A bottled water distributor wants to estimate the amount of water contained in 1-gallon bottles purchased from a nationally known water bottling company. The water bottling company's specications state that the standard deviation of the amount of water is equal to 0.03 gallon. A random sample of 50 bottles is selected, and the sample mean amount of water per 1-gallon bottle is 0.968 gallon. Complete parts (a) through (d). 3. Construct a 95% condence interval estimate for the population mean amount of water included in a 1-gallon bottle. 0.95968 5 p5 0.97632 (Round to ve decimal places as needed.) b. 0n the basis of these results, do you think that the distributor has a right to complain to the water bottling company? Why? |::l because a 1-gallon bottle containing exactly 1-gallon of water lies E the 95% condence interval. outside within A bottled water distributor wants to estimate the amount of water contained in 1-gallon bottles purchased from a nationally known water bottling company. The water bottling company's specications state that the standard deviation of the amount of water is equal to 0.03 gallon. A random sample of 50 bottles is selected, and the sample mean amount of water per 1-gallon bottle is 0.968 gallon. Complete parts (a) through (d). a. Construct a 95% condence interval estimate for the population mean amount of water included in a 1-gallon bottle. 0.95968 5 p5 0.97632 (Round to ve decimal places as needed.) b. 0n the basis of these results, do you think that the distributor has a right to complain to the water bottling company? Why? B because a 1-gallon bottle containing exactly 1-gallon of water lies |::l the 95% condence interval. NO, Yes, If X = 81, S = 23, and n = 81, and assuming that the population is normally distributed, construct a 90% condence interval estimate of the population mean, p. Click here to view p_age 1 of the table of critical values for the t distribution. Click here to view page 2 of the table of critical values for the t distribution. E> DsusD (Round to two decimal places as needed.) Assuming that the population is normally distributed, construct a 99% confidence interval for the population mean for each of the samples below. Explain why these two samples produce different confidence intervals even though they have the same mean and range. Sample A: 1 3 3 4 5 6 6 8 Full data set Sample B: 1 2 3 4 5 6 7 8 . . . Construct a 99% confidence interval for the population mean for sample A. 1.77 Sus 7.23 (Type integers or decimals rounded to two decimal places as needed.) Construct a 99% confidence interval for the population mean for sample B. 1.47 Sus 7.53 (Type integers or decimals rounded to two decimal places as needed.) Explain why these two samples produce different confidence intervals even though they have the same mean and range. O A. The samples produce different confidence intervals because their critical values are different. O B. The samples produce different confidence intervals because their standard deviations are different. O C. The samples produce different confidence intervals because their medians are different. O D. The samples produce different confidence intervals because their sample sizes are different.A survey of nonprofit organizations showed that online fundraising has increased in the past year. Based on a random sample of 50 nonprofits, the mean one-time gift donation in the past year was $29, with a standard deviation of $7. Complete parts a and b below. . . . a. Construct a 99% confidence interval estimate for the population one-time gift donation. Sus (Type integers or decimals rounded to two decimal places as needed.)The table below contains the amount that a sample of nine customers spent for lunch ($) at a fast-food restaurant. Complete parts a and b below. 4.79 5.26 5.73 6.29 7.15 7.55 8.23 8.57 9.19 . . . a. Construct a 95% confidence interval estimate for the population mean amount spent for lunch at a fast-food restaurant, assuming a normal distribution. The % confidence interval estimate is from $ 5.78 to $ 8.16 . (Round to two decimal places as needed.) b. Interpret the interval constructed in (a). Choose the correct answer below. O A. We have 95% confidence that the population mean amount in dollars spent for lunch at the fast-food restaurant is contained in the interval. O B. The mean amounts in dollars spent for lunch at the fast-food restaurant of 95% of all samples of the same size are contained in the interval. O C. We have 95% confidence that the mean amount in dollars spent for lunch at the fast-food restaurant for the sample is contained in the interval. O D. 95% of the sample data fall between the limits of this confidence interval.Determine the upper-tail critical value to /2 in each of the following circumstances. a. 1 - a = 0.95, n = 38 d. 1 - a = 0.95, n = 17 b. 1 - a = 0.90, n = 38 e. 1 - a = 0.99, n = 62 c. 1 - a = 0.95, n = 54 . . . a. t= (Round to four decimal places as needed.) b. t= (Round to four decimal places as needed.) c. t= (Round to four decimal places as needed.) d. t= (Round to four decimal places as needed.) e. t= (Round to four decimal places as needed.)A pharmaceutical company operates retail pharmacies in 10 eastern states. Recently, the company's internal audit department selected a random sample of n= 300 prescriptions issued throughout the system. The objective of the sampling was to estimate the average dollar value of all prescriptions issued by the company. The data collected were x= $13.48 and s = 2.50. Complete parts a and b below. <:> a. The 90% condence interval estimate for the true average sales value for prescriptions issued by the company is from $|:| to $D. (Round to the nearest cent- 2 decimal places. Use ascending order.) You are asked to interpret the meaning of this condence interval by choosing the correct answer below: 0 A. The company believes with 90% condence that the sample mean prescription amount is between these two amounts. 