Chapter 6 Confidence Intervals Sections 6-1 and 6-2 Confidence Intervals for Large and Small Samples - PowerPoint Presentation

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Chapter 6Confidence Intervals Sections 6-1 and 6-2 Confidence Intervals for Large and Small Samples

 VOCABULARY: Point Estimate – A single value estimate for a population parameter. The most unbiased point estimate of the population parameter is the sample mean. Interval Estimate - An interval, or range of values, used to estimate a population parameter. Level of Confidence - Denoted as c , it is the probability that the interval estimate contains the population parameter. Margin of Error - Sometimes also called the maximum error of estimate, or error tolerance. It is denoted as E , and is the greatest possible distance between the point estimate and the value of the parameter it is estimating. c- confidence interval - Is found by adding and subtracting E from the sample mean. The probability that the confidence interval contains µ is c .

 FORMULAS: Margin of Error = To use the formula, it is assumed that the population standard deviation is known. This is rarely the case, but when , the sample standard deviation s can be used in place of σ. The formula effectively becomes = InvNorm of , or (explained why on bottom of page 311)  

 FORMULAS: c- confidence interval - sample standard deviation: ; It is MUCH easier to use the STAT-Edit function of the calculator to find s. 

GUIDELINES (creating interval by hand): 1) Find the sample statistics and . is the sample size. is the sample mean. 2) Specify σ, if known. Otherwise, if , find the sample standard deviation s and use it as a point estimate for σ. 3) Find the critical value that corresponds to the given level of confidence. The three most commonly used confidence levels are 90%, 95%, and 99%. The corresponding values are: 90% -- 1.645; 95% -- 1.96; 99% -- 2.576 It would help to memorize these. You will be using them a lot. 4) Find the margin of error E. () 5) Find the left and right endpoints and form the confidence interval.  

GUIDELINES (creating interval with calculator): The TI-84 will actually create the interval for you!! 1) If you are given the actual data points (a list of numbers), enter them into L1 using the STAT – Edit menu. 2) STAT – Calc – 1 gives you the sample standard deviation, which you will need. Ignore everything else; the calculator will give it to you again. 3) STAT – Test – 7 (when you have EITHER n > 30 OR you know what is. 4) If you have the list of numbers in L1, choose Data and enter the sample standard deviation where it asks for Do NOT use the from the calculator!! Use the SAMPLE standard deviation. If you do NOT have a list of numbers, you will be given the mean, standard deviation, and sample size. Select Stats and enter the values asked for as given to you. 

GUIDELINES (creating interval with calculator):The TI-84 will actually create the interval for you!! 5) Enter the confidence level you are looking for at C-Level and Calculate. The calculator will give you the interval. From here, you can work backwards to find E, if necessary. Either take half of the difference between your endpoints, or find the difference between the mean and either endpoint.

We are going to walk through Example 4 on page 314, using the data points from Example 1 on page 310. 1) Enter the 50 data points into L1 on your calculator (STAT Edit). 2) STAT Calc 1 to find the sample standard deviation (we can use this because we have 50 data points; ). s = 5.01, = 12.4, and n = 50. 96 10 9 12 11 5 12 17 12 17 18 12 10 13 11 18 6 17 11 14 7 9 6 5 16 18 9 5 20 20 11 9 11 14 18 18 12 13 25 23 16 11 10 22 9 9 6 11 7

TO SOLVE BY HAND: The question asks for a 99% confidence interval for the population mean. To find E using the formula, , plug in what we know. is found by 2nd VARS 3 ((1-.99)/2, 0 ,1) = 2.576.To build the interval by hand, We can be 99% certain that the actual population mean is somewhere between 10.575 and 14.225.  

TO SOLVE ON THE CALCULATOR 1) Enter the 50 data points into L1 on your calculator (STAT Edit). 2) STAT Calc 1 to find the sample standard deviation (we can use this because we have 50 data points; ). s = 5.01 (don’t need mean or n; calculator handles those for us). 3) STAT TESTS 7 (Z-Interval) 4) Select Data, since you have the data entered into the calculator. 5) Enter 5.01 as the standard deviation, and .99 as the C-Level (level of confidence). 6) Select Calculate to get the interval. 

