Confidence Interval with t “Based on the sample, we are ____% confident that the population mean, 𝜇, is between _____ and ____.” 11/17/2018

Confidence Interval with t, instead of z Inputs Outputs A sample of 𝑛 items A list of 𝑛 data values measured in the sample The mean of the sample data, 𝑥 The population standard deviation, 𝜎, is unknown A chosen “Confidence Level”, like 90%, 95%, 99% “Margin of Error”, 𝐸=𝑡 𝛼/2 ∙ 𝒔 𝑛 A low-to-high confidence interval, centered at your sample mean: 𝑥 −𝐸 to 𝑥 +𝐸 “I’m ___% sure that the population mean, is in this interval.” 11/17/2018

When should you use 𝑡 instead of 𝑧? Anytime that the population standard deviation, 𝜎 is unknown, but we can use our sample standard deviation, 𝑠, in its place. ~ ~ ~ AND ~ ~ ~ At least one of these conditions is true: It’s a “large” sample, sample size 𝑛≥30 Or you know that the population is normally distributed 11/17/2018

What is this new 𝑡, anyway? When the population standard deviation 𝜎 is unknown, we have more uncertainty. 𝑡 is kind of like 𝑧 but it takes this extra uncertainty into account. So 𝑡 𝛼/2 will be a little bigger than the 𝑧 𝛼/2 . Other than that, it’s all going to work the same as the 𝑧 worked. Your challenge is to know when to use which. 11/17/2018

Example – Hours of studying Problem By-hand solution Sample of 𝑛=78 students surveyed Sample mean 𝑥 =15.0 hours of studying per week Suppose 𝜎= is unknown. But supppose our sample standard deviation 𝒔=𝟐.𝟑 Find the 95% confidence interval for hours studied. From page 361 of Beginning Statistics, by Warren, Denley, and Atchley, © 2008 Hawkes Learning Systems. Find 𝑡 𝛼/2 corresponding to 95% confidence interval and the degrees of freedom, 𝑑.𝑓.=𝑛−1 Find 𝐸= 𝑡 𝛼/2 ∙ 𝒔 𝑛 Form the confidence interval: 𝑥 −𝐸<𝜇< 𝑥 +𝐸 11/17/2018

Example – Hours of studying Details By-hand solution Table F: The t Distributions 95% in the middle 5%, or 0.05 in two tails “Degrees of Freedom”, 𝑑𝑓 = 𝑛 – 1 = 78−1 = 77 Column “Two tails 0.05” Row 𝑑𝑓=75 is closest 𝑡 𝛼/2 =1.992 Find 𝑡 𝛼/2 corresponding to 95% confidence interval. Find 𝐸= 𝑡 𝛼/2 ∙ 𝑠 𝑛 Form the confidence interval: 𝑥 −𝐸<𝜇< 𝑥 +𝐸 11/17/2018

Example – Hours of studying Details By-hand solution 𝐸=1.992∙ 2.3 78 𝐸=0.52 Confidence interval is 15−0.52<𝜇<15+0.52 14.48<𝜇<15.52 hours of studying per week Find 𝑡 𝛼/2 corresponding to 95% confidence interval. Find 𝐸= 𝑧 𝛼/2 ∙ 𝜎 𝑛 Form the confidence interval: 𝑥 −𝐸<𝜇< 𝑥 +𝐸 11/17/2018

What does it mean? Details Interpretation 𝐸=1.992∙ 2.3 78 𝐸=0.52 𝐸=1.992∙ 2.3 78 𝐸=0.52 Confidence interval is 15−0.52<𝜇<15+0.52 14.48<𝜇<15.52 hours of studying per week The true mean is within 0.52 hours, high or low, of our sample mean We’re 95% confident of that. We’re 95% confident that the true mean number of hours studied is between 14.48 and 15.52 hours/wk. 11/17/2018

Exact 𝑡 𝛼/2 values The printed table is limited. We had to take the closest row. Exact value using TI-84 invT(area to left, df) Compare to our closest row 1.992 11/17/2018

