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Chapter 10: Problem 20

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
According to the Chi-square test, we do not have enough evidence to suggest that the actual proportion of Americans not using the internet differs from the expected 11%

Step by step solution

01

State the Hypotheses

The Null Hypothesis, denoted by \(H_0\), is that 11% of Americans do not use the internet, meaning the proportion is identical to the reported one. The Alternative Hypothesis, denoted \(H_a\), is that the proportion is different from 11%.
02

Calculate the Expected Frequencies

Given that the sample size is 200 and assuming the Null Hypothesis is true, we'd expect 11% of these people not using the internet. Therefore, the expected frequency is \(0.11 * 200 = 22\).
03

Calculate the Observed Frequencies

The sample information mentions that 26 out of 200 people are not using the internet. This is our observed frequency.
04

Apply the Chi-square Test Formula

Using the formula \(\chi^2 = \sum \frac{(O-E)^2}{E}\), where \(O\) is the observed frequency and \(E\) is the expected frequency. Thus, we have \(\chi^2 = \frac{(26 - 22)^2}{22} = 0.7272\)
05

Compare Chi-square Value with Critical Value

For a Chi-square distribution, the critical value for a significance level of 0.05 (also known as the p-value) with 1 degree of freedom is 3.841. Our calculated Chi-square value (0.7272) is less than this, so we do not have sufficient evidence to reject the Null Hypothesis. Therefore, there is no evidence suggesting that the actual proportion of Americans not using the internet differs from 11%

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