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The report titled "Digital Democracy Survey" (2016, www .deloitte.com/us/tmttrends, retrieved December 16,2016 ) stated that \(31 \%\) of the people in a representative sample of adult Americans age 33 to 49 rated a landline telephone among the three most important services that they purchase for their home. In a representative sample of adult Americans age 50 to \(68,48 \%\) rated a landline telephone as one of the top three services they purchase for their home. Suppose that the samples were independently selected and that the sample size was 600 for the 33 to 49 age group sample and 650 for the 50 to 68 age group sample. Does this data provide convincing evidence that the proportion of adult Americans age 33 to 49 who rate a landline phone in the top three is less than this proportion for adult Americans age 50 to 68 ? Test the relevant hypotheses using \(\alpha=0.05\)

Short Answer

Expert verified
In conclusion, after comparing the calculated p-value with the given significance level \(\alpha = 0.05\), we can either reject or fail to reject the null hypothesis. If the p-value is less than \(\alpha\), it suggests that the proportion of adult Americans age 33 to 49 who rate a landline phone as one of the top three is indeed lower than this proportion for adult Americans age 50 to 68. If the p-value is greater than or equal to \(\alpha\), there is not enough evidence to support a difference in the proportions of the two age groups.

Step by step solution

01

State the Null and Alternative Hypotheses

Let \(p_1\) be the proportion of adult Americans age 33 to 49 who rate a landline phone in the top three and \(p_2\) be the proportion of adult Americans age 50 to 68 who rate a landline phone in the top three. - Null Hypothesis, \(H_0: p_1 - p_2 \geq 0\) - Alternative Hypothesis, \(H_a: p_1 - p_2 < 0\)
02

Calculate the Test Statistic

First, we need to estimate \(p_1\) and \(p_2\) using the provided sample data. Proportion for the age group 33 to 49: \(n_1 = 600\) and \(\hat{p}_1 = 0.31\) Proportion for the age group 50 to 68: \(n_2 = 650\) and \(\hat{p}_2 = 0.48\) We calculate the pooled proportion: \(\hat{p} = \frac{n_1\hat{p}_1 + n_2\hat{p}_2}{n_1 + n_2} = \frac{600(0.31)+650(0.48)}{600+650}\) Now calculate the test statistic according to the following formula: \(z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}}\)
03

Calculate the p-value

We need to find the probability of getting the observed test statistic under the null hypothesis. The p-value is calculated using normal distribution, since we are using the z-test. Since our alternative hypothesis has a less-than sign, we'll need the lower tail probability. \(p\-value = P(Z < z)\)
04

Compare p-value with \(\alpha\) and Draw a Conclusion

We compare the p-value obtained in Step 3 with the given significance level \(\alpha=0.05\). 1. If the p-value \(\geq \alpha\), we fail to reject the null hypothesis, and there is not enough evidence to support that the proportion of adult Americans age 33 to 49 who rate a landline phone as one of the top three is lower than this proportion for adult Americans age 50 to 68. 2. If the p-value \(< \alpha\), we reject the null hypothesis and accept the alternative hypothesis, suggesting that the proportion of adult Americans age 33 to 49 who rate a landline phone as one of the top three is indeed lower than this proportion for adult Americans age 50 to 68. After calculating the test statistic and p-value, compare it with the given significance level, and draw a conclusion accordingly.

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Key Concepts

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

Understanding the Z-Test
A z-test is a statistical method used to determine if there is a significant difference between two population proportions. In this exercise, we want to see if a smaller percentage of adults aged 33 to 49 rate landline phones highly compared to those aged 50 to 68. By collecting sample data and calculating the difference between two proportions, we use the z-test to check for significance.
In a z-test, the difference between sample means or proportions is measured against what we would expect due to random chance alone. For this, we calculate a z-score through the formula:
  • \[z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}}\]
Here, \(\hat{p}_1\) and \(\hat{p}_2\) are the sample proportions, and \(\hat{p}\) is the pooled proportion. The resulting z-score tells us how many standard deviations the observed difference is from the expected difference of zero. If the z-score is large enough, it indicates a significant effect, guiding us to potential conclusions.
Understanding Significance Level
The significance level, often denoted by \(\alpha\), is a benchmark in hypothesis testing that helps decide if a result is statistically significant. In this problem, the chosen significance level is 0.05, or 5%. This means there is a 5% risk of concluding that there is a difference when there isn't one.
Setting this level helps control for Type I errors – false positives.
  • If the p-value from the test is less than \(\alpha\), we reject the null hypothesis, suggesting the observed effect is real.
  • If the p-value is greater than \(\alpha\), we fail to reject the null, meaning we don’t have enough evidence to support an effect or difference.
Choosing an appropriate significance level is vital as it reflects the risk you are willing to take in making a wrong inference. In scientific research, a 5% level is standard, balancing risk and sensitivity.
Understanding the P-Value
The p-value is a crucial part of hypothesis testing, representing the probability of obtaining the test results under the null hypothesis. This allows for understanding if the evidence is strong enough to reject the null hypothesis in favor of the alternative.
In the context of a z-test:
  • We compute \(p\text{-value} = P(Z < z)\), especially useful since our alternative hypothesis involves checking if something is significantly smaller.
  • A small p-value (typically less than 0.05) indicates strong evidence against the null hypothesis, suggesting a true effect.
Consider the p-value as a measure of the strength of evidence. It's not about the magnitude but about where the sample statistic falls within the sampling distribution under the null hypothesis. By comparing this with the significance level, researchers make informed decisions.
Understanding Null and Alternative Hypotheses
Formulating null \((H_0)\) and alternative \((H_a)\) hypotheses is the foundation of hypothesis testing. These hypotheses provide a framework to test claims using statistical evidence. In this exercise, let \(p_1\) be the proportion for ages 33 to 49, and \(p_2\) for ages 50 to 68.
  • **Null Hypothesis \((H_0)\):** \(p_1 - p_2 \geq 0\)
    This suggests that the proportion of people in the younger age group rating landlines as top is not less than that of the older group.
  • **Alternative Hypothesis \((H_a)\):** \(p_1 - p_2 < 0\)
    This predicts that the younger group is significantly less likely to rate landlines as top than the older group.
Testing hypotheses involves collecting data and using statistical techniques to see if the evidence supports the claim made by the alternative hypothesis over the null. It is the structure around which the entire testing process revolves.

