Question 6You draw a sample from the population of a town (n = 312) and find that of this sample, 23% are highly educated and 27% are low-educated. What is the 80% confidence interval for the proportion of highly educated people in this town?1 point(0.21, 0.25) (0.20, 0.26) (0.25, 0.29) (0.23, 0.27)
Question
Question 6You draw a sample from the population of a town (n = 312) and find that of this sample, 23% are highly educated and 27% are low-educated. What is the 80% confidence interval for the proportion of highly educated people in this town?1 point(0.21, 0.25) (0.20, 0.26) (0.25, 0.29) (0.23, 0.27)
Solution
To calculate the 80% confidence interval for the proportion of highly educated people in the town, we first need to find the standard error (SE) of the proportion. The formula for the standard error of a proportion is sqrt[p(1-p)/n], where p is the proportion and n is the sample size.
In this case, p = 0.23 and n = 312.
SE = sqrt[0.23(1-0.23)/312] = 0.027
Next, we need to find the z-score that corresponds to an 80% confidence level. The z-score for an 80% confidence level is approximately 1.28 (you can find this value in a standard z-table or using a calculator).
The margin of error (ME) is then calculated as the z-score times the standard error.
ME = 1.28 * 0.027 = 0.03456
Finally, we calculate the confidence interval by subtracting the margin of error from the sample proportion for the lower limit and adding the margin of error to the sample proportion for the upper limit.
Lower limit = 0.23 - 0.03456 = 0.19544 Upper limit = 0.23 + 0.03456 = 0.26456
So, the 80% confidence interval for the proportion of highly educated people in this town is approximately (0.20, 0.26).
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