Sample Final-solutions

Sample Final-solutions - 1. From a large sample, one finds...

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Unformatted text preview: 1. From a large sample, one finds the 95% confidence interval for to be (60.92, 89.78). a) Based on this information, determine the 90% confidence interval for . x SE moe x * 96 . 1 43 . 14 92 . 60 35 . 75 35 . 75 2 78 . 89 92 . 60 = =- = = + = (z* is used, because n is large) 3622 . 7 = x SE 90% conf interval for = ] 4609 . 87 , 2391 . 63 [ ) 3622 . 7 ( 645 . 1 35 . 75 * = = x SE z x b) The t-value with 5 degrees of freedom and a right-tail area of 0.25 is t=0.727. If the degrees of freedom get larger and larger (but keep the right-tail area equal to 0.25), the t-value gets close to 0.674 (write a numerical value). 2. A random sample of size n=20 from a normal population gives the sample mean x =140 and the sample standard deviation s = 8. a) Construct a 98% confidence interval for the population mean. 5419 . 4 20 / ) 8 ( 539 . 2 / * , 19 = = = = n s t moe df 98% conf interval for = ] 5419 . 144 , 4581 . 135 [ = moe x b) What is the length of this confidence interval (i.e. two times the margin of error)? What is its center? 140 0838 . 9 = = center length c) If you construct a 95% confidence interval, would the length of this interval be smaller or larger than the length you have found in b). smaller, because the confidence level is smaller....
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Sample Final-solutions - 1. From a large sample, one finds...

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