Notice that we get the EXACT same interval we just got doing it by hand!! We can be 99% sure that the actual population mean is between 10.575 and 14.225. Notice that the calculator also tells us that the mean of the data we entered is 12.4, that the standard deviation of the data is 5.01 and that n is 50.

If we need to know what E is, simply find the distance between the interval endpoints and divide by 2. (14.225 – 10.575)/2 = 1.825. We could also simply find the distance between the sample mean and the endpoint(s) of the interval to find E.14.225 – 12.4 = 1.825 and 12.4 – 10.575 = 1.825Again, the same answer we got doing it by hand.

Look at Example 5 on page 315.   n = 20, = 22.9, σ = 1.5, and c = 90%NOTICE – Even though n < 30, we were given σ, so we can still run the z-TestThe question asks for a 90% confidence interval for the population mean.To find E using the formula, , plug in what we know. is found by 2nd VARS 3 ((1-.90)/2, 0 ,1) = 1.645. To build the interval by hand,  

Look at Example 5 on page 315.   n = 20, = 22.9, σ = 1.5, and c = 90%NOTE: as a general rule, round the interval to the same number of digits as the data you are given. Since the mean and standard deviation were given to you to one decimal point, round the interval to one decimal point.We can be 90% certain that the actual population mean is somewhere between 22.3 and 23.5. 

Look at Example 5 on page 315.   n = 20, = 22.9, σ = 1.5, and c = 90%On the calculator, STAT TESTS 7, select Stats (since you are going to provide the stats instead of the actual data points). Enter 1.5, 22.9, 20, and .9 and then calculate. 

Look at Example 5 on page 315. Again, as a general rule, we round our interval endpoints to the same number of decimal points as the data that is given to us. We were given 22.9 and 1.5, which have one decimal, so we should round our interval to one decimal place. SAME ANSWER we got doing it by hand! 

CALCULATING MINIMUM SAMPLE SIZE   How do you know how many experiments or trials are needed in order to achieve the desired level of confidence for a given margin of error? We take the formula for finding E and solve it for n. becomes . Remember, if you don’t know what σ is, you can use s, as long as . 

CALCULATING MINIMUM SAMPLE SIZE Example 6 on page 316-   c = 95%, (from Example 1), E = 1 (given). is found by going to 2nd VARS 3 and entering (1-.95)/2 for the area. Remember to use the positive of the value given to you. = 1.96  . If you want to be 95% certain that the true population mean lies within the interval created with an E of 1, you need AT LEAST 97 magazine advertisements in your sample. We round up, since 96 advertisements are not quite enough. REMEMBER!! Sample size is ALWAYS a WHOLE NUMBER!! You can’t do part of an experiment; you either get a data point or you don’t.  

Section 6-2 – Confidence Intervals for the Mean (Small Samples) Estimating Population Parameters   The t-distribution: What do we do if we don’t know the population standard deviation, and can’t find a sample size of 30 or more? If the random variable is normally distributed (or approximately normally distributed), you can use a t- distribution. t-distribution formula: Critical values of t are denoted as , just as critical z values are called . 

Is n ≥ 30? YES Use the normal distribution (STAT – TESTS – 7)NOIs the population normally, or approximately normally, distributed?NOYou can NOT use the normal or the t-distribution.YES Is σ known? NO Use the t- distribution (STAT – TESTS – 8) YES Use the normal distribution (STAT – TESTS – 7) How do we know when to use the t -distribution instead of the normal distribution? This flow chart comes straight from page 329 in your text book.

Properties of the t-distribution (Page 325)   1) The t-distribution is bell-shaped and symmetrical about the mean. 2) Degrees of freedom are equal to n-1. 3) The total area under the curve is 1, or 100%. 4) The mean, median, and mode of the t-distribution are equal to zero. 5) As the degrees of freedom increase, the t-distribution approaches the normal distribution. After 30 d.f., the t-distribution is very close to the standard normal z- distribution. Close enough, in fact, that we use the standard normal distribution for d.f. ≥ 29. (or n ≥ 30) We just did that in Section 6-1.

