Illowsky Homework 5

STAT 200Roman RomeroWeek 7 HomeworkIllowsky et al.70. The standard deviation of the chi-square distribution is twice the mean.False102. Do men and women select different breakfasts? The breakfasts ordered by randomly selected men and women at apopular breakfast place is shown in Table 11.55. Conduct a test for homogeneity at a 5% level ofsignificance.French Toast Pancakes Waffles OmelettesMen 47 35 28 53Women 65 59 55 60Null Hypothesis: Distribution for breakfast is the same for men and women.Alternate Hypothesis: Distribution for breakfast is different for men and women.Df: 3Frequency table: E= (row total)(column total)/overall totalTest Stat.= 4.01 using ∑(O-E)2/ Ep-value= 0.260 (Using CHIDIST(4.012533,3)The null hypothesis would not be rejected since the p-value>a.Use the following information to answer the next twelve exercises: Suppose an airline claims thatits flights are consistentlyon time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is nomore than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes.113. df = ________25-1=24117. Let α = 0.05Decision: ________Do not reject the null hypothesis

Week # Text Chapter Exercise page 4 1 Lane et al. 7 8 269 4 2 Lane et al. 7 11 270 4 3 Lane et al. 7 12 270 4 4 Illowsky et al. 6 60 361 4 5 Illowsky et al. 6 66 362 4 6 Illowsky et al. 6 76 363 4 7 Illowsky et al. 6 88 365 4 8 Illowsky et al. 7 62 400 4 9 Illowsky et al. 7 70 401 4 10 Illowsky et al. 7 96 406 # 1 (Lane Chap 7 # 8 – Page 269) (a) We can use a standard normal distribution table or a free online normal distribution calculator (for example, Stat Trek) to get the probability. If we use the standard normal distribution table to get the probability, we have to find a z-score corresponding to 65 first, and then find the cumulative area to the left of the z-scores. (1) Find a Z score for 65 using the formula ± ²³´ µ ± ¶·³¸¹ º ± »¼½¾¿ . (2) Find the area below a Z of -0.75 = 0.2266 Therefore, 22.66% of the vehicles are less than or equal to the speed limit. (b) We want to find the probability P(X<50). (1) Find a Z score for 50 using the formula ± ²³´ µ ± ·À³¸¹ º ± »Á½ÂÁ¿ . (2) Find the area below a Z of -2.625 = 0.0043 Therefore, .43% of the vehicles would be going less than 50 mph. (c) The new speed limit would be the 90 th percentile of the distribution. Z-score of 1.28 has a cumulative area from the left as 0.8997, which is closest to 0.90, so we use Z-score of 1.28 as the approximation. The speed corresponding to a Z-score of 1.28 is ´ Ã Ä Å ± ¾Æ à ÇƽÁÈÉ Ä ÇÈÉ ± ÈƽÁÊ The new speed limit is 81.24 mph. (d) The actual distribution of speeds has a longer tail extending in the positive direction, so it is positively skewed, or skewed to the right. # 2 (Lane Chap 7 # 11 - Page 270) (a) The lowest score in the top 30% is the 70 th percentile of the score distribution. Z-score of 0.52 has a cumulative area from the left as 0.6985, which is closest to 0.70, so we use 0.52 as the approximation. Therefore, the score corresponding to a Z-score of 0.52 is ´ Ã Ä Å ± Á¿ à Ǽ½¿ÁÉ Ä ÇÊÉ ± Á¾½¼È (b) The lowest score in the top 5% is the 95 th percentile of the score distribution. Z-score of 1.645 has a cumulative area from the left as 0.95. Hence the score corresponding to a Z-score of 1.645 is ´ Ã Ä Å ± Á¿ à ÇƽÂÊ¿É Ä ÇÊÉ ± Ëƽ¿È # 3 (Lane Chap 7 # 12 - Page 270) N = 25 and π = 0.5 for the binomial distribution, which has a mean of μ = Nπ = ( 25)(0.5) = 12.5 and a variance of σ 2 =

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