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- 1. #5.48 ACT scores of high school seniors. The scores of high school seniors on the ACT college entrance examination in 2003 h (a) What is the approximate probability that a single student randomly chosen from all those taking the test scores (b) Now taken as SRS of 25 students who took the test. What are the mean and standard deviation of the sample m (c) What is the approximate probability that the mean score of these students is 23 or higher? (d) Which of your two Normal probability calculations in (a) and (c) is more accurate? Why? Solutions: Given that, µ=20.8 and SD δ=4.8 (a) Required Prob: P( X >= 23 ) P( X >= 23 ) = 1 - P( X < 23 ) P( X < 23 ) = 0.6766 (by using excel normdist() function, click on the respective value for how it is P( X >= 23 ) = 1 - P( X < 23 ) = 1-0.6766 = 0.3234 (b) For this problem lets recollect that By the properties of means and variances of random variables, the mean and variance of the sample mean are the Note: Here x-bar also refer as M Now, when n=25 Expected value of M = 20.8 Standard deviation of M = σ / sqrt(n) = 4.8 / sqrt(25) = 4.8 / 5 = 0.96 ( c) Required Prob: P( sample mean(M) >= 23 ) P( M >= 23 ) = 1 - P( M < 23 ) P( M < 23 ) = 0.9890 (by using eMcel normdist() function, click on the respective value for how it is P( M >= 23 ) = 1 - P( M < 23 ) = 1-0.9890 = 0.0110 (d) Normal probability calculation in (c) is more accurate, because it is capturing the current sample's variability, whereas (a) captures overall population's variability
- 2. entrance examination in 2003 had mean µ=20.8 and SD δ=4.8. The distribution of scores is only roughly Normal. m all those taking the test scores 23 or higher? ndard deviation of the sample means score x of these 25 students? he respective value for how it is calculated) ance of the sample mean are the following: the respective value for how it is calculated) urrent sample's variability,
- 3. #5.52 A Lottery payoff. A $1 bet in a state lottery’s Pick 3 game pays $500 if the three-digit number you choose exactly matc the winning number, which is drawn at random. Here is the distribution of the payoff X: Payoff X $0 $500 Probability 1 0 Each day’s drawing is independent of other drawings. (a) What are the mean and SD of x? (b) Joe buys a Pick 3 ticket twice a week. What does the law of large numbers say about the average payoff Joe rece (c) What does the central limit theorem say about the distribution of Joe’s average payoff after 104 bets in a year? (d) Joe comes out ahead for the year if his average payoff is greater than $1 (the amount he spent each day on a tic What is the probability that Joe ends the year ahead? Solutions: (a) x $0 $500 Imp note: Generally there should be loss factor, here it sh P(x) 1 0 but considering as it is to avoide the confusion mean = Σxp(x) = 0*0.999 + 500*0.001 = 0.5 SD = sqrt ( V(x) ) V(x) = Σ x*x*p(x) - (mean*mean) = 0*0*0.999 + 500*500*0.001 - 0.5*0.5 = 249.75 SD = sqrt ( 249.75 ) = 15.803 (b) From the law of large numbers, the average payoff joe receives from his bets will be close population's average pa (c) The central limit theorem (CLT) states conditions under which the mean of a sufficiently large number of independent random variables, each with finite mean and variance, will be approximately normally (d) Required prob: P( X > 1) As here, n = 104 Mean of sample mean = 0.5 SD of sample mean = σ / sqrt(n) = 15.803 / sqrt(104) = 1.5496 therefore X follows normal dist with mean = 0.5 and SD=1.5496 P(X>1) = 1 - P(X<=1) = = 1 - 0.63 = 0.37
- 4. ber you choose exactly matches t the average payoff Joe receives from his bets? yoff after 104 bets in a year? nt he spent each day on a ticket). uld be loss factor, here it should be -1 in place of 0 t is to avoide the confusion lose population's average payoff will be approximately normally distributed
- 5. #5.60 Advertisements and brand image. Many companies place advertisements to improve the image of their brand rather t (a) There are 28 students in each group. Although individual scores are discrete, the mean score for a group of 28 w (b) What are the means and SD of the sample mean scores ӯ for the Journal group and ̅ for the Enquirer group? (c) We can take all 56 scores to be independent because students are not told each other’s scores. What is the dist between the mean scores in the two groups? (d) Find P (ӯ - x-bar ≥1 ). Solutions: (a) Because the central limit theorem (CLT) states conditions under which the mean of a sufficiently large number of independent random variables, each with finite mean and variance, will be approximately normally So based on this CLT,we could say that mean score for a group of 28 will be close to Normal (b) Given that, Journal's mean 4.8 and SD 1.5, and Enquirer's mean 2.4 and SD 1.6 here sample number, n=28 By the properties of means and variances of random variables, the mean and variance of the sample mean are the Now, for jounral when n=28 Expected value of ӯ of journal = 4.8 Standard deviation of ӯ = σ / sqrt(n) = 1.5 / sqrt(28) = 4.8 / 50.283 = Now, for Enquirer when n=28 Expected value of M of Enquirer = 2.4 Note: Here x-bar refers as M Standard deviation of M = σ / sqrt(n) = 1.6 / sqrt(28) = 4.8 / 5 = 0.302 (c) The distribution of y-bar - x-bar would be normal based central limit theorem and normal dist properties From CLT,we already know that y-bar and x-bar will be normal and form normal dist properties, if x1 and x2 follow normal dist then x1+x2 and x1-x2 also follows normal dist (d) for required prob, first we will need to calculate mean and SD of y-bar - x-bar
