Question

Who Is More Trusting: Internet Users or Non-users? In a randomly selected sample of 2237 US adults, 1754 identified themselves as people who use the Internet regularly while the other 483 indicated that they do not use the Internet regularly. In addition to Internet use, participants were asked if they agree with the statement "most people can be trusted." The results show that 807 of the Internet users agree with this statement, while 130 of the non-users agree. \({ }^{33}\) Find and clearly interpret a \(90 \%\) confidence interval for the difference in the two proportions.

Short answer

The 90% confidence interval for the difference in the proportion of 'trusting' individuals between Internet Users and Non-Internet Users is between 0.158 and 0.224.

Step-by-step solution

- Calculate the proportions

Firstly, the proportions of 'trusting' people in each group must be calculated. For Internet Users, this is \( \frac{807}{1754} = 0.460\), and for Non-Internet Users this is \( \frac{130}{483} = 0.269\). These represent the sample proportions, usually denoted \( \hat{p_1} \) and \( \hat{p_2} \).

- Calculate the standard error

The next step is to calculate the standard error of the difference in proportions. This is given by the formula: \( \sqrt{ \frac{\hat{p_1}(1-\hat{p_1})}{n_1} + \frac{\hat{p_2}(1-\hat{p_2})}{n_2} } \), where \( n_1 \) and \( n_2 \) are the sizes of the two groups. Substituting the numbers: \( SE = \sqrt{ \frac{0.460(1-0.460)}{1754} + \frac{0.269(1-0.269)}{483} } = 0.0198 \).

- Determine the z-score for 90% confidence

For a \(90\%\) confidence interval, the z-score is approximately \(1.645\) (this value provides the level of confidence that the population parameter lies within the interval calculated).

- Calculate the error margin

The margin of error can now be calculated by multiplying the z-score by the standard error. From this, \( E = 1.645 \times 0.0198 = 0.0326 \).

- Calculate the confidence interval

Lastly, the confidence interval is calculated by subtracting and adding the margin of error from the difference in sample proportions. With the calculated error margin and given sample proportions, \( CI = (0.460 - 0.269) - 0.0326 \quad to \quad (0.460 - 0.269) + 0.0326 = 0.158 to 0.224 \). This is the 90% confidence interval for the difference in the population proportions.

- Interpret the interval

This confidence interval can be interpreted as follows: There is a 90% level of confidence that the true difference in the proportion of 'trusting' people between internet users and non-internet users is between 0.158 and 0.224.