ELEMENTARY STATISTICS 3E William Navidi and Barry Monk ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. Confidence Intervals for a Population Mean, 𝜎 Unknown Section 8.2 ©McGraw-Hill Education. Objectives 1. Describe the properties of the Student’s 𝑡 distribution 2. Construct confidence intervals for a population mean when the population standard deviation is unknown ©McGraw-Hill Education. Objective 1 Describe the properties of the Student’s 𝑡 distribution ©McGraw-Hill Education. Student’s 𝑡 Distribution When constructing a confidence interval where we know the population 𝜎 standard deviation 𝜎, the confidence interval is 𝑥ҧ ± 𝑧𝛼/2 ∙ . The critical 𝑛 value is 𝑧𝛼/2 because the quantity 𝑥ഥ −𝜇 𝜎Τ 𝑛 has a normal distribution. It is rare that we would know the value of 𝜎 while needing to estimate the value of 𝜇. In practice, it is more common that 𝜎 is unknown. When we don’t know the value of 𝜎, we may replace it with the sample standard deviation 𝑠. However, we cannot then use 𝑧𝛼Τ2 as the critical value, because the quantity 𝑥ഥ −𝜇 𝑠Τ 𝑛 does not have a normal distribution. The distribution of this quantity is called the Student’s 𝒕 distribution. There are actually many different Student’s 𝑡 distributions and they are distinguished by a quantity called the degrees of freedom. When using the Student’s 𝑡 distribution to construct a confidence interval for a population mean, the number of degrees of freedom is 1 less than the sample size. ©McGraw-Hill Education. Student’s 𝑡 Distribution and the Standard Normal Student’s 𝑡 distributions are symmetric and unimodal, just like the normal distribution. However, they are more spread out. The reason is that 𝑠 is, on the average, a bit smaller than 𝜎. Also, since 𝑠 is random, whereas 𝜎 is constant, replacing 𝜎 with 𝑠 increases the spread. When the number of degrees of freedom is small, the tendency to be more spread out is more pronounced. When the number of degrees of freedom is large, 𝑠 tends to be close to 𝜎, so the 𝑡 distribution is very close to the normal distribution. ©McGraw-Hill Education. The Critical Value 𝑡𝛼Τ2 To find the critical value for a confidence interval, let 1 − 𝛼 be the confidence level. The critical value is then 𝑡𝛼Τ2 , because the area under the Student’s t distribution between −𝑡𝛼Τ2 and 𝑡𝛼Τ2 is 1−𝛼. The critical value 𝑡𝛼Τ2 can be found using a calculator. Press 2nd VARS then invT. 𝑖𝑛𝑣𝑇(𝑎𝑟𝑒𝑎 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡, 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚). ***Must be able to find critical t values on the calculator using 𝑖𝑛𝑣𝑇, a table is not allowed on the final*** ©McGraw-Hill Education. Example: Critical Value A simple random sample of size 10 is drawn from a normal population. Find the critical value 𝑡𝛼Τ2 for a 95% confidence interval. Solution: The sample size is 𝑛 = 10, so the number of degrees of freedom is 𝑛 – 1 = 9. We input the following into the calculator 𝑖𝑛𝑣𝑇(0.975, 9)). The critical value is 𝑡𝛼Τ2 = 2.262. [The table contains a limited amount of degrees of freedom so knowing how to use the calculator is important.] ©McGraw-Hill Education. Assumptions The assumptions for constructing a confidence interval for 𝜇 when 𝜎 is unknown are as follows. • We have a simple random sample. • The sample size is large (𝑛 > 30), or the population is approximately normal. When the sample size is small (𝑛 ≤ 30), we must check to determine whether the sample comes from a population that is approximately normal. A simple method is to draw a dotplot or boxplot of the sample. If there are no outliers, and if the sample is not strongly skewed, then it is reasonable to assume the population is approximately normal and it is appropriate to construct a confidence interval using the Student’s 𝑡 distribution. ©McGraw-Hill Education. Objective 2* Construct confidence intervals for a population mean when the population standard deviation is unknown *(By Hand) ***Must be able to calculate by Hand and Calculator*** ©McGraw-Hill Education. Confidence Interval When 𝜎 is Unknown If the assumptions are satisfied, the confidence interval for 𝜇 when 𝜎 is unknown is found using the following steps: Step 1: Compute the sample mean 𝑥ഥ and sample standard deviation, 𝑠, if they are not given. Step 2: Find the number of degrees of freedom 𝑛 – 1 and the critical value 𝑡𝛼Τ2 . Step 3: Compute the standard error 𝑠Τ 𝑛 and multiply it by the 𝑠 critical value to obtain the margin of error 𝑡𝛼Τ2 ∙ . 𝑛 Step 4: Construct the confidence interval: 𝑥ഥ ± 𝑡𝛼Τ2 ∙ Step 5: Interpret the result. ©McGraw-Hill Education. 𝑠 . 𝑛 Example 1: Confidence Interval A food chemist analyzed the calorie content for a popular type of chocolate cookie. Following are the numbers of calories in a sample of eight cookies. 