Confidence Intervals for the Mean

This section explains the need for confidence intervals and why a confidence interval is not the probability the interval contains the parameter. Then, it discusses how to compute a confidence interval on the mean when sigma is unknown and needs to be estimated. It also explains when to use t-distribution or a normal distribution. Next, it covers the difference between the shape of the t distribution and the normal distribution and how this difference is affected by degrees of freedom. Finally, it explains the procedure to compute a confidence interval on the difference between means.

Confidence Interval on the Mean

Answers


  1. Because you know the standard deviation of the population, you can use the normal distribution. If you use the normal distribution calculator, you will find that 90% of the area is within 1.65 standard deviations of the mean.

  2. You have to estimate the standard deviation of the population, so you need to use the t distribution. Because your sample size is 11, your degrees of freedom are 11 - 1 = 10. Looking at the t distribution table, you can see that 95% of the distribution falls within 2.23 standard deviations of the mean.

  3. First, the standard error of the mean is 6.5/5 = 1.3. Second, you know the population standard deviation, so you can use the normal distribution. 95% of the distribution lies within 1.96 standard deviations of the mean. Lower limit: 38 - (1.3)(1.96) = 35.45; Upper limit: 38 + (1.3)(1.96) = 40.55

  4. In this case, the population standard deviation is unknown, so you need to use the t distribution (df = N-1 = 9-1 = 8). From the t distribution table, you see that 99% of the distribution lies within 3.355 standard deviations of the mean when t has 8 degrees of freedom. The standard error of the mean = 4/3 = 1.333. Lower limit: 49 - (1.333)(3.355) = 44.53; Upper limit: 49 + (1.333)(3.355) = 53.47

  5. Use a statistical program to compute the 95% CI. To compute it by hand, you need the sample mean and standard error and the t value for 19 df. 14.25 + (2.093)(1.784) = 17.984