1. Section 8.2 Estimating Population Means (Large Samples) HAWKES LEARNING SYSTEMS math courseware specialists Copyright © 2008 by Hawkes Learning Systems/Quant Systems, Inc. All rights reserved.

2. σ unknown, but a large sample (n ≥ 30) or population is normally distributed Large sample: Use z, the normal distrib’n. Small sample: Use the Student t-distribution. z t TODAY

3. HAWKES LEARNING SYSTEMS math courseware specialists Criteria for estimating the population mean for large samples: Confidence Intervals 8.2 Estimating Population Means (Large Samples) All possible samples of a given size have an equal probability of being chosen. The size of the sample is at least 30 (n ≥ 30). The population’s standard deviation is unknown. When all of the above conditions are met, then the distribution used to calculate the margin of error for the population mean is the Student t-distribution. However, when n ≥ 30, the critical values for the t-distribution are almost identical to the critical values for the normal distribution at corresponding levels of confidence. Therefore, we can use the normal distribution to approximate the t-distribution.

4. HAWKES LEARNING SYSTEMS math courseware specialists Find the critical value: Confidence Intervals 8.2 Estimating Population Means (Large Samples) Find the critical value for a 95% confidence interval. To find the critical value, we first need to find the values for –z 0.95 and z 0.95. Since 0.95 is the area between –z 0.95 and z 0.95, there will be 0.05 in the tails, or 0.025 in one tail. Solution:

5. HAWKES LEARNING SYSTEMS math courseware specialists Critical Value, z c : Confidence Intervals 8.2 Estimating Population Means (Large Samples) Critical z-Values for Confidence Intervals Level of Confidence, c zczc 0.801.28 0.851.44 0.901.645 0.951.96 0.982.33 0.992.575

6. HAWKES LEARNING SYSTEMS math courseware specialists Critical Value, z c : Confidence Intervals 8.2 Estimating Population Means (Large Samples) Critical z-Values for Confidence Intervals Level of Confidence, c zczc 0.801.28 0.851.44 0.901.645 0.951.96 0.982.33 0.992.575 TI-84 confirmation of these values using invNorm(area to left)= z c = area in the middle So area in each tail = (1 – c) / 2 (1 – 0.90) / 2 =.05 (1 – 0.95) / 2 =.025 (1 – 0.99) / 2 =.005 Take absolute value of –z. Round Responsibly. (added content by D.R.S.)

7. HAWKES LEARNING SYSTEMS math courseware specialists Margin of Error, E, for Large Samples: Confidence Intervals 8.2 Estimating Population Means (Large Samples) When calculating the margin of error, round to one more decimal place than the original data, or the same number of places as the standard deviation. z c  the critical z-value s  the sample standard deviation n  the sample size This E is the value that we will + or – from the sample mean to build a Confidence Interval

8. Find the margin of error for a 99% confidence interval, given a sample of size 100 with a sample standard deviation of 15.50. Find the margin of error: HAWKES LEARNING SYSTEMS math courseware specialists n  100, s  15.50, c  0.99 z 0.99  Solution: Confidence Intervals 8.2 Estimating Population Means (Large Samples) 2.575

9. A survey of 85 homeowners finds that they spend on average $67 a month on home maintenance with a standard deviation of $14. Find the 95% confidence interval for the mean amount spent on home maintenance by all homeowners. Construct a confidence interval: HAWKES LEARNING SYSTEMS math courseware specialists c  0.95, n  85, s  14,  67 z 0.95  Solution: Confidence Intervals 8.2 Estimating Population Means (Large Samples) 1.96 $64.02 <  < $69.98 ($64.02, $69.98) 67 – 2.98 <  < 67 + 2.98

10. A survey of 85 homeowners finds that they spend on average $67 a month on home maintenance with a standard deviation of $14. Find the 95% confidence interval for the mean amount spent on home maintenance by all homeowners. Construct a confidence interval: HAWKES LEARNING SYSTEMS math courseware specialists Confidence Intervals 8.2 Estimating Population Means (Large Samples) Large sample, so we choose z. STAT, TESTS, Zinterval Use “Data” if data is in a List. We have “Stats” already given. We don’t know σ so use s. Sample mean and sample size. Confidence Level as a decimal. Highlight Calculate. Press ENTER. TI-84 easy way (added content by D.R.S)

11. TI-84 Solution, continued Inputs (as described on the previous slide.) Outputs: “We are 95% confident that the mean monthly maintenance cost is $64.02 < μ < $69.98”

12. What’s the smallest sample size I need to use? Confidence Interval scenario: – I want to estimate the mean of a population – It’s going to take time and cost money to collect a sample. – If sample is too small, my result is not accurate enough., – If sample is too big, I’ve worked too hard and spent too much for precision I didn’t need. There’s a way to find it out in advance, for z. (added content by D.R.S.)

13. HAWKES LEARNING SYSTEMS math courseware specialists Finding the Minimum Sample Size for Means: Confidence Intervals 8.2 Estimating Population Means (Large Samples) When calculating the sample size, round to up to the next whole number. z c  the critical z-value  the population standard deviation E  the margin of error To find the minimum sample size necessary to estimate an average, use the following formula: This applies when you qualify to use z, not t. always bump it up Example: 80.01 bumps up to n = 81 sample size needed

14. Determine the minimum sample size needed if you wish to be 99% confident that the sample mean is within two units of the population mean, given that   6.5. Assume that the population is normally distributed. Find the minimum sample size: HAWKES LEARNING SYSTEMS math courseware specialists c  0.99,   6.5, E = 2 z 0.99  You will need a minimum sample size of 71. Solution: Confidence Intervals 8.2 Estimating Population Means (Large Samples) 2.575 You can find z c from printed tables or by using TI-84 invNorm(, as described earlier.

15. The electric cooperative wishes to know the average household usage of electricity by its non-commercial customers. They believe that the mean is 15.7 kWh per day for each family with a variance of 3.24 kWh. How large of a sample would be required in order to estimate the average number of kWh of electricity used daily per family at the 99% confidence level with an error of at most 0.12 kWh? Find the minimum sample size: HAWKES LEARNING SYSTEMS math courseware specialists c  0.99,   1.8, E = 0.12, z 0.99  2.575 You will need a minimum sample size of 1492 families. Solution: Confidence Intervals 8.2 Estimating Population Means (Large Samples)