Question

Chpt 9 Q&A 815 xls - Compatibility Mode OFF BESU AutoSave Review View Data Formulas Draw Insert Page Layout Home ICH F Number 10 ~ A A 8 Wrap Text 6.99 Conditional Format Formatting as Table y $ % ) Merge & Center a. Av Paste X Cut Times New Roman LA Copy B IU Format X ✓ fx M O N L M60 K G F 1 H E D C A B Similar to those in the text. Assignment Chapter 9 Exercise 2: 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 A sample of 71 observations is taken from a normal population with a standard deviation of 5. The sample mean is 40. Determine the 95 percent confidence interval for the population mean. 40=1 Exercise 4: Suppose you want an 80 percent confidence level. What z value would you use? 7.3% 7.99 &$% 40 Exercise 8: 42 43 Prof. Participle is a professor of English. Recently she counted the number of misspelled words in a group of student essays. She noted the distribution of misspelled words per essay followed the normal distribution with a standard deviation of 2.44 words per essay. For her 10 A.M. section of 40 students, the mean numer of misspelled words was 6.05. Construct a 90 percent confidence interval for the mean number of misspelled worlds in the population of student essays. 6.05 Exercise 12: 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 The American Sugar Producers Associatin wants to estimate the mean yearly sugar consumption. A sample of 16 people reveals the mean yearly consumption to be 55 pounds with a standard deviation of 20 pounds. a. What is the best estimate of the population mean? c. For a 98 percent confidence interval, what is the value of r? Exercise 12 Ipart 2): 55+ d. Develop the 95 percent confidence interval for the populatin mean. e. Would it be reasonable to conclude that the population mean is 67 pounds? 60 61 62 63 64 65 66 Sheet1 Sheet2 Shao13 + Select destination and press ENTER or choose Paste A esc B IV A @ # 3 $ 4. % 5 & 7 1 2 6

Answer 1

$\\ \\ Answer:\ \ \ \ 40\pm 1.163$

We need to construct the 95% confidence interval for the population mean u. The following information is provided: Sample Mean à = 40 Population Standard Deviation (0) = 5 Sample Size (N) = 71 The critical value for a = 0.05 is Zc = 21-a/2 1.96. The corresponding confidence interval is computed as shown below: o о CI X - 2c X X + Zc x 9 ✓n 40 5 1.96 V71 5 40 + 1.96 V71 (38.837, 41.163) Therefore, based on the data provided, the 95% confidence interval for the population mean is 38.837 < p < 41.163, which indicates that we are 95% confident that the true population mean u is contained by the interval (38.837, 41.163).