113, 114, 111, 116, 115, 120, 118, 116 Find a 98% confidence interval for the mean number of calories in this type of cookie. Solution: We check the assumptions. We have a simple random sample. Because the sample size is small, the population must be approximately normal. A dotplot indicates that there is no evidence of strong skewness and no outliers, therefore we may proceed. Now, we find the sample mean and sample standard deviation. We have 𝑥ҧ = 115.375 and 𝑠 = 2.8253. ©McGraw-Hill Education. Example 1: Confidence Interval (Cont) The number of degrees of freedom is 𝑛 − 1 = 8 − 1 = 7. Using a calculator, 𝑖𝑛𝑣𝑇(0.01, 7), we find that the critical value corresponding to a level of 98% is 𝑡𝛼/2 = 2.998. The margin of error is: 𝑡𝛼/2 ∙ 𝑠 𝑛 = 2.998 ∙ The 98% confidence interval is: 𝑥ഥ ± 𝑡𝛼Τ2 ∙ 2.8253 8 𝑠 𝑛 Remember: 𝑥 = 115.375 𝑠 = 2.8253 𝑛=8 = 2.9947. = 115.375 ± 2.9947 or 112.4 < 𝜇 < 118.4 We are 98% confident that the mean number of calories per cookie is between 112.4 and 118.4. ©McGraw-Hill Education. Example 2: Confidence Interval A sample of 123 people aged 18–22 reported the number of hours they spent on the Internet in an average week. The sample mean was 8.20 hours, with a sample standard deviation of 9.84 hours. Assume this is a simple random sample from the population of people aged 18–22 in the U.S. Construct a 95% confidence interval for 𝜇, the population mean number of hours per week spent on the Internet by people aged 18–22 in the U.S. Solution: We have a simple random sample and the sample size is large. We may proceed. Note that 𝑥ҧ = 8.20 and 𝑠 = 9.84. The number of degrees of freedom is 𝑛 − 1 = 123 − 1 = 122. Using a calculator 𝑖𝑛𝑣𝑇(0.025, 122), the critical value corresponding to a level of 95% is 𝑡𝛼Τ2 = 1.980. ©McGraw-Hill Education. Example 2: Confidence Interval (Cont) Margin of error: 𝑡𝛼/2 ∙ 𝑠 𝑛 = 1.980 ∙ 9.84 123 The 95% confidence interval: 𝑥ഥ ± 𝑡𝛼Τ2 ∙ = 1.7567 𝑠 𝑛 Remember: 𝑥 = 8.20 𝑠 = 9.84 𝑛 = 123 𝑡𝛼/2 = 1.984 = 8.20 ± 1.7567 or 6.44 < 𝜇 < 9.96 We are 95% confident that the mean number of hours per week spent on the Internet by people 18 – 22 years old is between 6.44 and 9.96. ©McGraw-Hill Education. Objective 2** Construct confidence intervals for a population mean when the population standard deviation is unknown **(TI-84 PLUS) ***Must be able to calculate by Hand and Calculator*** ©McGraw-Hill Education. Confidence Intervals on the TI-84 PLUS The TInterval command constructs confidence intervals when the population standard deviation 𝜎 is unknown. This command is accessed by pressing STAT and highlighting the TESTS menu. If the summary statistics are given the Stats option should be selected for the input option. If the raw sample data are given, the Data option should be selected. ©McGraw-Hill Education. Example 1: Confidence Interval (TI-84 PLUS) A food chemist analyzed the calorie content for a popular type of chocolate cookie. Following are the numbers of calories in a sample of eight cookies. 113, 114, 111, 116, 115, 120, 118, 116 Find a 98% confidence interval for the mean number of calories in this type of cookie. Solution: We check the assumptions. We have a simple random sample. Because the sample size is small, the population must be approximately normal. A dotplot indicates that there is no evidence of strong skewness and no outliers, therefore we may proceed. ©McGraw-Hill Education. Example 1: Confidence Interval (TI-84 PLUS Continued) We enter the data into list L1 in the data editor. Then we press STAT and highlight the TESTS menu and select TInterval. Select Data as the input method and enter L1 in the List field, 1 in the Freq field, and 0.98 for the C-Level field. Select Calculate. The confidence interval is (112.4, 118.4). We are 98% confident that the mean number of calories per cookie is between 112.4 and 118.4. ©McGraw-Hill Education. Example 2: Confidence Interval (TI-84 PLUS) A sample of 123 people aged 18–22 reported the number of hours they spent on the Internet in an average week. The sample mean was 8.20 hours, with a sample standard deviation of 9.84 hours. Assume this is a simple random sample from the population of people aged 18–22 in the U.S. Construct a 95% confidence interval for 𝜇, the population mean number of hours per week spent on the Internet by people aged 18–22 in the U.S. Solution: We begin by checking the assumptions. We have a simple random sample and the sample size is large (greater than 30). The assumptions are satisfied. ©McGraw-Hill Education. Example 2: Confidence Interval (TI-84 PLUS Continued) We press STAT and highlight the TESTS menu and select Tinterval. Select Stats as the input method and enter 8.2 for ഥ field, 9.84 for the 𝒔 field, 123 for the 𝒏 field, the 𝒙 and 0.95 for the C-Level field. Now, select Calculate. The confidence interval is (6.4436, 9.9564). We are 95% confident that the mean number of hours per week spent on the Internet by people 18 – 22 years old is between 6.44 and 9.96. ©McGraw-Hill Education. Remember: 𝑥 = 8.20 𝑠 = 9.84 𝑛 = 123 You Should Know . . . • The properties of the Student’s 𝑡 distribution • Why we must determine whether the sample comes from a population that is approximately normal when the sample size is small (𝑛 ≤ 30) • How to construct and interpret confidence intervals for a population mean when the population standard deviation is unknown (By Hand and Calculator) ©McGraw-Hill Education. 
Purchase answer to see full attachment