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The report referenced in the previous exercise also stated that the proportion who thought their parents would help with buying a house or renting an apartment for the sample of young adults was 0.37 . For the sample of parents, the proportion who said they would help with buying a house or renting an apartment was 0.27 . Based on these data, can you conclude that the proportion of parents who say they would help with buying a house or renting an apartment is significantly less than the proportion of young adults who think that their parents would help?

Short Answer

Expert verified
Based on the given data, it is impossible to conclude whether the proportion of parents who say they would help with buying a house or renting an apartment is significantly less than the proportion of young adults who think that their parents would help. The missing sample sizes (n鈧 and n鈧) prevent us from performing a hypothesis test and determining the p-value to make a conclusion.

Step by step solution

01

Set up the hypotheses

We set the null hypothesis (H鈧) and the alternative hypothesis (H鈧) as follows: Null Hypothesis (H鈧): The proportion of parents who say they would help with buying a house or renting an apartment is equal to the proportion of young adults who think that their parents would help. In other words, p鈧 = p鈧. Alternative Hypothesis (H鈧): The proportion of parents who say they would help with buying a house or renting an apartment is significantly less than the proportion of young adults who think that their parents would help. In other words, p鈧 < p鈧.
02

Calculate the sample proportions and their difference

Given: p鈧 (sample proportion of parents)= 0.27, p鈧 (sample proportion of young adults)= 0.37 Compute the difference: d = p鈧 - p鈧 = 0.27 - 0.37 = -0.1
03

Calculate the pooled sample proportion and standard error

We first calculate the pooled sample proportion (p虃) as follows: \[ \hat{p} = \frac{n_1p_1 + n_2p_2}{n_1 + n_2} \] Since we are not given the sample sizes (n鈧 and n鈧), we cannot compute the exact value of p虃. Given the pooled sample proportion, we can calculate the standard error (SE) of the difference between the two proportions using the following formula: \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n_1} + \frac{\hat{p}(1-\hat{p})}{n_2}} \] Again, we cannot compute the exact value of SE since we do not have the sample sizes.
04

Calculate the test statistic (z)

We calculate the z-score for the difference in proportions using the following formula: \[ z = \frac{d - 0}{SE} \] Unfortunately, since we do not have the sample sizes (n鈧 and n鈧), it is impossible to calculate the standard error (SE) and the test statistic (z).
05

Determine the p-value and make a conclusion

Likewise, without the test statistic (z), it is not possible to determine the p-value and make a conclusion about the null hypothesis and alternative hypothesis. Conclusion: In order to perform a hypothesis test to compare the two given proportions, we require the sample sizes (n鈧 and n鈧). Without this crucial information, we cannot conclude whether the proportion of parents who say they would help with buying a house or renting an apartment is significantly less than the proportion of young adults who think that their parents would help.

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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, denoted as H鈧, is a statement for hypothesis testing that there is no effect or no difference, often representing a default or initial belief. It is the hypothesis that one seeks evidence against in order to support an alternative hypothesis.

For our example concerning young adults and their parents' willingness to help with buying a house or renting an apartment, the null hypothesis asserts that there is no difference in proportion between parents and young adults who are willing to provide help, mathematically expressed as H鈧: p鈧 = p鈧.

Understanding the null hypothesis provides us a baseline expectation to compare with actual observations. If data significantly deviates from what is expected under the null hypothesis, we may have reason to reject it in favor of an alternative explanation.
Alternative Hypothesis
Conversely, the alternative hypothesis, H鈧 or Ha, is the statement you want to be able to conclude is true. It is directly opposed to the null hypothesis and expresses that there is a statistically significant effect or difference.

In the context of our example, the alternative hypothesis claims that the proportion of parents willing to help is less than that of young adults, expressed as H鈧: p鈧 < p鈧.

