Q: (I shortened the question to only include the important details) Sample size/n = 1974, 354 respondents..... construct a 99.9% confidence interval for the proportion of people with a Master's degree.
My work:
354/1974=.179
1-.179=.821
sqrt (.179) (.821) / 1974 = .0086282858
margin of error = (3.29) x (.0086) =.028294
Calculation of intervals:
Lower limit = .179-.028294=.150706
Upper limit= .179-.028294= .207294
(.150706,.207294)
To construct a confidence interval for the proportion of people with a Master's degree, follow these steps:
1. Calculate the proportion of respondents with a Master's degree. In this case, you found that 354 out of 1974 respondents have a Master's degree, so the proportion is 354/1974 = 0.179.
2. Calculate the complement of the proportion. Subtract the proportion from 1 to get the proportion of respondents without a Master's degree. In this case, it is 1 - 0.179 = 0.821.
3. Calculate the standard error, which is the square root of (proportion * complement) divided by the sample size. In this case, it is sqrt(0.179 * 0.821) / 1974 = 0.0086282858.
4. Determine the critical value for the desired level of confidence. Since we want a 99.9% confidence interval, you can find this value by subtracting the desired confidence level from 1 and dividing the result by 2. In this case, it is (1 - 0.999) / 2 = 0.0005.
5. Calculate the margin of error by multiplying the critical value by the standard error. In this case, the critical value is approximately 3.29, so the margin of error is 3.29 * 0.0086282858 = 0.028294.
6. Finally, calculate the lower and upper limits of the confidence interval. Subtract the margin of error from the proportion to get the lower limit, and add the margin of error to the proportion to get the upper limit. In this case, the lower limit is 0.179 - 0.028294 = 0.150706, and the upper limit is 0.179 + 0.028294 = 0.207294.
Therefore, the 99.9% confidence interval for the proportion of people with a Master's degree is approximately (0.150706, 0.207294).