6.2 D & 6.2 CI

Exploring the T-distribution & Calculating Confidence Intervals for one Mean

There is a lot of difference between population and sample standard deviations, especially in small sample sizes. The t-distribution is a normal distribution which takes these differences into account when counting the number of standard deviations. The t-distribution takes into account the sample size n. We measure the sample size by using degrees of freedom = n-1. Think of t-scores like z-scores but with a different normal curve depending on the sample size.

Directions: Use the t-distribution calculator on StatCrunch to answer the following questions. Recall that the critical "z" values for 90%, 95% and 99% confidence are respectively.

1. For a sample size of n = 5, find the degrees of freedom. Use the calculator and the degrees of freedom to find the critical t-values for 90%, 95% and 99% confidence. How do they compare to the critical z-values?

2. For a sample size of n = 17, find the degrees of freedom. Use the calculator and the degrees of freedom to find the critical t-values for 90%, 95% and 99% confidence. How do they compare to the critical z-values?

3. For a sample size of n = 45, find the degrees of freedom. Use the calculator and the degrees of freedom to find the critical t-values for 90%, 95% and 99% confidence. How do they compare to the critical z-values?

4. What do you notice about the difference between the t-scores and the z-scores as the sample size increases?

Directions: For numbers 5 - 7, use the formula and your calculator to calculate the confidence interval estimate of the population mean.

5. A random sample of 650 high school students has a normal distribution. The sample mean average ACT exam score was 21 with a 3.2 sample standard deviation. Construct a 99% confidence interval estimate of the population mean average ACT exam.

6. A random sample of 200 adults found that they had a sample mean temperature of 98.2 degrees and a standard deviation of 1.8 degrees. Construct a 95% confidence interval estimate of the population mean body temperature of adults. Does the confidence interval indicate that normal body temperature could be 98.6 degrees?

7.A random sample of 315 adults found that the sample mean amount or credit card debt was $435 with a standard deviation of $106. Construct a 90% confidence interval estimate of the population mean amount of credit card debt.