Can someone help me how to solve it.
A new medically approved wheelchair was considered to be an improvement over a standard wheelchair by 325 out of 450 patients in a survery.
20) What is the margin of error for the survey? Round your ansewer to the nearest tenth of a percent.
160
thanks, tori
To find the margin of error for the survey, we need to calculate the standard error. The standard error is a measure of how much the responses in a survey are likely to vary from the true population values.
First, we need to calculate the sample proportion, which is the proportion of patients who considered the new wheelchair to be an improvement. This is obtained by dividing the number of patients who found it to be an improvement by the total number of patients surveyed:
Sample Proportion (p) = Number of patients who found improvement / Total number of patients surveyed
p = 325 / 450
Next, we calculate the standard error using the following formula:
Standard Error (SE) = sqrt((p * (1 - p)) / n)
where:
p is the sample proportion
n is the sample size (number of patients surveyed)
SE = sqrt((p * (1 - p)) / n)
SE = sqrt((325/450 * (1 - 325/450)) / 450)
Once we have the standard error, we can calculate the margin of error by multiplying the standard error by a critical value based on the desired confidence level. The critical value represents the number of standard errors we are willing to tolerate for our desired level of confidence.
For simplicity, let's assume a 95% confidence level, which is commonly used in surveys. The critical value for a 95% confidence level is approximately 1.96.
Margin of Error (ME) = critical value * Standard Error
ME = 1.96 * SE
Now, we can substitute the calculated standard error into the margin of error formula:
ME = 1.96 * calculated SE
Once you calculate the margin of error, you can round it to the nearest tenth of a percent as the question requests.