35%of adult Americans are regular voters. A random sample of 250 adults in a medium-size college town were surveyed, and it was found that 110 were regular voters. Estimate the true proportion of regular voters with 90% confidence and comment on your results...
Can you show you work please???
To estimate the true proportion of regular voters in a college town with 90% confidence, we can use a confidence interval. Since we have a sample size of 250 and 110 regular voters, we can calculate the sample proportion, denoted by p-hat.
Step 1: Calculate the sample proportion (p-hat):
p-hat = (number of regular voters in sample) / (sample size)
= 110 / 250
= 0.44
Step 2: Calculate the standard error (SE):
SE = sqrt[(p-hat * (1 - p-hat)) / n]
= sqrt[(0.44 * (1 - 0.44)) / 250]
= sqrt(0.000968)
= 0.0311 (rounded to four decimal places)
Step 3: Calculate the margin of error (ME):
ME = Z * SE
= Z * 0.0311
Here, Z is the z-score corresponding to the desired confidence level. For a 90% confidence level, the z-score is approximately 1.645.
ME = 1.645 * 0.0311
= 0.0511 (rounded to four decimal places)
Step 4: Calculate the confidence interval:
Confidence Interval = p-hat ± ME
= 0.44 ± 0.0511
The lower bound of the confidence interval = 0.44 - 0.0511
The upper bound of the confidence interval = 0.44 + 0.0511
The confidence interval for the true proportion of regular voters with 90% confidence is (0.3889, 0.4911).
Comment: Based on the survey data, we can be 90% confident that the true proportion of regular voters in the college town falls within the range of 0.3889 to 0.4911. This indicates that the sample estimate of 44% regular voters is relatively close to the true proportion. However, it is important to note that there is still some level of uncertainty due to sampling variability.
Sure! To estimate the true proportion of regular voters, we will use the formula for calculating a confidence interval. Here's how you can calculate it step-by-step:
Step 1: Convert the sample proportion to a standard error.
The formula for calculating the standard error is:
SE = √((p̂ * (1 - p̂)) / n)
Where p̂ is the sample proportion and n is the sample size.
p̂ = 110/250 = 0.44
SE = √((0.44 * (1 - 0.44)) / 250)
SE = √((0.44 * 0.56) / 250)
SE = √(0.2464 / 250)
SE = √0.0009856
SE ≈ 0.0314
Step 2: Determine the z-score for the desired confidence level.
For a 90% confidence level, we need to find the z-score that corresponds to the two-tailed confidence level of 0.10. Looking up this value in the standard normal distribution table gives us a z-score of approximately 1.645.
Step 3: Calculate the margin of error.
The margin of error is equal to the product of the standard error and the z-score.
Margin of Error = z * SE
Margin of Error = 1.645 * 0.0314
Margin of Error ≈ 0.0517
Step 4: Calculate the confidence interval.
The confidence interval is the range within which we estimate the true proportion of regular voters lies.
Confidence Interval = p̂ ± Margin of Error
Confidence Interval = 0.44 ± 0.0517
Confidence Interval ≈ (0.3883, 0.4917)
Therefore, with 90% confidence, we can estimate that the true proportion of regular voters in the college town is between 0.3883 and 0.4917.
Comment: Since the confidence interval does not include the extreme values of 0 or 1, it suggests that the estimate is plausible. However, it's important to note that this estimate is specific to the college town surveyed and may not necessarily represent the true proportion of regular voters across all adult Americans.
Confidence interval using proportions:
CI90 = p + or - (1.645)(√pq/n)
...where + or - 1.645 represents 90% confidence using a z-table.
Note: p = 110/250 (convert to a decimal); q = 1 - p; and n = 250 (sample size).
I'll let you take it from here.