Question

Many people have misconceptions about how profitable small, consistent investments can be. In a survey of 1010 randomly selected U.S. adults (Associated Press, October 29,1999 ), only 374 responded that they thought that an investment of \(\$ 25\) per week over 40 years with a \(7 \%\) annual return would result in a sum of over \(\$ 100,000\) (the correct amount is \(\$ 286,640\) ). Is there sufficient evidence to conclude that less than \(40 \%\) of U.S. adults are aware that such an investment would result in a sum of over \(\$ 100,000\) ? Test the relevant hypotheses using \(\alpha=.05\).

Short answer

Yes, there is enough evidence to support the claim that less than 40% of U.S. adults know that investing $25 per week for 40 years at 7% annual return would yield more than $100,000.

Step-by-step solution

Formulate the null and alternative hypotheses

The null hypothesis \(H_0\) would be that 40 percent (.40) of people knows that the regular small investment over the 40 years would result in > $100,000. Formally, \(H_0: p = 0.40\). The alternative hypothesis \(H_A\) is that less than 40% of people know this, so formally, \(H_A: p < 0.40\).

Compute the sample proportion

To find the sample proportion, divide the number who responded affirmatively by the total number of respondents. It’s \(p' = 374/1010 = 0.37\). It shows that 37% (less than 40%) of surveyed population have this knowledge.

Check the sampling distribution

The sampling distribution should be approximately normally distributed. This can be tested using these inequalities: \(np_0 > 5\) and \(n(1 - p_0) > 5\). Both \(1010 * 0.40 = 404\) and \(1010 * 0.60 = 606\) are more than 5. So, the approximation of a normal distribution can be used.

Calculate Standard Error and Z-Score

We calculate the standard deviation (Standard Error - SE) of the sampling distribution by the formula: SE = \(\sqrt{[p_0 * (1 - p_0)]/n}\) where \(p_0\) is the population proportion under \(H_0\) and n is the total survey count. So, SE = \(\sqrt{ [(0.40 * 0.60) / 1010] } = 0.0156\). Then, we find the Z-score which is the number of standard deviations the observed sample proportion \(p'\) is from the expected population proportion \(p_0\). So, Z = \((p' - p_0) / SE = (0.37 - 0.40) / 0.0156 = -1.923\).

Making a decision

Then, check Z-table for the probability associated with -1.923, which is 0.0274. We see that the p-value is 0.0274 which is less than our chosen alpha level of 0.05. This means we reject the null hypothesis. That is, we have sufficient evidence to support the claim that less than 40% of U.S. adults are aware that such an investment would result in a sum of over $100,000.