0 B. The company believes with 90% condence that the true mean prescription amount is between these two amounts. 0 C. There is a 0.90 probability that the true mean prescription amount is between these two values. 0 D. The company believes that the true mean prescription amount falls between these two values 90% of the time. A pharmaceutical company operates retail pharmacies in 10 eastern states. Recently, the company's internal audit department selected a random sample of n= 300 prescriptions issued throughout the system. The objective of the sampling was to estimate the average dollar value of all prescriptions issued by the company. The data collected were x= $13.48 and s = 2.50. Complete parts a and b below. 0 A. The company believes with 90% condence that the sample mean prescription amount is between these two amounts. 0 B. The company believes with 90% condence that the true mean prescription amount is between these two amounts. 0 C. There is a 0.90 probability that the true mean prescription amount is between these two values. 0 D. The company believes that the true mean prescription amount falls between these two values 90% of the time. b. One of its retail outlets recently reported that it had monthly revenue of $7,144 from 535 prescriptions. Are such results to be expected? Should that retail outlet be audited? (Round to the nearest cent as needed.) When the population mean is at the upper limit of the 90% condence interval computed in part a, the upper limit of the 90% condence interval for the expected total monthly revenue for 535 prescriptions would be $|:|. When the population mean is at the lower limit of the 90% condence interval computed in part a, the lower limit of the 90% condence interval for the expected total monthly revenue for 535 prescriptions would be $|:|. Since this outlet reported sales of $7,144 from 535 prescriptions, there is |:1l reason to believe that this is out of line. The retail outlet E] be audited. Use your calculator to nd critical values such that the following statements are true. Your task is to determine the critical values below. a. P( - \According to recent study, 162 of 486 technology CEOs from around the world responded that technological advances will transform their business and 76 responded that resource scarcity and climate change will transform their business. Complete parts (a) through (c) below. <:> a. Construct a 90% condence interval estimate for the population proportion of tech CEOs who indicate technological advances as one of the global trends that will transform their business. 0.2982 511:5 0.3685 (Round to four decimal places as needed.) b. Construct a 90% condence interval estimate for the population proportion of tech CEOs who indicate resource scarcity and climate change as one of the global trends that will transform their business. DSusD (Round to four decimal places as needed.) You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 40 home theater systems has a mean price of $110.00. Assume the population standard deviation is $17.40. . . . Construct a 90% confidence interval for the population mean. The 90% confidence interval is (,). (Round to two decimal places as needed.) Construct a 95% confidence interval for the population mean. The 95% confidence interval is (,). (Round to two decimal places as needed.) Interpret the results. Choose the correct answer below. O A. With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%. O B. With 90% confidence, it can be said that the sample mean price lies in the first interval. With 95% confidence, it can be said that the sample mean price lies in the second interval. The 95% confidence interval is wider than the 90%. O C. With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is narrower than the 90%You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 40 business days, the mean closing price of a certain stock was $111.13. Assume the population standard deviation is $10.67. . . . The 90% confidence interval is (Round to two decimal places as needed.) The 95% confidence interval is 7). (Round to two decimal places as needed.) Which interval is wider? Choose the correct answer below. O The 90% confidence interval O The 95% confidence intervalYou are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 40 business days, the mean closing price of a certain stock was $111.13. Assume the population standard deviation is $10.67. . . . The 95% confidence interval is (],). (Round to two decimal places as needed.) Which interval is wider? Choose the correct answer below. The 90% confidence interval O The 95% confidence interval Interpret the results. O A. You can be 90% confident that the population mean price of the stock is between the bounds of the 90% confidence interval, and 95% confident for the 95% interval. O B. You can be 90% confident that the population mean price of the stock is outside the bounds of the 90% confidence interval, and 95% confident for the 95% interval. O C. You can be certain that the population mean price of the stock is either between the lower bounds of the 90% and 95% confidence intervals or the upper bounds of the 90% and 95% confidence intervals. O D. You can be certain that the closing price of the stock was within the 90% confidence interval for approximately 36 of the 40 days, and was within the 95% confidence interval for approximately 38 of the 40 days