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CHAPTER 8.2 CHAPTER 8 ESTIMATION PART 2 – Estimating When Is Unknown ADVANCED PROBABILITY AND STATISTICS: 16.0 – Students know basic facts concerning the relation between the mean and the standard deviation of a sampling distribution and the mean and the standard deviation of the population distribution. 17.0 – Students determine confidence intervals for a simple random sample from a normal distribution of data and determine the sample size required for a desired margin of error. OBJECTIVE(S): Students will learn the definition of a Student’s t distribution, degrees of freedom. Students will learn how to calculate the margin of error. Students will learn how to find the critical value knowing a confidence level. Students will learn how to estimate the mean when is unknown. Students will learn which distribution you should use when estimating the mean . Student’s t Distribution In order to use the normal distribution to find confidence intervals for a population mean , we need to know the value of _____, the population __________________ ___________________________. Typically when _____ is unknown, _____ is unknown as well. When we use ____ to approximate _____, the sampling distribution for _____ follows a new distribution called a ____________________ ____ ________________________. Assume that x has a normal distribution with mean _____. For samples of size n with sample mean x and sample standard deviation s, the t variable t = _____________ has a Student’s t distribution with degrees of freedom d.f. = ____________. Properties of a Student’s t distribution 1.) The distribution is symmetric about the mean 0. 2.) The distribution depends on the degrees of freedom. 3.) The distribution is bell-shaped, but has thicker tails than the standard normal distribution. CHAPTER 8.2 4.) As the degrees of freedom increases, the t distribution approaches the standard normal distribution. Area Under the t Curve Between _____ and ______ Area = c tc 0 tc P (______ < t < _______) = ____ 1.) Find the critical value tc for a 0.99 confidence level for a t distribution with sample size n = 5. STEP 1: Find the column with c heading ___________. STEP 2: Compute the number of degrees of freedom: d.f. = ___________= ______________ = ______. STEP 3: Read down the under the heading c = ________ until we reach the row headed by ____ (under d.f.). The entry is _______________. Therefore, t 0.99 = ________. If the degrees of freedom d.f. you need are not in the table, use the closest d.f. in the table that is _________________. This procedure results in a critical value tc that is more conservative in the sense that it is larger. The resulting confidence interval will be longer and have a probability that is slightly higher than ____. CHAPTER 8.2 2.) Find the critical value tc for a 0.90 confidence level for a t distribution with sample size n = 9. STEP 1: Find the column with c heading ___________. STEP 2: Compute the number of degrees of freedom: d.f. = ___________= ______________ = ______. STEP 3: Read down the under the heading c = ________ until we reach the row headed by ____ (under d.f.). The entry is _______________. Therefore, t 0.90 = ________. 3.) Find the critical value tc for a 0.95 confidence level for a t distribution with sample size n = 9. Maximal Margin of Error, E E= Confidence Interval for with Unknown A c confidence interval for is an interval computed from sample data in such a way that c is the probability of generating an interval containing the actual value of . P (__________ < ____ < ___________) = __ How to find a confidence interval for with unknown Let x be a random variable appropriate to your application. Obtain a simple random sample (of size n, n>1) of x values from which you compute the sample mean x and the sample standard deviation s. If you can assume that x has a normal distribution or simply a mound-shaped symmetric distribution, then any sample size n will work. If you cannot assume this, then use a sample size of n 30 . Confidence Interval for with Unknown ________________ < < ______________ where x = sample mean of a simple random sample CHAPTER 8.2 E= ___________ c = ________________ ________________ (___< c < ____) tc = critical value for confidence level c and degrees of freedom d.f. = _________ 4.) Suppose an archaeologist discovers only seven fossil skeletons from a previously unknown species of miniature horse. Reconstruction of the skeletons of these seven miniature horses show the shoulder heights (in centimeters) to be 45.3 47.1 44.2 46.8 46.5 45.5 47.6 The mean is x _______ and the sample standard deviation is s _______. Let be the mean shoulder height (in centimeters) for this entire species of miniature horse, and assume that the population of shoulder heights is approximately normal. Find a 99% confidence interval for , the mean shoulder height of the entire population of such horses. n = ______ d.f. = _________ = ___________ = ______ c = ______ therefore, t 0.99 ______________ E = = tc s n _______________ CHAPTER 8.2 The 99% confidence interval is xE < < < < < < xE The archaeologist can be ______ confident that the interval from ___________ to _____________ is an interval that contains the population mean for shoulder height of this species of miniature horse. CALCULATOR Press the STAT key and select TESTS, then use 7:TInterval. Select Stats and then below inputs, x , n, and C-Level. Highlight Calculate and hit enter. Output: TInterval (_______________, ______________) x = _________ s = ______ n = ______ 5.) A company has a new process for manufacturing large artificial sapphires. In a trial run, 37 sapphires are produced. The mean weight for these 37 gems is x =6.75 carats, and the sample standard deviation is s = 0.33 carat. Let be the mean weight for the distribution of all sapphires produced by the new process. a. What is d.f. for this setting? b. Find E. CHAPTER 8.2 c. Find a 95% confidence interval for . d. Interpret the confidence interval in the context of the problem. Which distribution should you use for x Examine problem statement is known Use normal distribution with margin of error E zc n is unknown Use Student’s t distribution with margin of error E tc s n d.f. = n - 1