CH89-Assignment

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Statistics

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Apr 3, 2024

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Module 9 (Chapters 8 and 9) Assignment Problem 1 Quality Ratings of Airports. The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maximum possible rating is 10. Suppose a simple random sample of 50 business travelers is selected and each traveler is asked to provide a rating for the Miami International Airport. The ratings obtained from the sample of 50 business travelers follow. Develop a 95% confidence interval estimate of the population mean rating for Miami. The interval estimate for the population mean rating in Miami is (5.74, 6.93). To calculate this interval estimate, you'll need to determine the sample size and the mean of the provided data, which consists of 50 business travelers. The mean can be obtained by averaging the ratings of these travelers. Subsequently, the standard deviation for the ratings should be calculated using either the Excel formula =STDEV or the formula √(∑(x−¯x) ( x − x ¯ ) 2 /n). Given that the confidence coefficient is 0.95, as stated in the problem, the level of significance is 1 - 0.95. With this information, you can determine the margin of error. This margin of error is then added and subtracted from the standard deviation to establish the interval estimate. http://tavana.us/rutgers/assignments/CH89-Assignment-Problem1-Data.xlsx 6 4 6 8 7 7 6 3 3 8 10 4 8 7 8 7 5 9 5 8 4 3 8 5 5 4 4 4 8 4 5 6 2 5 9 9 8 4 8 9 9 5 9 7 8 3 10 8 9 6 Problem 2 Telemedicine. Health insurers are beginning to offer telemedicine services online that replace the common office visit. Wellpoint provides a video service that allows subscribers to connect with a physician online and receive prescribed treatments (Bloomberg Businessweek). Wellpoint claims that users of its LiveHealth Online service saved a significant amount of money on a typical visit. The data shown below ($), for a sample of 20 online doctor visits, are consistent with the savings per visit reported by Wellpoint. Assuming the population is roughly symmetric, construct a 95% confidence interval for the mean savings for a televisit to the doctor as opposed to an office visit. The interval estimate for the mean savings from a visit to the doctor is (60.54, 81.46). To calculate the interval estimate, you begin by determining the sample size, which in this case is 20, and finding the average of the data. The standard deviation for the savings should be calculated using either the Excel formula =STDEV or the formula √(∑(x−¯x) ( x − x ¯ ) 2 /n). The confidence coefficient is 0.95, as specified in the problem. The level of significance is calculated as 1 minus the confidence coefficient. As this is a two-tailed test, you divide the level of significance by 2. To determine the margin of error, the degrees of freedom need to be calculated, which is derived by subtracting 1 from the sample size. This value is then used to find the T-value for the problem. In this particular case, the margin of error can be found by multiplying the T-value by the sum of the standard deviation divided by the square root of the sample size. Finally, this margin of error is added and subtracted from the standard deviation to establish the interval estimate. http://tavana.us/rutgers/assignments/CH89-Assignment-Problem2-Data.xlsx
92 105 56 76 93 78 53 34 83 49 48 74 93 82 40 55 40 96 73 100 Problem 3 Federal Tax Returns. According to the IRS, individuals filing federal income tax returns prior to March 31 received an average refund of $1056 in 2018.Consider the population of “last-minute” filers who mail their tax return during the last five days of the income tax period (typically April 10 to April 15). a. A researcher suggests that a reason individuals wait until the last five days is that on average these individuals receive lower refunds than do early filers. Develop appropriate hypotheses such that rejection of H 0 will support the researcher’s contention. The Null Hypothesis always represents the status quo. In this context, the Null Hypothesis states that the number of individuals filing federal income tax is equal to the average refund of $1056. The Alternative Hypothesis, on the other hand, serves as the opposite of the Null Hypothesis. In this case, the Alternative Hypothesis suggests that the number of individuals filing federal income tax is less than the average refund of $1056. b. For a sample of 400 individuals who filed a tax return between April 10 and 15, the sample mean refund was $910. Based on prior experience a population standard deviation of σ =$1600 may be assumed. What is the p -value? The P-value is 0.034000515. To calculate the P-value, the first step is to determine the Z-test value. The formula to find the Z-test value is (Mean – Null) / (Standard Deviation divided by the Square Root of the sample size). Once the Z-value is obtained, the next step involves using the standard normal distribution to find the corresponding P-value. c. At α=.05, what is your conclusion? With an alpha of 0.05, the P-Value is less than 0.05 (0.034 < 0.05). Therefore, we should reject the null hypothesis. If the p-value were greater than 0.05, it would imply that everyone received a refund of $1056, but this was not the case. The researcher's assertion that individuals wait until the last five days due to expecting lower refunds seems to be accurate. d. Repeat the preceding hypothesis test using the critical value approach. Applying the critical value approach, we utilize the normal standard inverse formula to derive the z- value at 0.05, which equals -1.644853627. Next, we compare this critical value to our obtained Z-value of -1.825. Observing that -1.644853627 < -1.825, we conclude that the null hypothesis should still be rejected. Problem 4 Cost of Telephone Surveys. Fowle Marketing Research, Inc., bases charges to a client on the assumption that telephone surveys can be completed in a mean time of 15 minutes or less. If a longer mean survey time is necessary, a premium rate is charged. A sample of 35 surveys provided the survey times shown in
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