The TI-84 will do the t-distribution interval for you!! In the interest of time, we are not going to cover doing these by hand.   STAT TESTS 8 (T-Interval) Same as with the z-interval, if you have the data points, select data. If you have the statistics, select stats. Enter the appropriate numbers and select Calculate.

EXAMPLE 1 (Page 326) Find the critical value for a 95% confidence level when the sample size is 15. To do this, go to 2 nd VARS 4 (invT) instead of invNorm. The area is (this is the same formula we used to find the area for in Section 6-1). If n = 15, then the degrees of freedom are (n – 1) = 14 2nd VARS 4, (1-.95)/2, 14, Enter gives us a of -2.145, but we will use the positive of that, just like we did with the .  

EXAMPLE 2 (Page 327) You randomly select 16 coffee shops and measure the temperature of coffee sold at each. The sample mean temperature is 162.0 0 F with a sample standard deviation of 10.0 0 F. Find the 95% confidence interval for the mean temperature. Assume the temperatures are approximately normally distributed. We MUST use the t-distribution for this; the sample size is less than 30 and we don’t know what σ is, but we do know that the distribution is approximately normal.

EXAMPLE 2 (Page 327) You randomly select 16 coffee shops and measure the temperature of coffee sold at each. The sample mean temperature is 162.0 0 F with a sample standard deviation of 10.00 F. Find the 95% confidence interval for the mean temperature. Assume the temperatures are approximately normally distributed. STAT TESTS 8, select Stats, and enter the values for , s, n, and c. 

EXAMPLE 2 (Page 327) You randomly select 16 coffee shops and measure the temperature of coffee sold at each. The sample mean temperature is 162.0 0 F with a sample standard deviation of 10.00 F. Find the 95% confidence interval for the mean temperature. Assume the temperatures are approximately normally distributed. The 95% confidence interval is from 156.7 to 167.3. We round to one decimal because we were given and the standard deviation rounded to one decimal. SO, we are 95% confident that the actual population mean of coffee temperature in ALL coffee shops is between 156.70 F and 167.30 F. 

EXAMPLE 3 (Page 328) You randomly select 20 mortgage institutions and determine the current mortgage interest rate at each. The sample mean rate is 6.22%, with a sample standard deviation of 0.42%. Find the 99% confidence interval for the population mean mortgage interest rate. Assume the interest rates are approximately normally distributed. We MUST use the t-distribution for this; the sample size is less than 30 and we don’t know what σ is, but we do know that the distribution is approximately normal. STAT TESTS 8, select Stats, and enter the values for , s, n, and c. 

EXAMPLE 3 (Page 328) You randomly select 20 mortgage institutions and determine the current mortgage interest rate at each. The sample mean rate is 6.22%, with a sample standard deviation of 0.42%. Find the 99% confidence interval for the population mean mortgage interest rate. Assume the interest rates are approximately normally distributed. The 99% confidence interval is from 5.95 to 6.49. We round to two decimals because we were given rounded to two decimals. SO, we are 99% confident that the actual population mean mortgage interest rate for ALL mortgage institutions is between 5.95% and 6.49%. 

Now, find the 90% and 95% confidence intervals for the population mean mortgage interest rate. What happens to the widths of the intervals as the confidence levels change? 90% interval is 6.06 to 6.38 95% interval is 6.02 to 6.42 99% interval is 5.95 to 6.49 The more confident we need to be, the wider the interval must become. Put another way, the higher c is, the wider the interval will be.

EXAMPLE 4 (Page 329) 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 population mean construction costs? Explain your reasoning. Although n is less than 30, we can still use the normal distribution because we know that the distribution is normal and we know what σ is.

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 with a sample standard deviation of 2.5 beats per minute. Assuming 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. Because n < 30, the distribution is normal, and we do not know what σ is, we should use the t-distribution on this one.

ASSIGNMENTS Classwork: Pages 317-318; #2-34 Evens Classwork: Page 330; #1-16 All Homework: Pages 318-323; #35-40 All, #45-67 Odd Homework: Pages 331-332; #17-28 All