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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

The article "Public Acceptability in the UK and the USA of Nudging to Reduce Obesity: The Example of Reducing Sugar-Sweetened Beverages" (PLOS One, June 8,2016 ) describes a survey in which each person in a representative sample of 1082 adult Americans was asked about whether they would find different types of interventions acceptable in an effort to reduce consumption of sugary beverages. When asked about a tax on sugary beverages, 459 of the people in the sample said they thought that this would be an acceptable intervention. These data were used to test \(H_{0}: p=0.5\) versus \(H_{a^{*}}: p<0.5\) and the null hypothesis was rejected. a. Based on the hypothesis test, what can you conclude about the proportion of adult Americans who think that taxing sugary beverages is an acceptable intervention in an effort to reduce consumption of sugary beverages? b. Is it reasonable to say that the data provide strong support for the alternative hypothesis? c. Is it reasonable to say that the data provide strong evidence against the null hypothesis?

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?

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. Suppose that you plan to select a random sample of 400 students at your college. You will ask each student in the sample if they are a Facebook user and if they update their status at least two times a day. You plan to use the resulting data to decide if there is evidence that the proportion for your college is different from the proportion reported in the article for the college in California. What hypotheses should you test?

In a survey of 1000 women age 22 to 35 who work full-time, 540 indicated that they would be willing to give up some personal time in order to make more money (USA TODAY, March 4,2010 ). The sample was selected to be representative of women in the targeted age group. a. Do the sample data provide convincing evidence that a majority of women age 22 to 35 who work fulltime would be willing to give up some personal time for more money? Test the relevant hypotheses using \(\alpha=0.01\) b. Would it be reasonable to generalize the conclusion from Part (a) to all working women? Explain why or why not.

For which of the following \(P\) -values will the null hypothesis be rejected when performing a test with a significance level of \(0.05 ?\) a. 0.001 d. 0.047 b. 0.021 e. 0.148 c. 0.078

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