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Downloading Music. A husband and wife, Stan and Lucretia, share a digital audio player that has a feature that randomly selects which song to play. A total of 2444 songs have been loaded into the player, some by Stan and the rest by Lucretia. They are interested in determining whether they have loaded different proportions of songs into the player. Suppose that when the player was in the randomselection mode, 26 of the first 40 songs selected were songs loaded by Lucretia. Let \(p\) denote the proportion of songs that were loaded by Lucretia. a. State the null and alternative hypotheses to be tested. How strong is the evidence that Stan and Lucretia have loaded different proportions of songs into the player? Make sure to check the conditions for the use of this test. b. Are the conditions for the use of the large-sample confidence interval met? If so, estimate with \(95 \%\) confidence the proportion of songs that were loaded by Lucretia.

Short Answer

Expert verified
The evidence is weak for different proportions (p-value > 0.05). The 95% CI for Lucretia's proportion is [0.497, 0.803].

Step by step solution

01

State the Hypotheses

The null hypothesis, \( H_0 \), is that the proportion of songs loaded by Lucretia is \( p = 0.5 \). The alternative hypothesis, \( H_a \), is that the proportion \( p e 0.5 \).
02

Check Conditions for Hypothesis Testing

The conditions for a hypothesis test about a proportion are: the sample should be random, the sample size should be large enough so that both \( np \) and \( n(1-p) \) are greater than 5. Here \( n = 40 \) and the observed value is \( p = 26/40 = 0.65 \). So, \( np = 40 \times 0.65 = 26 \) and \( n(1-p) = 40 \times 0.35 = 14 \), both greater than 5.
03

Calculate Test Statistic

The test statistic for this hypothesis test is calculated using the formula \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \). Here \( \hat{p} = 0.65 \), \( p_0 = 0.5 \), and \( n = 40 \). Calculate \( z = \frac{0.65 - 0.5}{\sqrt{\frac{0.5(1-0.5)}{40}}} = \frac{0.15}{0.079} \approx 1.898 \).
04

Determine P-value and Conclusion

Using standard normal distribution tables, find the p-value for \( z = 1.898 \). The two-tailed p-value is approximately 0.057, which is greater than 0.05. Therefore, we do not reject the null hypothesis at the 5% significance level, indicating weak evidence against \( H_0 \).
05

Check Conditions for Confidence Interval

The conditions are the same as for the hypothesis test. Here, \( np = 26 \) and \( n(1-p) = 14 \), both of which are greater than 5, so the conditions are satisfied.
06

Calculate Confidence Interval

The 95% confidence interval for \( p \) is given by \( \hat{p} \pm z^* \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( z^* = 1.96 \) for a 95% confidence level. Calculate the interval: \( 0.65 \pm 1.96 \cdot \sqrt{\frac{0.65 \cdot 0.35}{40}} \approx 0.65 \pm 0.153 \), or \([0.497, 0.803]\).

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Key Concepts

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

Null Hypothesis
The null hypothesis is a fundamental concept in hypothesis testing. It is a statement or assumption that there is no effect or no difference. In the context of the problem, the null hypothesis, denoted as \(H_0\), assumes that the proportion of songs loaded by Lucretia is equal to 0.5. Essentially, this means that Stan and Lucretia have loaded an equal number of songs into the player. The null hypothesis serves as a starting point. It is the hypothesis that we initially assume to be true and then test against.For hypothesis testing, this assumption will be either rejected or not rejected based on the sample evidence. If the null hypothesis holds, it suggests no significant difference exists in the proportions. But remember, failing to reject the null hypothesis does not mean we have proved it true. It only suggests there's not enough evidence against it in this particular sample.
Alternative Hypothesis
The alternative hypothesis offers a statement that contradicts the null hypothesis. It represents the researcher's claim that there is a statistically significant effect or difference. For our example, the alternative hypothesis, denoted as \(H_a\), proposes that the proportion of songs loaded by Lucretia is not equal to 0.5. This is a 'not equal to' statement, which indicates that Lucretia's proportion of songs could be either greater or less than 0.5. It's a two-tailed test, meaning the difference can go in either direction. The goal of hypothesis testing is typically to gather evidence to support the alternative hypothesis.Testing this hypothesis involves calculating a test statistic, which in this case was a \(z\)-score of approximately 1.898. The \(p\)-value associated with this statistic tells us the probability of observing our data if the null hypothesis were true. Since the calculated \(p\)-value is greater than the typical significance level of 0.05, the result supports the null hypothesis. However, it doesn't strongly support it, indicating that there's some evidence in favor of the alternative hypothesis.
Confidence Interval
A confidence interval provides a range of values within which the true parameter is expected to fall, with a certain degree of confidence. In our exercise, we calculated a 95% confidence interval for Lucretia's proportion of songs, which ranges from 0.497 to 0.803. This interval gives us a range of plausible values for Lucretia's share of the songs on the player. Confidence intervals are useful because they provide more information than a simple hypothesis test. They not only tell us whether or not to reject the null hypothesis, but also give an understanding of the potential size of the effect. The width of the interval reflects the uncertainty of our estimate; a narrower interval indicates more precision. We are 95% confident that the true proportion of songs loaded by Lucretia falls within this range. Importantly, since the interval includes 0.5, it suggests that there is a possibility that the proportions are actually equal, consistent with the null hypothesis result.
Proportion Testing
Proportion testing is used when you want to make inferences about a population proportion based on a sample proportion. In this scenario, we are interested in Lucretia's proportion of songs on the digital player.The test begins by estimating the sample proportion, \(\hat{p}\), which in our case was 0.65. Then, it requires checking certain conditions:
  • The sample should be random.
  • The sample size must be large enough, so both \(np\) and \(n(1-p)\) are greater than 5.
These conditions are met, as both calculations resulted in values greater than 5, making proportion testing appropriate.The test involves calculating a \(z\)-score, which helps in determining the likelihood of observing our data under the null hypothesis. This \(z\)-score is compared against a threshold determined by our chosen significance level. The idea is to see whether the observed proportion could have occurred by random chance under the initial assumption of equality. The calculated \(p\)-value then helps make a decision about the null hypothesis. This entire process helps determine if there's enough evidence to claim a significant difference in proportions.

