Student t distribution is usually referred to as the t distribution.
Students usually relate to this quite well if they think of it as a group of exams during a semester and what would be needed on the final exam in order to make a certain average (assuming each exam carried the same weight).
Emphasize the difference in how Table A-3 is set up versus that of Table A-2. In Table A-3, the critical values are found in the body of the table. In Table A-2, the critical values (z scores) are found in the left column and across the top row. Table A-3 is set up for only certain common values of - a limitation of the table.
page 317 of text
Same as exercise #15 on page 321.
Transcript
1.
6-3: Estimating Populations/Small Samples
Assumptions:
n 30
The sample is a simple random sample.
The sample is from a normally distributed population.
The single value of a point estimate does not reveal how good the estimate is …. Thus we will construct a confidence interval.
A sample size n = 15 is a simple random sample selected from a normally distributed population. Find the critical value t /2 corresponding to a 95% level of confidence.
Degree of Confidence = ________% z = 0 /2 = /2 = t /2 = -t /2 =
9.
Find the critical values corresponding to a __% degree of confidence.
90%; n = 22; df = ___; t /2 = ___
99%; n = 12; df = ___; t /2 = ___
95%; n = 7; df = ___; t /2 = ___
90%; n = 25; df = ___; t /2 = ___
(Table: slide 10) Degree of Confidence = ________% z = 0 /2 = /2 = t /2 = -t /2 =
A study of 12 Dodge Vipers involved in collisions resulted in repairs averaging $26,227 and a standard deviation of $15,873. Find the 95% interval estimate of , the mean repair cost for all Dodge Vipers involved in collisions. (The 12 cars’ distribution appears to be bell-shaped.)
Find a confidence interval (95%) for “Heart Rates for Shoveling Snow for a Sample of 10 Individuals”.
n=10; x-bar = 175; s = 15.
1. t /2 = 2. =
14.
Example: Find the 90% confidence interval for given the following: x = 78.8, s = 12.2, n = 27 t /2 = 1.706 E = t 2 s = (1.706)(12.2) = 4.01 n 27 74.79 < < 82.81 x - E < µ < x + E
15.
Example: A study of 20 car tune-up times yield a mean of 100 minutes with a standard deviation of 15 minutes. Find the 80%confidence interval for x = 100, s = 15, n = 20 t /2 = 1.328 E = t 2 s = (1.328)(15) = 4.45 n 20 95.55 < < 104.45 x - E < µ < x + E
16.
Example: The manger of a restaurant finds that for 7 randomly selected days, the number of customers served are listed below. Find the 90%confidence interval for x = 303.4, s = 31.7, n = 7 t /2 = 1.943 E = t 2 s = (1.943)(31.7) = 23.3 n 7 280.1 < < 326.7 289, 326, 264, 318, 306, 269, 352 x - E < µ < x + E
17.
Example: In crash tests of 15 Honda Odyssey minivans, collision repair costs are found to have a distribution that is roughly bell-shaped with a mean of $1786 and a standard deviation of $937. Construct the 99%confidence interval for the mean repair costs x = 1786, s = 937, n = 15 t /2 = 2.977 E = t 2 s = (2.977)(937) = 720 n 15 $1066 < < $2506 x - E < µ < x + E
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