1. Sample Means W~N(980, 1002) μ=980 σ / √n =10.010 P(W>1000)=0.0229 We only expect 2 samples in every 100 to be this big

2. Aims: To know what a confidence interval is and be able to establish a CI for the mean of a population.

3. Name: To know what a “confidence interval” for the mean is. Describe: How to find a ___% confidence interval for the mean of a population from a sample mean if the variance of the population is known. Explain: How this process must be modified in cases where the population variance is not known.

4. We know the means of sample size n from a population follow a Normal Distribution with a of N(μ, σ² / n ) Since it is often impossible to find the mean of the population (since the population cannot be accurately measured) we use a sample mean to estimate it. Chances are the sample mean will not be the exact mean so we illustrate this by giving an interval around the estimate in which we are fairly sure the mean lies.

5. This is what Confidence Intervals are about. Suppose we know that a population generally has a standard deviation 14.6 A sample of 20 values is taken... We want to use these to estimate the mean of the whole population. 178151628102726435 22491165024461455

6. We know the standard deviation is 14.6 We know that 22.5 is approximately the mean (as this is the mean of our sample). We know that this value is somewhere in the X distribution with mean μ and standard deviation 14.6 / √20 We use this to decide our interval.

7. What we do now is choose a confidence bound (Common ones are 90%, 95% and 99%) to state how confident we are that the mean is in the interval. Note: More confidence=Larger Spread of Values

8. So our sample mean is 22.5 and the Standard error is 14.6 / √20 = 3.265 This diagram helps show how confidence intervals work. The Normal curves represent the potential distribution s of sample means and the blue line the mean of our actual sample that could be bigger than or smaller than the actual mean in the middle of the curves.

9. Lets do a 95% confidence interval. So for 95% we want to leave out 2.5% either side. The number of standard deviations above/below can be found by checking 0.975 in the percentage points table which gives 1.96 So we use the mean of 22.5 and standard error of 0.8660 to form the interval 22.5- 3.265x1.96 ≤ μ ≤ 22.5+ 3.265x1.96 16.101 ≤ μ ≤ 28.899 or (16.101, 28.899) We can be 95% confident the mean is between these values (as 95% of the time it should) {NB the mean of the data I sampled was 25.5}

10. 1) You need a sample mean and the population variance/standard deviation (or an unbiased estimate for this) 2) Establish the Standard Error (Divide the population standard deviation by root n) 3) Calculate the multiplier for your confidence interval. From percentage points table. E.g. 95%  1.96 (0.975 as we want 2.5% either side) 4) Calculate the Confidence Interval (Sample Mean)-(MultiplierxStandard Error)≤μ≤ (Sample Mean)+(MultiplierxStandard Error)

11. Sample Mean 88.4 Standard Error 11 / √5 = 4.919 99%  Multiplier = 2.5758 88.4-2.5758x4.919 ≤ μ ≤ 88.4-2.5758x4.919 75.729 ≤ μ ≤ 101.071 Often written (75.7,101.1)