Example 7-5: Using Table F (From Bluman slides, © McGraw Hill) Find the tα/2 value for a 95% confidence interval when the sample size is 22. Degrees of freedom are d.f. = 21. Bluman, Chapter 7

Compare that to invT 95% confidence interval, df = 21 The table says 𝑡 𝛼/2 =2.080 TI-84 invT gives exact value 2.079613837 Compare to 𝑧 𝛼/2 value 1.959963986 You can see how 𝑡 is “wider”, building in the added uncertainty because 𝛼 is unknown. 11/17/2018

Example – Hours of studying Problem TI-84 Solution Sample of 𝑛=78 students surveyed Sample mean 𝑥 =15.0 hours of studying per week Suppose 𝜎 is unknown, but we have 𝑠=2.3. Find the 95% confidence interval for hours studied. From page 361 of Beginning Statistics, by Warren, Denley, and Atchley, © 2008 Hawkes Learning Systems. STAT, TESTS, 8:TIinterval Note Inpt: Stats Highlight Calculate Press ENTER 11/17/2018

Example – Hours of studying Problem TI-84 Solution Sample of 𝑛=78 students surveyed Sample mean 𝑥 =15.0 hours of studying per week Suppose 𝜎 is unknown, but we have 𝑠=2.3. Find the 95% confidence interval for hours studied. From page 361 of Beginning Statistics, by Warren, Denley, and Atchley, © 2008 Hawkes Learning Systems. 11/17/2018

How big of a sample do I need? Calculations with z What about with t? “I want a ____% confidence level.” (which determines the 𝑧 𝛼/2 value) with a margin of error that’s no larger than 𝐸.” 𝑛= 𝑧∙𝜎 𝐸 2 and bump up All of the formula’s inputs are conveniently known in advance! We don’t do this with t. It’s too complicated for us right now. Trouble spots: 𝑛 and 𝑡 𝛼/2 are interdependent (via d.f.) We don’t know 𝑠 until we have the sample. But we were depending on this to give us the sample size to take! 11/17/2018

When you have only the raw data Many book problems are nice Raw data only more real-life Textbook problems are nice to you, usually They usually just tell you the 𝑥 , the 𝑛, the 𝑠, and the desired confidence interval %. They do this to save time They do this so you can focus on the big picture, finding the confidence interval You’re doing your own real-life statistical research All you have is the raw data, a bunch of measurements. But if you have only the raw data, you have to calculate the 𝑥 and the 𝑛 and the 𝑠. Book tells you only 𝜎 and which confidence level % And then apply the formula. 11/17/2018

When you have raw data and TI-84 Put the data into a TI-84 list, such as L1. If there are frequencies, put them into list L2. Choose Inpt: Data, instead of Stats Tell it which List (like 2ND 1 for L1) If no frequencies, keep Freq:1 C-Level decimal as usual. Highlight Calculate Press ENTER. 11/17/2018

Example 7-3: Credit Union Assets (from Bluman © McGraw Hill) The following data represent a sample of the assets (in millions of dollars) of 30 credit unions in southwestern Pennsylvania. Find the 90% confidence interval of the mean. (Assume that the population is not normally distributed and that 𝜎 is not known, so use t.) The data: (see, 𝑛 = 30) 12.23 16.56 4.39 2.89 1.24 2.17 13.19 9.16 1.42 73.25 1.91 14.64 11.59 6.69 1.06 8.74 3.17 18.13 7.92 4.78 16.85 40.22 2.42 21.58 5.01 1.47 12.24 2.27 12.77 2.76 Bluman, Chapter 7

Example 7-3: Credit Union Assets Step 4: Substitute in the formula. (BUT TRY TI-84 LIST INSTEAD) Recall Bluman’s 𝒛 version got this: One can be 90% confident that the population mean of the assets of all credit unions is between $6.752 million and $15.430 million, based on a sample of 30 credit unions. Our TI-84 𝑡 version – compare to the Zinterval result: Bluman, Chapter 7