Thank you in advance for anyone able to help me with this...
In a sample of 67 randomly selected students, 27 favored the amount being budgeted for next year's intramural and interscholastic sports. Construct a 98% confidence interval for the proportion of all students who support the proposed budget amount. (Give your answers correct to three decimal places.)
Lower Limit
Upper Limit
To construct a confidence interval for the proportion of all students who support the proposed budget amount, we can use the formula:
Confidence Interval = sample proportion ± (critical value * standard error)
Step 1: Calculate the sample proportion
We are given that 27 out of 67 students favored the budget amount. So, the sample proportion can be calculated as:
Sample proportion = favorable outcomes / total number of outcomes
= 27 / 67
≈ 0.403
Step 2: Determine the critical value
The critical value is based on the desired confidence level and the sample size. Since we want a 98% confidence interval, we need to find the critical value corresponding to this confidence level.
To find the critical value, we can refer to a standard normal distribution table or use statistical software. For a 98% confidence level, the critical value is approximately 2.33.
Step 3: Calculate the standard error
The standard error measures the variability in the sample proportion and can be calculated using the formula:
Standard error = sqrt((sample proportion * (1 - sample proportion)) / sample size)
Plugging in the values we know:
Standard error = sqrt((0.403 * (1 - 0.403)) / 67)
≈ 0.058
Step 4: Calculate the confidence interval
Using the formula mentioned earlier:
Confidence Interval = sample proportion ± (critical value * standard error)
Confidence Interval = 0.403 ± (2.33 * 0.058)
= 0.403 ± 0.135
The lower limit of the confidence interval can be found by subtracting the margin of error from the sample proportion:
Lower Limit = sample proportion - margin of error
= 0.403 - 0.135
≈ 0.268
The upper limit of the confidence interval can be found by adding the margin of error to the sample proportion:
Upper Limit = sample proportion + margin of error
= 0.403 + 0.135
≈ 0.538
Therefore, the 98% confidence interval for the proportion of all students who support the proposed budget amount is approximately 0.268 to 0.538.