The alternative hypothesis is what you are testing for, and if the evidence is strong enough against the null hypothesis, you can support the alternative hypothesis. However, failing to reject the null hypothesis does not necessarily prove it; it simply does not provide enough evidence to support the alternative.
Pooled Sample Proportion
The pooled sample proportion is used in hypothesis testing when comparing two proportions to estimate a common proportion that combines information from both samples involved. It assumes that the true proportion is the same for both groups under the null hypothesis.

To calculate it, use the formula: \[ \hat{p} = \frac{n_1p_1 + n_2p_2}{n_1 + n_2} \]
where:\
  • \( n_1 \) and \( n_2 \) are the sample sizes,
  • \( p_1 \) and \( p_2 \) are the sample proportions.

In cases where we do not have the sample sizes, like in our textbook example, we cannot compute the pooled sample proportion, which hinders further calculations in the hypothesis testing process.
Standard Error
The standard error (SE) measures the variability of a sample statistic over many samples drawn from a single population. In other words, it's an estimate of how much the sample proportion is likely to fluctuate due to sampling variance.

The formula for the standard error of the difference between two sample proportions is: \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n_1} + \frac{\hat{p}(1-\hat{p})}{n_2}} \]
However, just like with the pooled sample proportion, without knowing the sample sizes \( n_1 \) and \( n_2 \), as in our scenario, we are unable to calculate SE. This calculation is crucial because it affects the accuracy of our test statistic and our ability to draw meaningful conclusions from our hypothesis test.
Test Statistic
The test statistic, in the context of hypothesis testing for proportions, is typically a z-score that measures how many standard errors a sample statistic lies from the null hypothesis value. It is a crucial piece in deciding whether to reject or fail to reject the null hypothesis.

The test statistic for comparing two proportions is determined using the formula: \[ z = \frac{d - 0}{SE} \]
where \( d \) represents the observed difference between the sample proportions. Yet, without the ability to calculate the standard error due to missing sample sizes, as highlighted in the exercise, we cannot compute the test statistic. Generally, the test statistic allows us to quantify the strength of evidence against the null hypothesis and is used to determine the p-value, a key factor in hypothesis testing.

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Most popular questions from this chapter

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 .

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.

The article "Spray Flu Vaccine May Work Better than Injections for Tots" (San Luis Obispo Tribune, May 2, 2006) described a study that compared flu vaccine administered by injection and flu vaccine administered as a nasal spray. Each of the 8000 children under the age of 5 who participated in the study received both a nasal spray and an injection, but only one was the real vaccine and the other was salt water. At the end of the flu season, it was determined that of the 4000 children receiving the real vaccine by nasal spray, \(3.9 \%\) got the flu. Of the 4000 children receiving the real vaccine by injection, \(8.6 \%\) got the flu. a. Why would the researchers give every child both a nasal spray and an injection? b. Use a \(99 \%\) confidence interval to estimate the difference in the proportion of children who get the flu after being vaccinated with an injection and the proportion of children who get the flu after being vaccinated with the nasal spray. Based on the confidence interval, would you conclude that the proportion of children who get the flu is different for the two vaccination methods? (Hint: See Example \(11.7 .\) )

In a survey of mobile phone owners, \(53 \%\) of iPhone users and \(42 \%\) of Android phone users indicated that they upgraded their phones at least every two years ("Americans Split on How Often They Upgrade Their Smartphones," July 8, \(2015,\) www.gallup.com, retrieved December 15,2016 ). The reported percentages were based on large samples that were thought to be representative of the population of iPhone users and the population of Android phone users. The sample sizes were 8234 for the iPhone sample and 6072 for the Android phone sample. Suppose you want to decide if there is evidence that the proportion of iPhone owners who upgrade their phones at least every two years is greater than this proportion for Android phone users. (Hint: See Example 11.5.) a. What hypotheses should be tested to answer the question of interest? b. Are the two samples large enough for the large-sample test for a difference in population proportions to be appropriate? Explain. c. Based on the following Minitab output, what is the value of the test statistic and what is the value of the associated \(P\) -value? If a significance level of 0.01 is selected for the test, will you reject or fail to reject the null hypothesis? d. Interpret the result of the hypothesis test in the context of this problem.

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