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In 2018 Pew Research reported that \(11 \%\) of Americans do not use the Internet. Suppose in a random sample of 200 Americans, 26 reported not using the Internet. Using a chi-square test for goodness-of-fit, test the hypothesis that the proportion of Americans who do not use the Internet is different from \(11 \%\). Use a significance level of \(0.05\).

Short Answer

Expert verified
Using a chi-square test for goodness-of-fit, we do not have evidence at the 0.05 significance level to conclude that the proportion of Americans who do not use the Internet is different from 11%.

Step by step solution

01

Set up the hypotheses

The null hypothesis \(H_0\) is that the proportion of Americans who do not use the Internet is 11%, and the alternative hypothesis \(H_1\) is that the proportion is different from 11%.
02

Calculate the expected frequencies

If the null hypothesis is true, we can expect 11% of the 200 randomly sampled Americans to not use the Internet, which translates to \(0.11 * 200 = 22\) individuals.
03

Calculate the test statistic

We use the chi-square formula: \(χ^2 = Σ [ (O_i - E_i)^2 / E_i ]\), where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency. Substituting the observed frequency (26 individuals) and the expected frequency (22 individuals), we find \(χ^2 = (26 - 22)^2 / 22 = 0.7273\).
04

Calculate the p-value and make a decision

The p-value associated with a chi-square statistic of 0.7273 and 1 degree of freedom (calculated as observed categories - 1) can be found using a chi-square distribution table or a statistical software. It's approximately 0.3939. Because this p-value is greater than the significance level of 0.05, we do not reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
When performing a chi-square test for goodness-of-fit, our starting point is the null hypothesis, which is essentially the claim we're aiming to test. In the context of our Internet usage problem, the null hypothesis (\( H_0 \) ) posits that the true proportion of Americans not using the Internet is consistent with the presumed rate of 11%.

While the null hypothesis maintains the status quo, it’s the job of the data (through the chi-square test) to challenge this position. If the collected data show a significant departure from what's expected under the null hypothesis, it can lead to the null being rejected. However, if the data do not show a significant difference, the null continues to stand. It's important to note that not rejecting the null hypothesis isn't the same as confirming it's true; it merely indicates that there is not enough evidence to support a change.
Expected Frequencies
Expected frequencies are cornerstone figures in the chi-square test. They represent the predicted counts of observations in each category if the null hypothesis were true. In the exercise involving whether the proportion of Americans who don't use the Internet strays from 11%, the expected frequency is calculated based on the assumption that this percentage holds up in the larger population.

To determine the expected frequency, we take the total number of observations (in this case, a sample of 200 Americans) and multiply it by the expected proportion (11%). As a result, we anticipate that if the null hypothesis is accurate, about 22 out of 200 individuals would not be using the Internet (\(0.11 \times 200 = 22\)). The expected frequency allows us to gauge how the observed data compares to what we would predict under the null hypothesis, providing the basis for the next steps in our chi-square analysis.
Statistical Significance
Statistical significance is the determination of whether the observed differences between the expected and actual results are due to random chance or are strong enough to be considered noteworthy. In chi-square tests, statistical significance is evaluated against a pre-defined threshold, known as the alpha level.

Common alpha levels include 0.05, 0.01, or 0.10, with 0.05 being the most widely used. This alpha level represents the probability of rejecting the null hypothesis when it is actually true (known as a Type I error). If the p-value that we calculate from our chi-square test is less than the alpha level, it suggests that the observed data are highly unlikely to have occurred if the null hypothesis was true, leading us to reject the null hypothesis in favor of the alternative. Conversely, if the p-value is greater than the alpha level, we do not have enough evidence to reject the null, which is the conclusion we reached in our Internet usage example.
P-Value
The p-value is a crucial measure in hypothesis testing that helps us decide whether to reject or support the null hypothesis. It quantifies the probability of observing a test statistic as extreme as, or more extreme than, the value calculated from the sample data, assuming the null hypothesis is true. In our example, the p-value is the probability of finding a chi-square value of 0.7273 or higher with our Internet usage data.

The p-value allows us to assess the strength of the evidence against the null hypothesis. A small p-value indicates strong evidence against the null hypothesis, suggesting that the observed result is unusual under the null hypothesis. In the case of the Internet usage study, a p-value of approximately 0.3939 is considerably larger than the alpha threshold of 0.05, leading us to maintain the null hypothesis that the proportion of Americans who do not use the Internet is indeed close to 11%.

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Most popular questions from this chapter

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