Question

For this term, we will create confidence intervals to estimate a population value using the general formula:

sample estimator +/- (reliability factor)(standard error of the estimator)

Recall that the (reliability factor) x (standard error of the estimator)= margin of error (ME) for the interval. The ME is a measure of the uncertainty in our estimate of the population parameter. A confidence interval has a width=2ME.

A 95% confidence interval for the unobserved population mean(µ), has a confidence level = 1-α = 0.95. The confidence level is the probability that the confidence interval contains µ .  

If 1-α = 0.95, then the level of significance is α = 0.05, and the reliability factor (assuming the population variance is unknown and  must be estimated with the sample variance, s2) will be the t value with n-1 degrees of freedom that gives α/2=0.025 probability in the right-tail of a t distribution. (s/√n) is the standard error of the sample mean.

If the sample size n=25, the sample mean = 10 and the sample standard deviation s=6, create the 95% confidence interval for the population mean and interpret it.

• A.

  The confidence interval is [7.52 ,12.48] and 95% of the population values fall in this interval.

• B.

  The confidence interval is [7.52 ,12.48] and there is a 95% chance that the population mean has a value that falls within this interval.

• C.

  The confidence interval is [7.95 ,12.05] and there is a 95% chance that the population mean has a value that falls within this interval.

• D.

  The confidence interval is [7.95,12.05] and 95% of the population values lie within this interval.

Answer 1