Confidence Interval with z “Based on the sample, we are ____% confident that the population mean, 𝜇, is between _____ and ____.” 11/20/2018

Confidence Interval with 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 known A chosen “Confidence Level”, like 90%, 95%, 99% “Margin of Error”, 𝐸=𝑧 𝛼/2 ∙ 𝜎 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.” 11/20/2018

When can you do this legally? Anytime that the population standard deviation, 𝜎 is known, “Known” may mean “previous studies indicate that 𝜎 is…” ~ ~ ~ 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 11/20/2018

95% Confidence Interval of the Mean from Bluman’s slides © McGraw Hill

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=2.3 hours is known. Find the 95% confidence interval for hours studied. From page 361 of Beginning Statistics, by Warren, Denley, and Atchley, © 2008 Hawkes Learning Systems. Find 𝑧 𝛼/2 corresponding to 95% confidence interval. Find 𝐸= 𝑧 𝛼/2 ∙ 𝜎 𝑛 Form the confidence interval: 𝑥 −𝐸<𝜇< 𝑥 +𝐸 11/20/2018

Example – Hours of studying Details By-hand solution 95% in the middle 0.9500 area in the middle 1.0000 – 0.9500 = 0.0500 in two tails 0.0500 / 2 = 0.0250 in each tail What z has 0.0250 area to its left? z = -1.96 So use 𝑧=1.96, positive Find 𝑧 𝛼/2 corresponding to 95% confidence interval. Find 𝐸= 𝑧 𝛼/2 ∙ 𝜎 𝑛 Form the confidence interval: 𝑥 −𝐸<𝜇< 𝑥 +𝐸 11/20/2018

Example – Hours of studying Details By-hand solution 𝐸=1.96∙ 2.3 78 𝐸=0.51 Confidence interval is 15−0.51<𝜇<15+0.51 14.49<𝜇<15.51 hours of studying per week Find 𝑧 𝛼/2 corresponding to 95% confidence interval. Find 𝐸= 𝑧 𝛼/2 ∙ 𝜎 𝑛 Form the confidence interval: 𝑥 −𝐸<𝜇< 𝑥 +𝐸 11/20/2018

What does it mean? Details Interpretation 𝐸=1.96∙ 2.3 78 𝐸=0.51 𝐸=1.96∙ 2.3 78 𝐸=0.51 Confidence interval is 15−0.51<𝜇<15+0.51 14.49<𝜇<15.51 hours of studying per week The true mean is within 0.51 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 14.49 and 15.51 hours/wk. 11/20/2018

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=2.3 hours is known. 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, 7:Zinterval Note Inpt: Stats Highlight Calculate Press ENTER 11/20/2018

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=2.3 hours is known. Find the 95% confidence interval for hours studied. From page 361 of Beginning Statistics, by Warren, Denley, and Atchley, © 2008 Hawkes Learning Systems. 11/20/2018

90% vs. 95% vs. 99% Confidence Recall: Margin of Error is 𝐸= 𝑧 𝛼/2 ∙ 𝜎 𝑛 The Level of Confidence determines 𝑧 𝛼/2 If Level of Confidence is 90%, 𝑧 𝛼/2 =1.645 If Level of Confidence is 95%, 𝑧 𝛼/2 =1.96 If Level of Confidence is 98%, 𝑧 𝛼/2 =2.33 If Level of Confidence is 99%, 𝑧 𝛼/2 =2.575 11/20/2018

90% vs. 95% vs. 99% Confidence Recall: Margin of Error is 𝐸= 𝑧 𝛼/2 ∙ 𝜎 𝑛 The Level of Confidence determines 𝑧 𝛼/2 If you choose a higher Level of Confidence %, The 𝑧 𝛼/2 value is higher. Which causes a higher Margin of Error. Which makes the confidence interval wider. 11/20/2018

90% vs. 95% vs. 99% Confidence Again with 𝑛=78, 𝑥 =15.0, 𝜎=2.3 90% confidence interval narrow confidence interval, (14.572,15.428) 95% confidence interval medium confidence interval, (14.49, 15.51) 99% confidence interval Wider confidence interval, (14.329, 15.671) 11/20/2018

How big of a sample do I need? Inputs Calculations “I want a ____% confidence level.” (which determines the 𝑧 𝛼/2 value) The population standard deviation is 𝜎. “I want the margin of error to be no bigger than 𝐸.” 𝑛= 𝑧∙𝜎 𝐸 2 Always bump up 𝑛 to the next highest integer. Bump, don’t round. (Unless 𝑛 just happened to come out to an exact integer, very rare.) 11/20/2018

Sample Size Example 𝑛= 𝑧∙𝜎 𝐸 2 𝑛= (1.96)∙(3.25) (0.5) 2 Inputs Calculations “How many credit cards do you have?” Suppose you know that the standard deviation 𝜎=3.25 cards. And you can tolerate an error of 0.50 (half a card). And you want a 95% confidence interval. (Taken from Hawkes, page 362) 𝑛= 𝑧∙𝜎 𝐸 2 𝑛= (1.96)∙(3.25) (0.5) 2 𝑛=162.3. Bump it up! We need a sample of 163 people. 11/20/2018

Vocabulary: “Point Estimate” Our sample mean, 𝑥 , is a point estimate of the population mean, 𝜇. 11/20/2018

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 𝜎, 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. Book tells you only 𝜎 and which confidence level % And then apply the formula. 11/20/2018

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 It still asks for population 𝜎. Tell it which List (like 2ND 1 for L1) If no frequencies, keep Freq:1 C-Level decimal as usual. Highlight Calculate Press ENTER. 11/20/2018

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 normally distributed.) The data: 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) 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. Bluman, Chapter 7