Question

The article referenced in the previous exercise also reported that \(38 \%\) of the 1,200 social network users surveyed said it was OK to ignore a coworker's friend request. If \(p=0.38\) is used as an estimate of the proportion of all social network users who believe this, is it likely that this estimate is within 0.05 of the actual population proportion? Use what you know about the sampling distribution of \(\hat{p}\) to support your answer.

Short answer

No, it's not likely that the estimate of 0.38 is within 0.05 of the actual population proportion, given the large Z-scores and assuming a 95% confidence level.

Step-by-step solution

Define the method

This problem requires hypothesis testing, and in this context, we are using the Z-score formula. The Z-score represents how many standard deviations an observation or datum is from the mean.

Mean and Standard Deviation

Using the given proportion \(p = 0.38\) as the population mean and standard deviation \(\sigma = \sqrt{p(1-p)/n}\), we get the mean \(\mu = 0.38\) and standard deviation \(\sigma = \sqrt{(0.38)(1-0.38)/1200} = 0.0141\). It is important to note that this only applies if the sampling distribution is approximately normal. This will be checked in the next step.

Checking the Normal condition

To use the Normal model, we need \(np\) and \(n(1-p)\) both to be at least 10.\nThe number expected with the event is \(1200 \times 0.38 = 456\), and the number expected without the event is \(1200 \times (1-0.38) = 744\). Since both these numbers are greater than 10, the sampling distribution is approximately normal, and we can proceed with calculating the Z-score.

Calculate the Z-scores

Given the interval within 0.05 of the actual proportion, this corresponds to two intervals of interest, \(p+0.05\) and \(p-0.05\).\nFor \(p+0.05 = 0.43\), \(Z = (0.43 - 0.38) / 0.0141 = 3.55\). The Z-score for \(p-0.05 = 0.33\) is \(Z = (0.33 - 0.38) / 0.0141 = -3.55\)

Interpret the Z-scores

The Z-scores for both upper and lower limits are beyond the usual threshold for a statistical test (±1.96 for a 95% confidence level). This means that the estimate 0.38 is not within 0.05 of the true population proportion, assuming a 95% confidence level.