A polling organization wants to estimate the proportion of voters who support candidate A in an upcoming election. They conduct a survey of 1000 randomly selected voters, and find that 540 of them indicate support for candidate A. Assume that the conditions for applying the Central Limit Theorem are met.
(a) Compute a 95% confidence interval for the true proportion of voters who support candidate A.
(b) Interpret the meaning of this 95% confidence interval in the context of the poll.
(c) Discuss how increasing the sample size in future polls would affect the width of the confidence interval and the precision of the estimated proportion. Explain the statistical reasoning behind this effect.
Question
From a random sample of 1000 voters, 540 indicated support for candidate A. Calculate a 95% confidence interval for the true proportion of supporters using the formula \(CI = p \pm z \sqrt{\frac{p(1-p)}{n}}\), where \(p\) is the sample proportion and \(z=1.96\). Round your answer to two decimal places.
Hint
Hint
Hint
Use \(p = \frac{540}{1000} = 0.54\), \(n = 1000\), and \(z = 1.96\) in the formula \(CI = p \pm 1.96 \sqrt{\frac{p(1-p)}{n}}\).
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Complete Answer
[0.51, 0.57]
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Question
Interpret the 95% confidence interval you computed in part (a) in the context of the polling data. Explain what the 95% confidence level means.
Hint
Hint
Hint
Your interpretation should include the range of proportions and the meaning of ‘95% confidence’ in terms of repeated sampling.
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We are 95% confident that the true proportion of all voters who support candidate A lies between 0.51 and 0.57 (or 51% and 57%). This means that if we were to take many random samples of 1000 voters and construct a 95% confidence interval for each sample, approximately 95% of these intervals would contain the true proportion of voters supporting candidate A.
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Question
Discuss how increasing the sample size would affect the width of the confidence interval and the precision of the estimate. Explain the statistical reasoning.
Hint
Hint
Hint
Consider the formula for the margin of error and how the sample size \(n\) influences it. Relate this to the width of the confidence interval and the precision of the estimate.
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Increasing the sample size would decrease the width of the confidence interval and increase the precision of the estimate. Statistically, as the sample size \(n\) increases, the standard error \(\sqrt{\frac{p(1-p)}{n}}\) decreases because \(n\) is in the denominator. Since the margin of error is directly proportional to the standard error (\(ME = z^* \times SE\)), a smaller standard error leads to a smaller margin of error. A smaller margin of error results in a narrower confidence interval, indicating a more precise estimate of the true population proportion. With a narrower interval, we have a more refined range within which we expect the true proportion to lie, thus increasing the precision of our estimate.
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