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Unformatted text preview: Introduction Week Nine. Goals for this week. 1. Go over assignment due this week. 2. Chapter Nine Notes 3. HS Exercise #6.8, p. 267 Stat 601, B. D. McCullough, Fall 2009 HS Ch. 7, #23 All three variables have a significant effect and there is a significant interaction between mortgage size and credit history. Mortgage size has no effect on approval time when credit history is good but significantly increases the approval times when the credit history is bad. A sample of 5 mortgages is minimally adequate. Since you are selecting from a database, data acquisition should be inexpensive and you should be able to get a larger sample quite easily. A sample of 2030 would provide a good estimate of the average approval times. Stat 601, B. D. McCullough, Fall 2009 HS Ch. 7, #23 A sample of at least 50 and preferably 100200 are needed to give you a good estimate of the shape of the distribution of approval times. When data are plentiful and inexpensive you should take advantage of the situation. Another important response would be percent of loans approved and how it varies as a function of these three factors. Data collected without the aid of a statistical design can have many problems such as those discussed in Chapter 6. You should carefully assess the quality and pedigree of the data obtained from such a database. Stat 601, B. D. McCullough, Fall 2009 HS Ch 8 #4 If the race is close, then we can assume that the proportion of votes each candidate will receive will be about .5. So let us calculate a confidence interval for the proportion using a sample of size 1000 and a sample proportion of .5. The formula for the 95% confidence interval for a proportion (p. 497) is p ± 1 . 96 radicalbigg p (1 p ) n recalling the n is the sample size and p estimates π . We get a confidence interval of about .47 to .53, or +/ .03. Stat 601, B. D. McCullough, Fall 2009 HS Ch 8 #4 If we desire a confidence interval with a width of +/ .01, we would use the sample size formula for an infinite population, (see top of page 363) assuming a proportion of .5, which gives a required sample size of 9604. All of the above is based on a 95% confidence level. n = 4(1 . 96 2 ) π (1 π ) W 2 = 4(1 . 96 2 )0 . 5(1 . 5) . 02 2 = 9604 Stat 601, B. D. McCullough, Fall 2009 HS ch. 8, #7 see page 363 for the relevant formula. Plugging in, we find ± 5 : . 95 σ to 1 . 05 σ W / 2 σ = 0 . 05 ⇒ n = 807 . 9 ± 30 : . 70 σ to 1 . 30 σ W / 2 σ = 0 . 30 ⇒ n = 28 . 9 ± 50 : . 50 σ to 1 . 50 σ W / 2 σ = 0 . 50 ⇒ n = 12 . 7 Stat 601, B. D. McCullough, Fall 2009 HS ch. 8, #8 First we calculate: ¯ x = 136 . 5 , s = 287 . 7 n = 30. 95% CI for average overrun over a long time period, assuming it was stable Relevant formula: Confidence Interval for the Average page 337 ¯ x ± ts / √ n What is ” t ”? df = n1 so t = 2 . 05 (Appendix I page 511) and √ 30 = 5 . 5 136 . 5 ± 2 . 05 × 287 . 7 / 5 . 5 136 . 5 ± 2 . 05 × (52 . 3) 136 . 5 ± 107 . 2 [ 243 . 7 , 29 . 3] Stat 601, B. D. McCullough, Fall 2009 HS Ch 8 #8...
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This note was uploaded on 05/14/2010 for the course FINANCE 32159 taught by Professor Adamyore during the Fall '09 term at Drexel.
 Fall '09
 ADAMYORE
 Finance

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