In a survey of 1037 adults, ages 65 and over, 643 were concerned about getting the flu.
a) Find the point estimate for the population proportion , p , of those concerned about getting the flu.
b) Construct a 90% Confidence Interval for the population proportion.
c) How big a sample size would be needed to allow for a 2% margin of error?
the only thing this is, is useless
a) To find the point estimate for the population proportion (p), we need to divide the number of people concerned about getting the flu (643) by the total number of adults surveyed (1037).
Point estimate (p) = Number of people concerned / Total sample size
= 643 / 1037
Using a calculator or performing the division, we find that the point estimate for the population proportion (p) is approximately 0.6199 or 61.99%.
b) To construct a 90% confidence interval for the population proportion, we can use the formula:
Confidence Interval = Point Estimate ± (Z * Standard Error)
where Z represents the z-score associated with the desired level of confidence and Standard Error is calculated using the formula:
Standard Error = sqrt((p * (1 - p)) / n)
First, we need to find the z-score associated with a 90% confidence level. Typically, this can be found by looking up the value in a standard normal distribution table or by using a statistical software. For a 90% confidence level, the z-score is approximately 1.645.
Now, we can plug in the values:
Standard Error = sqrt((0.6199 * (1 - 0.6199)) / 1037)
Calculating the standard error, we get:
Standard Error ≈ 0.0147
With the z-score and standard error, we can now construct the confidence interval:
Confidence Interval = 0.6199 ± (1.645 * 0.0147)
Calculating the confidence interval, we get:
Confidence Interval ≈ 0.5942 to 0.6456
Therefore, the 90% confidence interval for the population proportion is approximately 0.5942 to 0.6456, or 59.42% to 64.56%.
c) To determine the sample size needed to allow for a 2% margin of error, we can use the following formula:
Sample size (n) = (Z^2 * p * (1 - p)) / (margin of error)^2
First, we need to determine the z-score associated with the desired level of confidence. For a 95% confidence level, the z-score is approximately 1.96.
Plugging in the values:
Sample size (n) = (1.96^2 * 0.6199 * (1 - 0.6199)) / (0.02^2)
Calculating the sample size, we get:
Sample size (n) ≈ 2339.12
So, to have a sample size that allows for a 2% margin of error, we would need a sample size of approximately 2339.
Use a proportional confidence interval formula.
CI90 = p + or - (z-value)(√ pq/n)
Find z-value using a z-table representing 90%. p = 643/1037 (convert to a decimal). q = 1 - p. n = 1037 (sample size). Plug values into the formula and calculate the interval.
Formula to find sample size:
n = [(z-value)^2 * p * q]/E^2
... where n = sample size, z-value is the same as above, p and q are the same as above, ^2 means squared, * means to multiply, and E = .02 (for 2%).
Plug values into the formula and calculate n.
I hope this will help get you started.