statistically significant due to the p-value being above 0.05.
type II error
A Type II error occurs in hypothesis testing when a true alternative hypothesis is incorrectly not supported. It's the mistake of concluding that there is no effect or difference, when, in fact, an effect or difference exists.

In simpler terms, it means failing to reject a false null hypothesis. This error can have several causes, such as inadequate sample sizes or low statistical power.

The context of our exercise suggests a potential Type II error when we don't reject the null hypothesis, concluding there's no difference in sleep deprivation rates between the two states when there might actually be one.

Minimizing Type II error generally involves using larger sample sizes and ensuring the test has enough statistical power to detect a real effect.
null hypothesis
The null hypothesis, often denoted as \( H_0 \), is a statement used in statistics that proposes no effect or no difference between groups. It serves as a starting point for hypothesis testing, providing a baseline that you seek to test against alternative hypotheses.

In the given exercise, the null hypothesis was that the sleep deprivation rates in California and Oregon were equal. This hypothesis assumes that any observed variations in the sample proportions are due to random sampling error rather than a true difference.

Testing this involved assuming that any discrepancy between the two states' sleep deprivation rates could just be random chance. A failure to reject the null hypothesis in this case suggests that there wasn't enough statistical evidence to conclude that these rates truly differ between California and Oregon residents.

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Most popular questions from this chapter

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?

A survey on 1,509 high school seniors who took the SAT and who completed an optional web survey shows that \(55 \%\) of high school seniors are fairly certain that they will participate in a study abroad program in college. \(^{19}\) (a) Is this sample a representative sample from the population of all high school seniors in the US? Explain your reasoning. (b) Let's suppose the conditions for inference are met. Even if your answer to part (a) indicated that this approach would not be reliable, this analysis may still be interesting to carry out (though not report). Construct a \(90 \%\) confidence interval for the proportion of high school seniors (of those who took the SAT) who are fairly certain they will participate in a study abroad program in college, and interpret this interval in context. (c) What does "90\% confidence" mean? (d) Based on this interval, would it be appropriate to claim that the majority of high school seniors are fairly certain that they will participate in a study abroad program in college?

A survey of 2,254 American adults indicates that \(17 \%\) of cell phone owners browse the internet exclusively on their phone rather than a computer or other device. \(^{59}\) (a) According to an online article, a report from a mobile research company indicates that 38 percent of Chinese mobile web users only access the internet through their cell phones. \(^{60}\) Conduct a hypothesis test to determine if these data provide strong evidence that the proportion of Americans who only use their cell phones to access the internet is different than the Chinese proportion of \(38 \%\). (b) Interpret the p-value in this context. (c) Calculate a \(95 \%\) confidence interval for the proportion of Americans who access the internet on their cell phones, and interpret the interval in this context.

Suppose that \(90 \%\) of orange tabby cats are male. Determine if the following statements are true or false, and explain your reasoning. (a) The distribution of sample proportions of random samples of size 30 is left skewed. (b) Using a sample size that is 4 times as large will reduce the standard error of the sample proportion by one-half. (c) The distribution of sample proportions of random samples of size 140 is approximately normal. (d) The distribution of sample proportions of random samples of size 280 is approximately normal.

CA vs. OR, Part I. 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.
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