We considered the differences between the temperature readings in January 1 of 1968 and 2008 at 51 locations in the continental US in Exercise 5.19. The mean and standard deviation of the reported differences are 1.1 degrees and 4.9 degrees respectively.(a) Calculate a 90% confidence interval for the average difference between the temperature measurements between 1968 and 2008.lower bound: degrees(please round to two decimal places)upper bound: degrees(please round to two decimal places)
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9, the sample size (n) is 50, and the critical t-value for a 90% confidence interval with 50 degrees of freedom is approximately 1.676606. Margin of error = t(α/2) * (σ / √n) Margin of error = 1.676606 * (4.9 / √50) Margin of error ≈ 1.150 Show more…
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We considered the differences between the temperature readings in January 1 of 1968 and 2008 at 51 locations in the continental US in Exercise 5.19. The mean and standard deviation of the reported differences are 1.1 degrees and 4.9 degrees respectively.(a) Calculate a 90% confidence interval for the average difference between the temperature measurements between 1968 and 2008. lower bound: (please round to two decimal places) (answer is not -6.9) upper bound: (please round to two decimal places) (answer is not 9.2)
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The recorded high temperatures (in degrees Fahrenheit) in June for a sample of 17 US cities are listed below. Assuming the temperatures are normally distributed, find the 92% confidence interval (including units) of the population mean. Give the distribution used and the mean and standard deviation of the sample rounded to the nearest hundredth (two decimal places). Give all digits provided by the calculator for the confidence interval. 78 105 91 80 101 93 92 78 70 91 101 71 81 84 102 70 86 Distribution used: Confidence interval: Mean: Standard deviation:
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