Chapter TwelveInference about a Population668Overview• In Chapter 10 we studied how to construct confidence intervals for the population mean when the variance was assumed to be known.• Here we study the more realistic case where we do not know the true variance and we want to construct a confidence interval for the population mean. • We also study how to construct a confidence interval for the variance itself.669Inference about the Population Mean when the Variance is Unknown• In Chapter 10, when the variance is known, using the Central Limit Theorem we noted that was approximately distributed as a Standard Normal random variable (this would hold exactly if X is Normal).• From here , we concluded that the following was approximately true://where /where Z is a Standard Normal.670• From here, we concluded that a (1‐)% confidence interval for can be constructed as://• Now suppose is unknown.• We can replace it with the sample standard deviation 671• And now we use the statistic• The key question is: What is the sampling distribution of this statistic?• RESULT: If X is Normally distributed, then the statistic t has a Student t distribution with n‐1 degrees of freedom.• We studied the t distribution before. This is why we did it…672• Thus, if X is Normal, then ,,where ⁄,where T has a t‐distribution with n‐1 degrees of freedom.• From here, we conclude that a (1‐)% confidence interval for can be constructed as:,,673• REMARK: When we studied the properties of the t‐distribution, we stated that when the number of degrees of freedom is large, the t‐distribution becomes closer and closer to the Standard Normal distribution. • Thus, if n is large (that is, if we have a large sample size), then ⁄,/and it does not really matter numerically if we use the Standard Normal tables or the t‐distribution tables. • In other words, as the sample size n becomes larger, the statistic becomes approximately distributed as a Standard Normal. 674• What if X is NOT Normally distributed?• As long as the distribution of X is “not too far awa y from being Normal”, we can still construct a (1‐)% confidence interval for as:,,• Even if the distribution of X is very “far awa y from being Normal” the Central Limit Theorem provides us reassurance that the above confidence interval will become an increasingly good approximation for relatively large sample sizes n. 675• Thus, whether or not X is Normally distributed, whenever is unknown, we will construct a (1‐)% confidence interval for as:,,keeping in mind that, if X is not Normally distributed, the accuracy of this confidence interval will depend on how large the sample size n is. 676Hypothesis Testing the Population Mean when the Variance is Unknown• We proceed as we described in Chapter 10, replacing the z‐statistic ∗with the t‐statistic ∗• And we replace Zα and Zα/2with tα,n‐1and tα/2,n‐1677• Analogously to Chapter 11, for a given significance level : • We reject H0: µ = µ* in favor of H1: µ > µ* ift > tα,n‐1• We reject H0: µ = µ* in favor of H1: µ < µ* ift <‐ tα,n‐1• We reject H0: µ = µ* in favor of H1: µ ≠ µ* if|t| > tα/2,n‐1678• The p‐values are also obtained as in Chapter 11…• Let T be a t‐random variable with n‐1 degrees of freedom and ‘t’ be the value obtained for our test‐statistic in the data observed. Then: • If H0: µ = µ* vs. H1: µ > µ*. p‐value = • If H0: µ = µ* vs. H1: µ < µ*. p‐value = • If H0: µ = µ* vs. H1: µ ≠ µ*. p‐value679• Example: A statistician randomly sampled 1,500 observations and found and . Test at the 5% significance level: against • The first step, once again, it to remember that it suffices to test against • We reject H0: µX= 15 in favor of H1: µX> 15 ift > t0.05,n‐1• OK, so first let us find tα,n‐1 in Table 4 in Appendix B…680• In the problem, we have n=1,500. So the number of degrees of freedom we need to look up is n‐1=1,499. • The largest number of degrees of freedom in Table 4 is 200. Anything larger than that is considered “infinity”, meaning that the distribution is already practically equal to a Standard Normal. • We have t0.05,n‐1 = 1.645. • Next step is to compute the test‐statistic… 681• We have∗• Thus, we have t < t0.05,n‐1(since ‐1.54 is smaller than 1.645). Therefore, we fail to reject the null hypothesis H0.682• What is the p‐value of this test? •
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