396 Chapter 8 Confidence Intervals Once you have determined which type of parameter you are estimating, proceed as follows: ∙ Population mean: There are two methods for constructing a confidence interval for a population mean, the z method (Section 8.1) and the t method (Section 8.2). To determine which method to use, we must determine whether the population standard deviation is known, whether the population is approximately normal, and whether the sample size is large (n > 30). The following diagram can help you make the correct choice. Is σ known? Yes No Use z method Yes (Section 8.1) Use z method (Section 8.1) Is the population approximately normal? Consult a Is the population approximately normal? statistician Use t method (Section 8.2) Consult a statistician Yes Yes No Yes Is n > 30? Is n > 30? Use t method (Section 8.2) No No No ∙ Population proportion: To construct a confidence interval for a population proportion, use the method described in Section 8.3. ∙ Population standard deviation or variance: To construct a confidence interval for a population standard deviation or variance, use the method described in Section 8.4. EXAMPLE 8.21 Determining which method to use A random sample of 41 pumpkins harvested from a pumpkin patch has a mean weight of ̄x = 8.53 pounds with a sample standard deviation of s = 1.32 pounds. Construct a 95% confidence interval for the mean weight of pumpkins from this patch. Determine the type of parameter that is to be estimated and construct the confidence interval. Solution We are asked to find a confidence interval for the mean weight; this is a population mean. We consult the diagram to determine the correct method. We must first determine whether �� is known. There is no information given about ��, so �� is unknown. We follow the ‘‘No’’ path. Next we must determine whether n > 30. There are 41 pumpkins in the sample, so n > 30. We follow the ‘‘Yes’’ path, and find that we should use the t method described in Section 8.2. To construct the confidence interval, there are 41 − 1 = 40 degrees of freedom. Because the confidence level is 95%, the critical value is t0.025 = 2.021. The 95% confidence interval is 8.53 ± 2.021 1.32 √ 41 8.11 < �� < 8.95
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