Chapter 8 – Statistical Inference: Confidence IntervalsUnderstanding how sample statistics vary gives us a jumping point for drawing inference about the population from them. We can get a probability about our sample using the sampling distribution and use that to generate an interval or make a decision (Chapter 9) about the state of the population.Thought Exercise: Flip a coin 10 times. You get H 8 times. What is the true probability of getting heads with that coin? Can you claim the coin is fair? How many times do you have to flipthe coin and how many Hs do you have to get before you start to doubt its fairness?Definition: a point estimate is a single number from the sample that is our best guess for the population parameterDefinition: an interval estimate is a range of values that may or may not contain the parameterGood point estimators (such as ^p and ´x) have several desirable properties:1. they are unbiased (the mean of their sampling distribution is the parameter they estimate2. they have a smaller standard error compared to other estimatorsWe use these nice point estimates to help create a confidence interval, which is based on the point estimate and its standard error.For confidence intervals about population means and proportions, the form of a confidence interval is:point estimate ± (critical value)(standard error)which is the same aspoint estimate ± margin of errorHow do we find the critical value? That depends on the sampling distribution of the point estimate. Recall the Central Limit Theorem, which allows us to use the standard normal distribution for sufficiently large samples when estimating p and µ.For a population proportion, if np and n(1-p) are both at least 15 we can use z-scores to find the critical values, given a specific (pre-determined) confidence level.For example, if we want a 95% confidence interval (which is common), this corresponds to z-scores that bound the middle 95% of the standard normal distribution.Example: You have a sample of 100 Skittles and 23 of them are red. Find a 95% CI for the true proportion of red Skittles.x = 23; n = 100; α = 0.05^p=xn=23100=0.23s . e .{^p}=√^p(1−^p)n=√(.23) (.77)/100=0.042 (we substitute in the sample proportion for the standard error because we don’t know the population proportion)A 95% CI for p: ^p± zα2√^p(1−^p)n→.23 ± 1.96(0.042)→ .23± 0.0825A 95% CI for p: 95% (0.1475, 0.3125)To interpret this, remember the parameter is a fixed target: “There is 95% confidence that the interval (0.1475, 0.3125) contains the true proportion of red Skittles candies.” z = -1.96 z = 1.96We do something similar to get a confidence interval for the population mean. If the standard deviation of the population is known, a CI for µ is: ´x ± zα2σ /√nThis rarely happens in practice. If you don’t know the mean of a population, you almost definitely don’t know the standard deviation. In that case, we estimate σ with s, but then we now longer can use z critical values. We have to use the T distributions.tν=(´x −μ)s√n This is a lot like the z scores we calculated before, but because s is a function of the sample mean´x, the numerator and denominator are no longer independent. This is reflected by the parameter υ (nu), the degrees of freedom of the T distribution, which tells us how many independent pieces of information we have available. By using the mean to calculate the standard deviation, we lose one piece of information, so we lose one degree of freedom.Therefore, υ = n – 1 in this case.To use the T distribution, the population distribution (that is, the distribution of X, our random variable), must be approximately normal.Example: You have a sample of 10 female huskies, with an average weight of 41.5 lbs and a standard deviation of .6 lbs Assuming the population standard deviation of weights is unknown, construct and interpret a 90% CI for the true average weight of female huskies.pe ± cv(se)´x ± tν,α2(s√n)41.5 ± 1.833(.6√10)41.5 ± 0.348 90% CI for µ: (41.152, 41.848)lbsWe are 90% confident that the interval (41.152, 41,848)lbs contains the true average weight for female