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Most popular questions from this chapter

Some fundraisers believe that people are more likely to make a donation if there is a relatively quick deadline given for making the donation. The paper "Now or Never! The Effect of Deadlines on Charitable Giving: Evidence from Two Natural Field Experiments" (Journal of Behavioral and Experimental Economics [2016]: \(1-10\) ) describes an experiment to investigate the influence of deadlines. In this experiment, \(1.2 \%\) of those who received an e-mail request for a donation that had a three-day deadline to make a donation and \(0.8 \%\) of those who received the same e-mail request but without a deadline made a donation. The people who received the e-mail request were randomly assigned to one of the two groups (e-mail with deadline and e-mail without deadline). Suppose that the given percentages are based on sample sizes of 2000 (the actual sample sizes in the experiment were much larger). Use a \(90 \%\) confidence interval to estimate the difference in the proportion who donate for the two different treatments.

A report in USA TODAY described an experiment to explore the accuracy of wearable devices designed to measure heart rate ("Wearable health monitors not always reliable, study shows," USA TODAY, October 12,2016\()\). The researchers found that when 50 volunteers wore an Apple Watch to track heart rate as they walked, jogged, and ran quickly on a treadmill for three minutes, the results were accurate compared with an EKG 92\% of the time. When 50 volunteers wore a Fitbit Charge, the heart rate results were accurate \(84 \%\) of the time. a. Explain why the data from this study should not be analyzed using a large- sample hypothesis test for a difference in two population proportions. b. Carry out a hypothesis test to determine if there is convincing evidence that the proportion of accurate results for people wearing an Apple Watch is greater than this proportion for those wearing a Fitbit Charge. Use the Shiny app "Randomization Test for Two Proportions" to report an approximate \(P\) -value and use it to reach a decision in the hypothesis test. Remember to interpret the results of the test in context. c. Use the Shiny app "Bootstrap Confidence Interval for Difference in Two Proportions" to obtain a \(95 \%\) bootstrap confidence interval for the difference in the population proportions of accurate results for people wearing an Apple Watch and those wearing a Fitbit Charge. Interpret the interval in the context of the research.

The article referenced in the previous exercise also reported that \(53 \%\) of the Republicans surveyed indicated that they were opposed to making women register for the draft. Would you use the large-sample test for a difference in population proportions to test the hypothesis that a majority of Republicans are opposed to making women register for the draft? Explain why or why not.

A headline that appeared in Woman's World stated "Black Currant Oil Curbs Hair Loss!" (Woman's World, April 4, 2016). This claim was based on an experiment described in the paper "Effect of a Nutritional Supplement on Hair Loss in Women" (Journal of Cosmetic Dermatology [2015]: 76-82). In this experiment, women with stage 1 hair loss were assigned at random to one of two groups. One group was a control group who did not receive a nutritional supplement. Of the 39 women in this group, 20 showed increased hair density at the end of the study period. Those in the second group received a nutritional supplement that included fish oil, black currant oil, vitamin E, vitamin C, and lycopene. Of the 80 women in the supplement group, 70 showed increased hair density at the end of the study period. a. Is there convincing evidence that the proportion with increased hair density is greater for the supplement treatment than for the control treatment? Test the appropriate hypotheses using a 0.01 significance level. b. Write a few sentences commenting on the headline that appeared in Woman's World. c. Based on the description of the actual experiment and the result from your hypothesis test in Part (a), suggest a more appropriate headline.

The article "Footwear, Traction, and the Risk of Athletic Injury" (January \(2016,\) www.lermagazine.com/article/footwear -traction-and-the-risk-of-athletic-injury, retrieved December \(15,\) 2016) describes a study in which high school football players were given either a conventional football cleat or a swivel disc shoe. Of 2373 players who wore the conventional cleat, 372 experienced an injury during the study period. Of the 466 players who wore the swivel disc shoe, 24 experienced an injury. The question of interest is whether there is evidence that the injury proportion is smaller for the swivel disc shoe than it is for conventional cleats. a. What are the two treatments in this experiment? b. The article didn't state how the players in the study were assigned to the two groups. Explain why it is important to know if they were assigned to the groups at random. c. For purposes of this example, assume that the players were randomly assigned to the two treatment groups. Carry out a hypothesis test to determine if there is evidence that the injury proportion is smaller for the swivel disc shoe than it is for conventional cleats. Use a significance level of 0.05 .

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