Question

In a survey on supernatural experiences, 722 of 4,013 adult Americans reported that they had seen a ghost (“What Supernatural Experiences We've Had," USA Today, February 8,2010 ). Assume that this sample is representative of the population of adult Americans. a. Use the given information to estimate the proportion of adult Americans who would say they have seen a ghost. b. Verify that the conditions for use of the margin of error formula to be appropriate are met. c. Calculate the margin of error. d. Interpret the margin of error in context. e. Construct and interpret a \(90 \%\) confidence interval for the proportion of all adult Americans who would say they have seen a ghost. f. Would a \(99 \%\) confidence interval be narrower or wider than the interval calculated in Part (e)? Justify your answer.

Short answer

The estimated proportion of adult Americans who have seen a ghost is around 0.18. The margin of error calculated under the 90% confidence level is detailed in step 3 of the solution. The 90% confidence interval for the true proportion of Americans who have seen a ghost is then determined in step 5. Lastly, a 99% confidence interval would be wider than a 90% confidence interval. This is because the higher the level of confidence, the larger the margin of error.

Step-by-step solution

Calculate the Proportion

Length of the sample \(n = 4013\) and those who have seen a ghost are 722. So, the proportion \(p = \frac{722}{4013}\).

Verify Conditions for Margin of Error Formula

Check the conditions: \(np \geq 10\) and \(n(1-p) \geq 10\). If both are true, the margin of error formula can be used.

Calculate Margin of Error

The margin of error is calculated using the formula \(Z*\sqrt{\frac{p(1-p)}{n}}\). For a Z of 1.645 (which corresponds to 90% confidence), substitute the calculated value of p and n into the formula to find the margin of error.

Interpret Margin of Error

The calculated margin of error gives a range to predict where the true proportion of the adult population who have seen a ghost lies.

Calculate 90% Confidence Interval

A 90% confidence interval is given as \(p ± \) margin of error.

Comparison of 90% and 99% Confidence Intervals

A 99% confidence interval would be wider than a 90% confidence interval. This is because to be more confident that the true proportion lies within the interval, we need to allow for more possiblities therefore a wider interval.

Key concepts

Margin of Error

Understanding the margin of error is essential when interpreting survey results. It represents the extent to which the sample might differ from the population. To put it simply, it's a buffer zone, giving us a range within which we can expect the true population parameter to fall.

In the provided example regarding sightings of ghosts, we would first affirm that the sample size is large enough for us to use this concept. Once that's established, we can calculate the margin of error using the formula you've learned: \(Z * \sqrt{\frac{p(1-p)}{n}}\). This calculation incorporates a 'Z-score', which is linked to our confidence level—in this case, 90%. The resulting margin of error can then be communicated as the 'plus-or-minus' figure in your final confidence interval estimate.

Proportion Estimation

Proportion estimation involves calculating an estimate of the population proportion (\(p\))) based on sample data. When analyzing data, like how many adult Americans have seen a ghost, we translate the count of positive responses (seeing a ghost) into a proportion by dividing it by the total number of people surveyed. This proportion serves as a point estimate of the true population proportion. It's crucial to remember that this estimation can have some degree of error, hence the inclusion of the margin of error in confidence intervals.

Z-score

The Z-score is a statistical measurement describing a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. When constructing confidence intervals for proportion estimation, Z-scores help to identify the range that covers the true population parameter with a certain level of confidence. Specifically, for a 90% confidence interval, we use the Z-score associated with the central 90% of the normal distribution, which is 1.645. This score is crucial because it adjusts our margin of error to ensure that the interval is neither too wide nor too narrow for our given level of confidence.

Sample Size

Sample size is a pivotal factor in determining the reliability and precision of our estimation. A larger sample size generally decreases the margin of error, leading to a narrower and more precise confidence interval. Hence, ensuring the adequacy of sample size is a critical initial step. In our ghost sighting example, the large sample of 4,013 adult Americans meets the necessary requirements for proportion estimation and credibility of the margin of error. A small sample size might not adequately represent the population and could increase the risk of a wider margin of error, reducing the confidence we could have in the results.

Confidence Level

The confidence level expresses the degree of certainty we have that the true population parameter lies within our calculated interval. Commonly used confidence levels include 90%, 95%, and 99%, translating to Z-scores that stretch our interval to capture the true population parameter within their respective ranges.

It's a balancing act: higher confidence levels produce wider intervals, as we've seen in the comparison between a 90% and 99% confidence interval for the ghost sighting data. Students should take note that opting for extremely high confidence levels can sometimes result in intervals that are too broad to provide useful information, whereas lower levels, while offering a narrower interval, come with reduced certainty.