The z-score associated with 98% is 2.33. If the sample p-hat is 0.60 and the standard deviation is 0.02, find the lower limit of the 98% confidence interval.
The Correct Answer and Explanation is:
To calculate the lower limit of the 98% confidence interval for a sample proportion (p^\hat{p}p^), we can use the formula:Confidence Interval=p^±z×σp^\text{Confidence Interval} = \hat{p} \pm z \times \sigma_{\hat{p}}Confidence Interval=p^±z×σp^
Where:
- p^=0.60\hat{p} = 0.60p^=0.60 (the sample proportion),
- z=2.33z = 2.33z=2.33 (the z-score associated with a 98% confidence level),
- σp^=0.02\sigma_{\hat{p}} = 0.02σp^=0.02 (the standard deviation of the sample proportion).
Now, we are asked to find the lower limit of the confidence interval. This is the value given by:Lower Limit=p^−z×σp^\text{Lower Limit} = \hat{p} – z \times \sigma_{\hat{p}}Lower Limit=p^−z×σp^
Plugging in the values:Lower Limit=0.60−(2.33×0.02)\text{Lower Limit} = 0.60 – (2.33 \times 0.02)Lower Limit=0.60−(2.33×0.02)Lower Limit=0.60−0.0466\text{Lower Limit} = 0.60 – 0.0466Lower Limit=0.60−0.0466Lower Limit=0.5534\text{Lower Limit} = 0.5534Lower Limit=0.5534
So, the lower limit of the 98% confidence interval is approximately 0.5534.
Explanation:
To construct a confidence interval for a sample proportion, we use the formula p^±z×σp^\hat{p} \pm z \times \sigma_{\hat{p}}p^±z×σp^, where σp^\sigma_{\hat{p}}σp^ is the standard deviation of the sample proportion, and zzz is the z-score corresponding to the desired confidence level. Since we know the sample proportion (p^\hat{p}p^) and standard deviation, we multiply the z-score (2.33 for 98% confidence) by the standard deviation, then subtract this value from the sample proportion to get the lower limit. This ensures that we account for the margin of error on the lower end of the interval.
