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Chapter 9: Problem 108

According to an article on PCMag.com, Facebook users spend an average of 190 minutes per month checking and updating their Facebook page (Source: http://www.pcmag.com/article2/ \(0,2817,2342757,00\) asp). A random sample of 55 college students aged 18 to 22 years with Facebook accounts resulted in a sample mean time and sample standard deviation of \(219.50\) and \(69.30\) minutes per month, respectively. a. Using \(\alpha=.025\), can you conclude that the average time spent per month checking and updating their Facebook pages by all college students aged 18 to 22 years who have Facebook accounts is higher than 190 minutes? Use the critical value approach. b. Find the range of the \(p\) -value for the test of part a. What is your conclusion with \(\alpha=.025\) ?

Short Answer

Expert verified
To come up with a short answer, follow the steps and perform the calculations to test the hypothesis. If the test statistic is larger than the critical value and the p-value is less than or equal to the \(\alpha\) level, conclude that the mean time spent per month by college students aged 18 to 22 years on Facebook is over 190 minutes. Otherwise, the conclusion will be the reverse.

Step by step solution

01

Set Up the Hypothesis

The null hypothesis (\(H_0\)): The average time college students aged 18 to 22 spend on Facebook is equal to 190 minutes: \(\mu = 190\). The alternative hypothesis (\(H_a\)): the average time college students aged 18 to 22 spend on Facebook is greater than 190 minutes: \(\mu > 190\).
02

Calculate the Test Statistic

Standardize the sample mean by subtracting the population mean and dividing by the standard error (standard deviation divided by the square root of the sample size). This can be represented as \(t = (219.50 - 190) / (69.30 / \sqrt{55})\).
03

Find the Critical Value

The critical value is found by checking a t-table for a one-tailed test at a degree of freedom of 54 (\(n-1\)) and \(\alpha = 0.025\). The critical value will be compared to the test statistic.
04

Compare and Make a Decision

If the test statistic is greater than the critical value, reject the null hypothesis in favour of the alternative hypothesis. If not, fail to reject the null hypothesis.
05

Find the p-value range

The p-value for the test is found by determining the probability that a t statistic is more extreme than the calculated test statistic, given the degree of freedom (54). The p-value will be compared to the \(\alpha\) level.
06

Compare p-value and \(\alpha\)

If the p-value is less than or equal to the \(\alpha\) level, the null hypothesis is rejected in favour of the alternative hypothesis. If not, the null hypothesis is not rejected.
07

Conclude

The final conclusion will be based on the results from the previous steps.

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Key Concepts

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

t-test
The t-test is a statistical method used to determine if there is a significant difference between the means of two groups. It's particularly useful when the sample size is small and the population standard deviation is unknown. In the context of the problem, a t-test examines whether the average monthly Facebook usage by college students significantly differs from the known average of 190 minutes. There are different types of t-tests, such as:
  • One-sample t-test: compares the mean of a single group to a known value or theoretical expectation.
  • Independent two-sample t-test: compares the means of two independent groups.
  • Paires t-test: compares means from the same group at different times.
For this problem, we use a one-sample t-test. This determines whether the sample mean of 219.50 minutes per month is statistically greater than the population mean of 190 minutes.
null hypothesis
The null hypothesis, denoted as \(H_0\), is the statement being tested in a hypothesis test. It typically represents a default position that there is no effect or no difference. In this problem, the null hypothesis is that the mean monthly time spent by college students on Facebook is equal to 190 minutes: \(\mu = 190\).The null hypothesis serves as a starting point for any test. It's often contrasted with the alternative hypothesis. We aim to test whether evidence from the data supports rejecting the null hypothesis. If our test shows significant evidence, we may reject \(H_0\). Otherwise, we fail to reject it, indicating insufficient evidence to support the alternative.
alternative hypothesis
The alternative hypothesis, denoted as \(H_a\), is the statement that indicates the presence of a meaningful effect or difference from the null hypothesis. In this scenario, the alternative hypothesis proposes that the average time college students spend on Facebook is greater than 190 minutes per month: \(\mu > 190\).Our goal in hypothesis testing is to examine whether our sample data provides sufficient evidence to support this alternative hypothesis over the null hypothesis. In cases where the test statistic and p-value show strong results against the null hypothesis, we may accept the alternative hypothesis as a more plausible explanation of what is observed.
p-value
The p-value in hypothesis testing is the probability of observing a test statistic as extreme as, or more extreme than, the statistic computed from the sample data, given that the null hypothesis is true. It helps assess the strength of the evidence against the null hypothesis.In this problem, the calculated p-value indicates how likely it is to find an average Facebook use greater than 190 minutes if, in fact, the population mean is 190. A low p-value (typically less than or equal to the given significance level, \(\alpha\)) suggests strong evidence against the null hypothesis. For example, if the p-value is less than \(\alpha = 0.025\), we reject the null hypothesis.
critical value
The critical value is a threshold applied in hypothesis testing. It is a cut-off point that defines the boundary for rejecting the null hypothesis. To find it, we look at a t-distribution table using the chosen significance level \(\alpha\) and the degrees of freedom (which is the sample size minus one).For this exercise, with \(\alpha = 0.025\) and 54 degrees of freedom, the critical value helps to determine the region where the null hypothesis should be rejected. If the calculated test statistic exceeds the critical value, it indicates that the result is statistically significant, leading us to reject the null hypothesis in favor of the alternative. Otherwise, we fail to reject it.

