Find a 99% confidence interval for A given that 105, o = 24, n = 18
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A: ACCORDING TO GIVEN QUESTION WE HAVE ;- 98% confidence interval for widget width is 14.3<μ <…
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Q: A random sample of n1 = 16 communities in western Kansas gave the following information for people…
A: Given: A random sample of n1 = 16 communities in western Kansas gave the following information for…
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Q: Give a 99% confidence interval, for µ₁ −µ2 given the following information. n1 n₁ = 50, x1 = 2.03,…
A: Given: x¯1 = 2.03s1 = 0.4n1 = 50x¯2 = 2.52s2 = 0.98n2 = 45
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- A consumer magazine wants to compare the lifetimes of ballpoint pens of three different types. The magazine takes a random sample of pens of each type in the following table. Brand 1 Brand 2 Brand 3 260 181 238 218 240 257 184 162 241 219 218 213 Test the claim that there is a difference in mean lifetime for the three brands of ballpoint pens, using α=0.01. Round you answer to 3 decimal places.15 people were asked how many hours they spend on homework a week. Data Set: 4,5,6,7,9,11,12,12,14,15,17,19,20,21,23 n = 15 x bar = 13.6428571 s = 5.78601953 With 93% confidence, the amount of hours that are typically spent working on homework per week is between ---- hours and ---- hours. Need : t= SE= ME=Random samples of 324 are taken from a large population and studied. It was found that x = 7.41. If 22.96% of all sample means were greater than 288.6834. What is μ? μ =
- A student was asked to find a 99% confidence interval for widget width using data from a random sample of size n = 24. Which of the following is a correct interpretation of the interval 12 to 33.6? The mean width of all widgets is between 12 and 33.6, 99% of the time. We know this is true because the mean of our sample is between 12 and 33.6. We are 99% confident that the mean width of a randomly selected widget will be between 12 and 33.6. There is a 99% chance that the mean of a sample of 24 widgets will be between 12 and 33.6. We are 99% confident that the mean width of all widgets is between 12 and 33.6. Which is true?6.2.13 Use the given confidence interval to find the margin of error and the sample mean. (14.2,24.0) The sample mean is (Type an integer or a decimal.)A random sample of n1 = 16 communities in western Kansas gave the following information for people under 25 years of age. x1: Rate of hay fever per 1000 population for people under 25 97 91 122 130 94 123 112 93 125 95 125 117 97 122 127 88 A random sample of n2 = 14 regions in western Kansas gave the following information for people over 50 years old. x2: Rate of hay fever per 1000 population for people over 50 94 111 100 95 110 88 110 79 115 100 89 114 85 96 What is the value of the sample test statistic? (Test the difference μ1 − μ2. Round your answer to three decimal places.)
- A sample of 46 observations is taken from a normal population with a standard deviation of 15. The sample mean is 57. Determine the 99% confidence interval for the population mean 38 39 40 4k / Walk jate # 1 ។ 4 5 4 Round to three decimals Co De M 2 Which Will MAfind a 95% confidence interval for the true proportion of grapes that are used for raisins if a sample of 340 bins of grapes 25.1% are used for raisins. We are 95% confident the true proportion of grapes used for raisins is between ____ and ———Which one of the following statements is true? Select one: O a. In large samples, there are not many discrepancies between the outcomes of the F test and the LM test O b. Degrees of freedom of the unrestricted model is necessary for using the LM test. O c. The LM test can be used to test hypotheses with single restrictions only and provides inefficient results for multiple restrictions O d. The LM statistic is derived based on the normality assumption
- A random sample of 67 lab rats are enticed to run through a maze, and a 95% confidence interval is constructed of the mean time it takes rats to do it. It is [2.3min, 3.1 min]. Which of the following statements is/are true? (More than one statement may be correct.) 95% of the lab rats in the sample ran the maze in between 2.3 and 3.1 minutes. 95% of the lab rats in the population would run the maze in between 2.3 and 3.1 minutes. Using the sample for these rats, about 95% of the time, the interval [2.3, 3.1] would contain the population mean. The true population mean is not known.A random sample of n1 = 16 communities in western Kansas gave the following information for people under 25 years of age. x1: Rate of hay fever per 1000 population for people under 25 100 92 122 127 93 123 112 93 125 95 125 117 97 122 127 88 A random sample of n2 = 14 regions in western Kansas gave the following information for people over 50 years old. x2: Rate of hay fever per 1000 population for people over 50 93 112 100 97 111 88 110 79 115 100 89 114 85 96 State the null and alternate hypotheses. H0: ?1 = ?2; H1: ?1 ≠ ?2 H0: ?1 > ?2; H1: ?1 = ?2 H0: ?1 = ?2; H1: ?1 > ?2 H0: ?1 = ?2; H1: ?1 < ?2 What sampling distribution will you use? What assumptions are you making? The standard normal. We assume that both population distributions are approximately normal with known standard deviations. The Student's t. We assume that both population distributions are approximately normal with known standard deviations. The Student's t. We assume that…6.05 Based on a random sample of size n = 40, the 95% confidence interval for the true mean weight in mg for beans of a certain type is (229.7,233.5). Obtain the 99% confidence interval.
