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Common Sense Media surveyed 1,000 teens and 1,000 parents of teens to learn about how teens are using social networking sites such as Facebook and MySpace ("Teens Show, Tell Too Much Online," San Francisco Chronicle, August 10,2009 ). The two samples were independently selected and were chosen to be representative of American teens and parents of American teens. When asked if they check online social networking sites more than 10 times a day, 220 of the teens surveyed said yes. When parents of teens were asked if their teen checked networking sites more than 10 times a day, 40 said yes. Use a significance level of 0.01 to determine if there is convincing evidence that the proportion of all parents who think their teen checks social networking sites more than 10 times a day is less than the proportion of all teens who check more than 10 times a day.

Short Answer

Expert verified
Yes, there is convincing evidence at the 0.01 level of significance to suggest that the proportion of all parents who think their teen checks social networking sites more than 10 times a day is less than the proportion of all teens who check these sites more than 10 times a day.

Step by step solution

01

State the Hypotheses

The null hypothesis is that the proportions are equal, and the alternative hypothesis is that the proportion of parents who think their teen checks social networking sites more than 10 times a day is less than the proportion of teens who check more than 10 times a day. So, \n\nNull Hypothesis \(H_0: p1 = p2\), \nAlternative Hypothesis \(H_a: p1 < p2\)
02

Calculate the Proportions and Test Statistic

First, calculate the proportions. For the teens, the proportion \(p1\) is 220 out of 1000, or 0.22. For the parents, the proportion \(p2\) is 40 out of 1000, or 0.04. Now, calculate the pooled proportion \(p\) which is equal to \((x1 + x2) / (n1 + n2)\) = (220 + 40) / (1000 + 1000) = 0.13. Then, calculate the test statistic using the formula for the z score in testing of two proportions: \(z = (p1 - p2) / sqrt [p(1-p)(1/n1 + 1/n2)]\). Substituting the given numbers, our test statistic becomes \(z = (0.22 - 0.04) / sqrt [0.13(1 - 0.13)(1/1000 + 1/1000)] = 15.94.
03

Determine the P-Value and Decision

Use a standard normal distribution (z-distribution) to find the p-value associated with your calculated z-test statistic. In this case, our z-value is so large that the p-value will be effectively 0. Comparing the p-value with our significance level of 0.01, since p-value (0) < \(\alpha\) (0.01), we reject the null hypothesis.

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Key Concepts

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

Proportions
In statistics, **proportions** help us understand the part of a whole that meets a particular criterion. For example, in surveys or polls, proportions allow us to express results as percentages or fractions.
In the exercise, proportions were calculated to compare behaviors between two groups: teens and parents. To find the proportion, we simply take the number of individuals meeting the criterion and divide it by the total number in the sample:
  • For teens: 220 out of 1000 said they check social sites frequently, giving a proportion of \( p_1 = \frac{220}{1000} = 0.22 \).
  • For parents: 40 out of 1000 thought their teens checked frequently, giving a proportion of \( p_2 = \frac{40}{1000} = 0.04 \).
Understanding these proportions is vital, as it sets the baseline for comparison and further statistical testing. Proportions such as these highlight the extent to which different groups engage in or perceive a specific behavior.
Significance Level
When conducting research, the **significance level** is crucial as it signifies the risk of rejecting a true null hypothesis. For this exercise, the significance level is set at 0.01. This means we are willing to accept a 1% chance of incorrectly rejecting the null hypothesis, implying that we could falsely find a difference that doesn't actually exist.
The implication here is caution; the smaller the significance level, the stronger the evidence must be to reject the null hypothesis. Choosing a significance level is an important step because it guides the interpretation of results:
  • If the calculated p-value is less than the significance level (0.01 in this situation), we reject the null hypothesis.
  • It's common to use levels like 0.01, 0.05, or 0.10, but the choice depends on the context and consequences of potential errors.
A significance level of 0.01 indicates the need for strong evidence to assert that parents perceive their teens check social networking less frequently than the teens actually do.
Z-Test
A **Z-Test** is a statistical method used to determine if there is a significant difference between sample means or proportions. It's applicable when dealing with large sample sizes and known population standard deviations.In the scenario provided, a Z-Test for two proportions is appropriate since we want to compare two independent sample proportions:
  • Null Hypothesis \( H_0: p1 = p2 \)
  • Alternative Hypothesis \( H_a: p1 < p2 \)
Using the formula:\[z = \frac{p1 - p2}{\sqrt{p(1-p)\left(\frac{1}{n1} + \frac{1}{n2}\right)}}\]We calculate the test statistic, where \( p \) represents the pooled proportion. The pooled proportion accounts for combined data from both groups, leading to more reliable statistical outcomes. In this exercise, the Z-Test helped demonstrate whether the proportion of perceived behaviors differs significantly between the groups.
Given the calculated z-value of 15.94, it's clear that there's a substantial difference, leading to the rejection of the null hypothesis, as the p-value is effectively 0, less than our significance level of 0.01.

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Most popular questions from this chapter

The article "More Teen Drivers See Marijuana as OK; It's a Dangerous Trend (USA Today, February 23,2012 ) describes two surveys of U.S. high school students. One survey was conducted in 2009 and the other was conducted in 2011 . In \(2009,78 \%\) of the people in a representative sample of 2,300 students said marijuana use is very distracting or extremely distracting to their driving. In \(2011,70 \%\) of the people in a representative sample of 2,294 students answered this way. Use the five-step process for estimation problems \(\left(\mathrm{EMC}^{3}\right)\) to construct and interpret a \(99 \%\) large- sample confidence interval for the difference in the proportion of high school students who believed marijuana was very distracting or extremely distracting in 2009 and this proportion in 2011 .

The article "College Graduates Break Even by Age 33" (USA Today, September 21,2010 ) reported that \(2.6 \%\) of college graduates were unemployed in 2008 and \(4.6 \%\) of college graduates were unemployed in \(2009 .\) Suppose that the reported percentages were based on independently selected representative samples of 500 college graduates in each of these two years. Construct and interpret a \(95 \%\) confidence interval for the difference in the proportions of college graduates who were unemployed in these 2 years.

"Smartest People Often Dumbest About Sunburns" is the headline of an article that appeared in the San Luis Obispo Tribune (July 19,2006\()\). The article states that "those with a college degree reported a higher incidence of sunburn than those without a high school degree-43\% versus \(25 \% . "\) Suppose that these percentages were based on independent random samples of size 200 from each of the two groups of interest (college graduates and those without a high school degree). a. Are the sample sizes large enough to use the largesample confidence interval for a difference in population proportions? b. Estimate the difference in the proportion of people with a college degree who reported sunburn and the corresponding proportion for those without a high school degree using a \(90 \%\) confidence interval. c. Is zero included in the confidence interval? What does this suggest about the difference in the two population proportions? d. Interpret the confidence interval in the context of this problem.

The report "Young People Living on the Edge" (Greenberg Quinlan Rosner Research, 2008 ) summarizes a survey of people in two independent random samples. One sample consisted of 600 young adults (ages 19 to 35 ), and the other sample consisted of 300 parents of young adults ages 19 to \(35 .\) The young adults were presented with a variety of situations (such as getting married or buying a house) and were asked if they thought that their parents were likely to provide financial support in that situation. The parents of young adults were presented with the same situations and asked if they would be likely to provide financial support in that situation. When asked about getting married, \(41 \%\) of the young adults said they thought parents would provide financial support and \(43 \%\) of the parents said they would provide support. Carry out a hypothesis test to determine if there is convincing evidence that the proportion of young adults who think their parents would provide financial support and the proportion of parents who say they would provide support are different.

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?

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