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The article "Facebook Use and Academic Performance Among College Students" (Computers in Human Behavior [2015]: \(265-272\) ) estimated that \(87 \%\) percent of students at a large public university in California who are Facebook users update their status at least two times a day. This estimate was based on a random sample of 261 students at this university. a. Does this sample provide convincing evidence that more than \(80 \%\) of the students at this college who are Facebook users update their status at least two times a day? Test the relevant hypotheses using \(\alpha=0.05\). b. Would it be reasonable to generalize the conclusion from the test in Part (a) to all college students in the United States? Explain why or why not.

Short Answer

Expert verified
The sample provides convincing evidence that more than 80% of the students at this college who are Facebook users update their status at least two times a day (p-value = 0.0001 < α = 0.05). However, it is not reasonable to generalize this conclusion to all college students in the United States, as the sample is from a specific university in California and may not represent the entire population of college students.

Step by step solution

01

Define the null and alternative hypotheses

First, we need to define the null hypothesis (H0) and the alternative hypothesis (Ha). In this case: - Null hypothesis (H0): The proportion of students who update their status at least two times a day is equal to 80% (p = 0.80). - Alternative hypothesis (Ha): The proportion of students who update their status at least two times a day is more than 80% (p > 0.80).
02

Set up the test statistic

We will use the test statistic Z to perform the hypothesis test, which can be calculated using the formula: \[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\] where: - \(\hat{p}\) is the sample proportion (0.87). - \(p_0\) is the proportion in the null hypothesis (0.80). - n is the number of samples (261).
03

Calculate the test statistic and find the p-value

Plugging in the values into the formula, we get: \[Z = \frac{0.87 - 0.80}{\sqrt{\frac{0.80(1 - 0.80)}{261}}} = 3.728\] Now, we need to find the p-value corresponding to this Z value. Since Ha states p>0.80, this is a right-tailed test. We can find the p-value using a Z-distribution table or a statistical software: p-value = P(Z > 3.728) ≈ 0.0001
04

Compare the p-value with the significance level and make a decision

Now we need to compare the p-value (0.0001) with the given significance level (α = 0.05). Since the p-value is less than α, we reject the null hypothesis.
05

Interpret the result

We are now able to answer part (a) of this exercise. The sample provides convincing evidence that more than 80% of the students at this college who are Facebook users update their status at least two times a day. For part (b), the question asks whether it would be reasonable to generalize this conclusion to all college students in the United States. While our test provided evidence supporting the claim for students at this particular university in California, we would need a more representative sample (including students from various regions, colleges, and demographic backgrounds) to confidently generalize the conclusion to all college students in the United States. Therefore, in this case, it is not reasonable to generalize the conclusion.

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Key Concepts

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

Z-test: Understanding the Basics
In statistics, a Z-test is a hypothesis test that is used to determine if there is a significant difference from a stated population proportion. It is particularly useful when dealing with large sample sizes. In our exercise, the sample involves 261 students, which provides a robust basis for applying the Z-test. The Z-test compares the sample proportion (in our case, 87%) with the hypothesized proportion from the null hypothesis (80%). The core idea is to see if the observed sample proportion is sufficiently far from the null hypothesis proportion due to reasons other than random chance. If our Z value is significantly high or low, it indicates that the sample proportion is not just a statistical fluke.
Decoding the p-value
The p-value is pivotal in hypothesis testing as it tells us the probability of observing our sample results, or something more extreme, if the null hypothesis is true. In simpler terms, it quantifies the evidence against the null hypothesis. For the exercise provided, a p-value of 0.0001 indicates that there's a very small probability of observing such a high sample proportion (like 87%), assuming the true proportion of students updating their status is just 80%. The smaller the p-value, the stronger the evidence against the null hypothesis. Here, our p-value is much less than our chosen significance level (0.05), leading us to reject the null hypothesis.
Significance Level and Decision Making
The significance level, often denoted as \( \alpha \), is the threshold we set to determine when to reject the null hypothesis. In most research, a 5% level (\( \alpha = 0.05 \)) is used, reflecting a willingness to accept a 5% chance of making a Type I error—incorrectly rejecting a true null hypothesis. In our hypothesis test, since the p-value (0.0001) is less than \( \alpha = 0.05 \), we reject the null hypothesis. This means there is considerable evidence that more than 80% of these college students update their status at least twice a day, based on our sample.
Generalization of Results: Caution Required
While our test results are convincing for students at this particular university, they cannot be safely generalized to all college students in the United States. Each university or college may have different student behaviors based on various factors like demographics, location, and culture. For broader generalization, a more diverse and representative sample across multiple institutions should be gathered. This includes varying geographical locations and demographics to ensure the findings are not biased by regional characteristics. Failing to consider this can lead to inaccurate generalizations that may misrepresent actual trends. Hence, in this exercise, a cautious approach is necessary when considering generalizations beyond the sample itself.

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Most popular questions from this chapter

One type of error in a hypothesis test is failing to reject a false null hypothesis. What is the other type of error that might occur when a hypothesis test is carried out?

One type of error in a hypothesis test is rejecting the null hypothesis when it is true. What is the other type of error that might occur when a hypothesis test is carried out?

Researchers at the University of Washington and Harvard University analyzed records of breast cancer screening and diagnostic evaluations ("Mammogram Cancer Scares More Frequent Than Thought," USA TODAY, April 16,1998 ). Discussing the benefits and downsides of the screening process, the article states that although the rate of falsepositives is higher than previously thought, if radiologists were less aggressive in following up on suspicious tests, the rate of false-positives would fall, but the rate of missed cancers would rise. Suppose that such a screening test is used to decide between a null hypothesis of \(H_{0}:\) no cancer is present and an alternative hypothesis of \(H_{a}:\) cancer is present. (Although these are not hypotheses about a population characteristic, this exercise illustrates the definitions of Type I and Type II errors.) a. Would a false-positive (thinking that cancer is present when in fact it is not) be a Type I error or a Type II error? b. Describe a Type I error in the context of this problem, and discuss the consequences of making a Type I error. c. Describe a Type II error in the context of this problem, and discuss the consequences of making a Type II error. d. Recall the statement in the article that if radiologists were less aggressive in following up on suspicious tests, the rate of false-positives would fall but the rate of missed cancers would rise. What aspect of the relationship between the probability of a Type I error and the probability of a Type II error is being described here?

The paper "Teens and Distracted Driving"" (Pew Internet \& American Life Project, 2009 ) reported that in a representative sample of 283 American teens age 16 to \(17,\) there were 74 who indicated that they had sent a text message while driving. For purposes of this exercise, assume that this sample is a random sample of 16- to 17 -year-old Americans. Do these data provide convincing evidence that more than a quarter of Americans age 16 to 17 have sent a text message while driving? Test the appropriate hypotheses using a significance level of 0.01 . (Hint: See Example 10.11 .)

Suppose that for a particular hypothesis test, the consequences of a Type I error are very serious. Would you want to carry out the test using a small significance level \(\alpha\) (such as 0.01 ) or a larger significance level (such as 0.10 )? Explain the reason for your choice.

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