So i have a test in stats tomorrow and im trying to work on my study guide but im stuck at this question : Construct a confidence interval for the proportion for each case 90 percent interval 861 out of 2500. please help.
To construct a confidence interval for a proportion, you need to follow a few steps. In this case, you have the following information:
Sample proportion (p̂) = 861/2500
Sample size (n) = 2500
Confidence level (C) = 90%
Now let's calculate the confidence interval using these steps:
1. Determine the critical value: Start by finding the critical value for your desired confidence level. For a 90% confidence level, the critical value can be found using a Z-table or calculator. The critical value is approximately 1.645.
2. Calculate the standard error: The standard error (SE) measures the variability between sample proportions. Use the formula SE = √((p̂(1-p̂))/n) to calculate the standard error. Plugging in the values we have, SE ≈ √((861/2500)(1-861/2500))/2500) ≈ 0.01466.
3. Calculate the margin of error: The margin of error is the range above and below the sample proportion within which the population parameter is likely to fall. The margin of error is calculated by multiplying the critical value by the standard error. In this case, the margin of error ≈ 1.645 * 0.01466 ≈ 0.02409.
4. Calculate the lower bound and upper bound: To construct the confidence interval, subtract and add the margin of error from the sample proportion. The lower bound is calculated as p̂ - margin of error, and the upper bound is calculated as p̂ + margin of error. In this case, the lower bound ≈ 861/2500 - 0.02409 ≈ 0.3269, and the upper bound ≈ 861/2500 + 0.02409 ≈ 0.3594.
Therefore, the confidence interval for the proportion, with a 90% confidence level, is approximately 0.3269 to 0.3594.
Remember, this is just an explanation of the process. Make sure to double-check your calculations to ensure accuracy.