A sample statistic, and margin of error are given. Find the corfidence interval likely to contain the population parameter of interest and anewer the question,
wears a00.
according to govemment records, 17% of the population of Elmwood had no health insurance. This year a poll revealed that 19.9% were without
insurance. The margin of enor was 2.2 percentage points. Find a corfidence interval for the true percentage without insurance this year. Can ws conduds that the
percentage without insurance has increased from ten years ago?
The sample statistic is 19.9%.
The margin of error is 2.2 percentage points.
To find the confidence interval, we can use the formula:
Confidence interval = Sample statistic ± (Margin of error x Critical value)
The critical value depends on the desired level of confidence. Let's assume a 95% confidence level, which corresponds to a critical value of approximately 1.96.
Confidence interval = 19.9 ± (2.2 x 1.96)
Confidence interval = 19.9 ± 4.312
Confidence interval = (15.588, 24.212)
The confidence interval for the true percentage without insurance this year is (15.588%, 24.212%).
To determine if the percentage without insurance has increased from ten years ago, we can compare the confidence interval to the 17% of the population without insurance from the government records.
Since the confidence interval (15.588%, 24.212%) does not include the value 17%, we can conclude with 95% confidence that the percentage without insurance has increased from ten years ago.