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Most popular questions from this chapter

Cannabidiol- (CBD-) based products are widely touted for their therapeutic benefits without any psychoactive effects. A 2019 Gallup Poll asked a random sample of 2543 U.S. adults if they personally use CBD products or not. Suppose that, in fact, \(15 \%\) of all U.S. adults have used CBD products. In repeated samples, the sample proportion \(\widehat{p}\) would follow approximately a Normal distribution with mean a. \(381.45\). b. \(0.15\). c. \(0.007\).

Based on the sample, the large-sample \(90 \%\) confidence interval for the proportion of all American adults who actively try to avoid drinking regular soda or pop is a. \(0.61 \pm 0.015\). b. \(0.61 \pm 0.025\). c. \(0.61 \pm 0.029\).

Book Reading. Although an increasing share of Americans are reading e-books on tablets and smartphones rather than dedicated e-readers, print books continue to be much more popular than books in digital format (digital format includes both e-books and audio books). A Pew Research Center survey of 1502 adults nationwide conducted January 8-February 7, 2019, found that 1081 of those surveyed had read a book in either print or digital format in the preceding 12 months. 28 (You may regard the 1081 adults in the survey who had read a book in the preceding 12 months as a random sample of readers.) a. What can you say with \(95 \%\) confidence about the percentage of all adults who had read a book in either print or digital format in the preceding 12 months? b. Of the 1081 surveyed who had read a book in the preceding 12 months, 105 had read only digital books. Among those adults who had read a book in the preceding 12 months, find a \(95 \%\) confidence interval for the proportion that had read digital books exclusively.

Black Raspberries and Cancer. Sample surveys usually contact large samples, so we can use the large-sample confidence interval if the sample design is close to an SRS. Scientific studies often use smaller samples that require the plus four method. For example, Familial Adenomatous Polyposis (FAP) is a rare inherited disease characterized by the development of an extreme number of polyps early in life and colon cancer in virtually \(100 \%\) of patients before the age of 40 . A group of 14 people suffering from FAP being treated at the Cleveland Clinic drank black raspberry powder in a slurry of water every day for nine months. The numbers of polyps were reduced in 11 out of 14 of these patients. 19 a. Why can't we use the large-sample confidence interval for the proportion \(p\) of patients suffering from FAP that will have the number of polyps reduced after nine months of treatment? b. The plus four method adds four observations: two successes and two failures. What are the sample size and the number of successes after you do this? What is the plus four estimate \(\tilde{p}\) of \(p\) ? c. Give the plus four \(90 \%\) confidence interval for the proportion of patients suffering from FAP who will have the number of polyps reduced after nine months of treatment.

Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. Only half of all mice complete one particular maze in less than 18 seconds. A researcher thinks that a loud noise will cause the mice to complete the maze faster. She measures the proportion of 40 mice that completed the maze in less than 18 seconds with noise as a stimulus. The proportion of mice that completed the maze in less than 18 seconds is \(\widehat{p}=0.7\). The hypotheses for a test to answer the researcher's question are a. \(H_{0}: p=0.5, H_{a}: p>0.5\). b. \(H_{0}: p=0.5, H_{a}: p<0.5\). c. \(H_{0}: p=0.5, H_{a}: p \neq 0.5\).

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