Question

In a recent poll, 180 people were asked if they liked sking, and 55% said they did. Find the margin of error of this poll, at the 90%confidence level.

As in the reading, in your calculations:
--Use z=1.645 for a 90% confidence interval
--Use z=2 for a 95% confidence interval
--Use z=2.576 for a 99% confidence interval

Answer 1

To construct a 90% confidence interval for a poll where 55% of 180 people liked skiing, calculate the standard error and then multiply by the z-score for 90% confidence level, yielding a margin of error of approximately 6.07%.

Step-by-step explanation:

The question involves constructing a 90% confidence interval for the proportion of people who like skiing based on a poll. First, we need to identify the percentage of people who liked skiing, which is 55%, or 0.55 in decimal form. With a sample size of 180 people, we proceed to calculate the standard error (SE) for the proportion:

SE = √[p(1-p)/n]
SE = √[0.55(1-0.55)/180]
SE = 0.03688

Using the provided critical value Z for a 90% confidence level, which is 1.645, we calculate the margin of error (MOE):

MOE = Z * SE
MOE = 1.645 * 0.03688
MOE = 0.0607

Therefore, the margin of error for this poll, at the 90% confidence level, is approximately 0.0607 or 6.07%.