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Most popular questions from this chapter

A tool manufacturing company claims that its top-of-the-line machine that is used to manufacture bolts produces an average of 88 or more bolts per hour. A company that is interested in buying this machine wants to check this claim. Suppose you are asked to conduct this test. Briefly explain how you would do so when \(\sigma\) is not known.

Professor Hansen believes that some people have the ability to predict in advance the outcome of a spin of a roulette wheel. He takes 100 student volunteers to a casino. The roulette wheel has 38 numbers, each of which is equally likely to occur. Of these 38 numbers, 18 are red, 18 are black, and 2 are green. Each student is to place a serics of five bets, choosing either a red or a black number before each spin of the wheel. Thus, a student who bets on red has an \(18 / 38\) chance of winning that bet. The same is true of betting on black. a. Assuming random guessing, what is the probability that a particular student will win all five of his or her bets? b. Suppose for each student we formulate the hypothesis test \(H_{0}:\) The student is guessing \(H_{1}:\) The student has some predictive ability Suppose we reject \(H_{0}\) only if the student wins all five bets. What is the significance level? c. Suppose that 2 of the 100 students win all five of their bets. Professor Hansen says, "For these two students we can reject \(H_{0}\) and conclude that we have found two students with some ability to predict." What do you make of Professor Hansen's conclusion?

In 2006, the average number of new single-family homes built per town in the state of Maine was \(14.325\) (www.mainehousing.org). Suppose that a random sample of 42 Maine towns taken in 2009 resulted in an average of \(13.833\) new single-family homes built per town, with a standard deviation of \(4.241\) new single-family homes. Using the \(5 \%\) significance level, can you conclude that the average number of new single-family homes per town built in 2009 in the state of Maine is significantly different from \(14.325\) ? Use both the \(p\) -value and critical-value approaches.

Consider \(H_{0}: \mu=40\) versus \(H_{1}: \mu>40\) a. A random sample of 64 observations taken from this population produced a sample mean of 43 and a standard deviation of \(5 .\) Using \(\alpha=.025\), would you reject the null hypothesis? b, Another random sample of 64 observations taken from the same population produced a sample mean of 41 and a standard deviation of 7 . Using \(\alpha=.025\), would you reject the null hypothesis?

According to the most recent Bureau of Labor Statistics survey on time use in the United States, the average U.S. man spends \(67.20\) minutes per day eating and drinking. Suppose that a survey of 43 Norwegian men resulted in an average of \(81.10\) minutes per day eating and drinking [Note: This value is consistent with the data in a report by the Organization for Economic Cooperation and Development (Source: http://economix.blogs.nytimes.com/2009/05/05/obesity-and- the-fastness-of-food/)]. Assume that the population standard deviation for all Norwegian men is \(18.30\) minutes. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the average daily time spent eating and drinking by all Norwegian men is higher than \(67.20\) minutes. What is your conclusion at \(\alpha=.05\) ? b. Test the hypothesis of part a using the critical-value approach. Use \(\alpha=.01\).
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