Presentation on theme: "6.2 Confidence Intervals for the Mean (Small Samples) Statistics Mrs. Spitz Spring 2009."— Presentation transcript:
6.2 Confidence Intervals for the Mean (Small Samples) Statistics Mrs. Spitz Spring 2009
Objectives/Assignment How to interpret the t-distribution and use a t-distribution table How to construct confidence intervals when n < 30 and is unknown Assignment: pp. 271-273 #1-26
The t-Distribution In many real-life situations, the population standard deviation is unknown. Moreover, because of various constraints such as time and cost, it is often not practical to collect samples of size 30 or more. So, how can you construct a confidence interval for a population mean given such circumstances? If the random variable is normally distributed (or approximately normal), the sampling distribution for is a t-distribution.
Note: Table 5 of Appendix B lists critical values of t for selected confidence intervals and degrees of freedom. Pg. A20
Concept: Degrees of freedom Suppose there are an equal number of chairs in a classroom as there are students: 25 chairs and 25 students. Each of the first 24 students to enter the classroom has a choice as to which chair he or she will sit in. There is no freedom of choice, however, for the 25 th student who enters the room.
Example 1: Finding Critical Values of t Find the critical value t c, for a 95% confidence when the sample size is 15. SOLUTION: Because n = 15, the degrees of freedom are: d.f. = n – 1 = 15 – 1 = 14 A portion of Table 5 is shown. Using d.f. = 14 and c = 0.95, you can find the critical value, t c, as shown by the highlighted areas in the table.
The graph shows the t-distribution for 14 degrees of freedom, c = 0.95 and t c = 2.145
Note: After 30 d.f., the t-values are close to the z-values. Moreover, the values in the table that show ∞ d.f. correspond EXACTLY to the normal distribution values.
Try it yourself #1: Find the critical value t c, for a 90% confidence when the sample size is 22. SOLUTION: Because n = 22, the degrees of freedom are: d.f. = n – 1 = 22 – 1 = 21 A portion of Table 5 is shown. Using d.f. = 21 and c = 0.90, you can find the critical value, t c, as shown by the highlighted areas in the table.
A. Identify the degrees of freedom: 21 B. Identify the level of confidence: 0.90 C. Use Table 5 to find t: 1.721 Three things to do:
Study Tip: Unlike the z-table, critical values for a specific confidence interval can be found in the column headed by c in the appropriate d.f. row. (The symbol ∝ will be explained in chapter 7.)
Confidence Intervals and t- Distributions Constructing a confidence interval using the t-distribution is similar to constructing a confidence interval using the normal distribution—both use a point estimate and a maximum error of estimate, E.
Another Study Tip Before using the t-distribution to construct a confidence interval, you should check that n < 30, is unknown, and the population is approximately normal.
Ex. 2: Constructing a Confidence Interval You randomly select 16 restaurants and measure the temperature of the coffee sold at each. The sample mean temperature is 162 〫F with a standard deviation of 10 〫F. Find the 95% confidence interval for the mean temperature. Assume the temperatures are approximately normally distributed.
SOLUTION: Because the sample size is less than 30, is unknown, and the temperatures are approximately normally distributed, you can use the t-distribution. Using n = 16, is 162, s = 10, c = 0.95 and d.f. =15, you can use Table 5 to find that t c =
Step 3: Maximum error of Estimate, E. Step 4: Left/Right endpoints by adding and subtracting from the mean.
Ex. 3: Constructing a Confidence Interval You randomly select 20 mortgage institutions and determine the current mortgage interest rate at each. The sample mean rate is 6.93% with a sample standard deviation of 0.42%. Find the 99% confidence interval for the mean mortgage rate. Assume the interest rates are approximately normally distributed.
Identify all your necessary info Sample size is less than 30, is unknown and the interest rates are approximately normally distributed so we can use the t-distribution. n=20, = 6.93, s = 0.42, c = 0.99, and d.f. = 19. Find the maximum error of estimate at the 99% confidence interval:
Step 3: Maximum Error of Estimate Step 4: Subtract/Add 0.269 to
Ex. 4: Choosing the Normal or t-Distribution You randomly select 25 newly constructed houses. The sample mean construction cost is $181,000 and the population standard deviation is $28,000. Assuming construction costs are normally distributed, should you use the normal distribution, the t-distribution or neither to construct a 95% confidence interval for the mean construction cost? Explain your reasoning.
Ex. 4: Choosing the Normal or t-Distribution SOLUTION: Because the population is normally distributed and the standard deviation is known, you should use the normal distribution.
Try it yourself #4 You randomly select 18 adult male athletes and measure the resting heart rate of each. The sample mean heart rate is 64 beats per minute. Assuming that the heart rates are normally distributed, should you use the normal distribution, the t-distribution or neither to construct a 90% confidence interval for the mean heart rate? Explain your reasoning.
Flowchart will help! Is n ≥ 30? NO, IT IS 18. Is population normally or approximately normally distributed? Yes, normally distributed. Is known? No, sample standard deviation only.