Are you annoyed with spam e-mail? Suppose a random sample of 200 Kent State students was asked this question of which 80% said that they are annoyed. From the provided information we can find the following:
sample percent = 80% (sample proportion = .80)
standard deviation (S.D.) = .03
a. Set up the calculation of a 95% confidence interval to estimate the population proportion of Penn State students who are annoyed by spam e-mail?
.8 ± .06 =(.50 to .56) or (50 to 56) %
I think this is totally wrong -- help!
To calculate the confidence interval for estimating the population proportion, you can use the formula:
Confidence Interval = Sample proportion ± (Z * Standard Error)
Here, the sample proportion is 80% (or 0.80), and the standard deviation (SD) is 0.03. However, to compute the standard error, we first need to find the critical value (Z) for a 95% confidence level.
The critical value (Z) can be found by using a standard normal distribution table or a calculator. For a 95% confidence level (which corresponds to a 2-sided interval), the critical value is approximately 1.96.
Now, the standard error (SE) is calculated by dividing the standard deviation by the square root of the sample size. In this case, since you have a sample of 200 students, the sample size is 200.
Standard Error (SE) = sqrt((Sample proportion * (1 - Sample proportion)) / Sample size)
= sqrt((0.80 * (1 - 0.80)) / 200)
= sqrt(0.16 / 200)
= 0.02
Now, you can substitute the values into the confidence interval formula:
Confidence Interval = 0.80 ± (1.96 * 0.02)
= (0.80 - 0.0392, 0.80 + 0.0392)
≈ (0.7608, 0.8392)
So, the 95% confidence interval for estimating the population proportion of Kent State students who are annoyed by spam e-mail is approximately 76.08% to 83.92%.
I apologize for the incorrect initial answer you provided. The correct calculation for the confidence interval is shown above.