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Unformatted text preview: MATH 2300 026 Homework #7 Name___________________________________ Due on 10th November 1. Assume that 20% of students at a university wear contact lenses. We randomly pick 200 students. Would it be unusual to obtain a sample proportion of 22%? Answer by calculating the appropriate z score. gG¡¢£¤¥¦G§ ¡¨G¡G¨¥¦G§ , ¡ = 0 .20 ©¤ª¡£« ¡¨G¡G¨¥¦G§ , ¡̂¬ ¦®«§ G¯©«¨®¤¥¦G§° = 0 .22 , § = 200 ± − ©²G¨« = ¡̂ − ¡ ³ ´ ¬µ ¶´ ° · = .22 − 0 .2 ³ ¸ .¹ ¬µ ¶¸ .¹ ° ¹¸¸ = 0 .7071 Zscore is within the scope of the distribution. (3 < Zscore < 3). Therefore it is not an unusual observation. A. No, z = 0.71 B. No, z = 25 C. Yes, z = 25 D. No, z = 0.68 E. Yes, z = 0.71 2. The closer p is to 0 or 1, the larger n must be in order for the sampling distribution of the sample proportion to A. have a standard error equal to º§¡¬1 − ¡° . B. be approximately normal. C. have a standard error equal to ³ ´¬µ¶´° · . D. be centered at ¡ . E. be approximately binomial. 3. Based on previous studies, researchers believe that 6% of children are born with a gene that is linked to a certain childhood disease. If the researchers test 950 newborns for the presence of this gene, would it be unlikely for them to find fewer than 25 children with the gene? Answer by calculating the appropriate z score. There are two methods to get the answer. Method I: Considering the distribution of X, which is binomial; ± − ©²G¨« = » − §¡ º §¡ ¬1 − ¡ ° = 25 − ¬950 ∗ 0 .06 ° º 950 ∗ 0 .06 ∗ ¬1 − 0 .06 ° = 25 − 57 √ 53 .58 = −4 .37 Method II: Considering the sampling distribution of sample proportion, ¼ ½ ; ¡̂ = 25 950 = 0.0263 ± − ©²G¨« = ¡̂ − ¡ ³ ´ ¬µ ¶´ ° · = .0263 − 0 .06 ³ ¸ .¸¾ ¬µ ¶¸ .¸¾ ° ¿À¸ = −4 .37 In both ways you will get the same zscore. Since Zscore is below 3, we can say that the sample proportion given is unusual. A. No, z = 4.37 B. No, z = 0.005 C. Yes, z = 6.59 D. No, z = 6.59 E. Yes, z = 4.37 4. Based on past experience, a bank believes that 8% of the people who receive loans will not make payments on time. The bank has recently approved 600 loans. Describe the sampling distribution model of the proportion of clients in this group who may not make timely payments. A. mean = 8%; standard error = 1.1% B. mean = 92%; standard error = 1.1% C. There is not enough information to describe the distribution. D. mean = 8%; standard error = 0.3% E. mean = 92%; standard error = 0.3% 5. A candy company claims that its jelly bean mix contains 21% blue jelly beans. Suppose that the candies are packaged at random in small bags containing about 400 jelly beans. Describe the sampling distribution model of the proportion of blue jelly beans in a bag....
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This note was uploaded on 12/12/2011 for the course STATS 2300 taught by Professor Abeysundara during the Fall '11 term at Texas Tech.
 Fall '11
 ABEYSUNDARA
 Statistics

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