- 6. using (b) and (c ), y-bar mean = 4.8 and var = square of 0.283 = 0.0804 x-bar mean = 2.4 and var = square of 0.302 = 0.0914 y-bar - x-bar follows normal dist with mean = 4.8-2.4 = 2.4 and var = 0.0804 + 0.0914 = SD = sqrt(0.1718) = P (ӯ - x-bar ≥1 ) = 1 - P (ӯ - x-bar<1) = 1 - 0.0004 = 0.9996
- 7. he image of their brand rather than to promote specific products. In a randomized comparative experiment, business students read ads th e mean score for a group of 28 will be close to Normal. Why? other’s scores. What is the distribution of the difference ӯ- a sufficiently e, will be approximately normally distributed ance of the sample mean are the following: Note: Here x-bar refers as M ormal dist properties st properties,
- 8. 4 and var = 0.0804 + 0.0914 = 0.1718 0.4145
- 9. iment, business students read ads that cited either the Wall Street Journal or the National Enquirer for important facts about a fictitious c
- 10. r important facts about a fictitious company. The students then rated the trustworthiness of the source on a 7-point scale. Suppose that i
- 11. ce on a 7-point scale. Suppose that in the population of all students’ scores for the Journal have mean 4.8 and SD 1.5, while scores for the
- 12. 4.8 and SD 1.5, while scores for the Enquirer have mean 2.4 and SD 1.6
- 13. #6.18 Mean OC in young women. Refer to the previous exercises. A biomarket for bone formations measured in the same study was osteocalcin (OC), measured in the blood. The units are nanograms per milliliter (ng/ml). For the 31 subjects in the study the mean was 33.4 ng/ml. Assume that the SD is known to be 19.6 ng/ml. report the 95% conf Confidence Interval Estimate for the Mean Data Population Standard Deviation 19.6 Sample Mean 33.4 Sample Size 31 Confidence Level 95% Intermediate Calculations Standard Error of the Mean 3.5203 Note: Click on corresponding cell, to know how it is calculate Z Value -1.9600 Interval Half Width 6.8996 Confidence Interval Interval Lower Limit 26.5004 Interval Upper Limit 40.2996
- 14. mations measured milliliter (ng/ml). 19.6 ng/ml. report the 95% confidence interval. cell, to know how it is calculated
- 15. #6.32 Accuracy of a laboratory scale. To assess the accuracy of a laboratory scale, a standard weight known to weigh 10 grams is weighed repeatedly. The scale readings are Normally distributed with unknown mean (this mean is 10 grams if the scale has no bias.) the SD of the scale readings is known to be 0 (a) The weight is measured five times. The main result is 10.0023 grams. Give a 98% confidence interval for the mean of rep (b) How many measurements must be averaged to get a margin of error of ±0.0001 with 98% confidence? (a) Confidence Interval Estimate for the Mean Data Population Standard Deviation 0 Sample Mean 10 Sample Size 5 Confidence Level 98% Intermediate Calculations Standard Error of the Mean 0.0001 Note: Click on corresponding cell, to know how it is calculat Z Value -2.3263 Interval Half Width 0.0002 Confidence Interval Interval Lower Limit 10.0021 Interval Upper Limit 10.0025 (b) Sample Size Determination Data Population Standard Deviation 0 Sampling Error 0 Confidence Level 98% Intemediate Calculations Z Value -2.33 Note: Click on corresponding cell, to know how it is calculat Calculated Sample Size 21.65 Result Sample Size Needed 22
- 16. scale readings is known to be 0.0002 gram. ce interval for the mean of repeated measurements of the weight. confidence? cell, to know how it is calculated cell, to know how it is calculated
- 17. #6.58 A two-sided test and the confidence interval. The P-value for a two-sided test of the null hypothesis (a) Does the 95% confidence interval include the value 30? Why? (b) Does the 90% confidence interval include the value 30? Why? Solutions: (a) No, it does not include the value 30 at 95% confidence interval Because generally, one rejects the null hypothesis if the p-value is smaller than or equal to the significance level Here, P-value 0.04 < 0.05 (significance level), so it is rejecting the null hypothesis. Means, it doesn't inlcude 30 (b) No, it does not include the value 30 at 90% confidence interval Because generally, one rejects the null hypothesis if the p-value is smaller than or equal to the significance level Here, P-value 0.04 < 0.1 (significance level), so it is rejecting the null hypothesis. Means, it doesn't inlcude 30
- 18. he null hypothesis H0:µ=30 is 0.04. n or equal to the significance level n or equal to the significance level
- 19. #6.66 Are the pine trees randomly distributed north to south? In example 6.1 we looked at the distribution of longleaf pine One way to formulate hypotheses about whether or not the trees are randomly distributed in the tract is to examine the averag north-south direction. The values range from 0 to 200, so if the trees are uniformly distributed in this direction, any difference fr values (100) should be due to chance variation. The sample means for the 584 trees in the tract is 99.74. A theoretical calcula assumption that the trees are uniformly distributed gives a SD of 58. Carefully state the null and alternative hypotheses in term Note that this requires that you translate the research question about the random distribution of the trees into specific statemen of a probability distribution. Test your hypotheses, report your results, and write a short summary of what you have found. Solution: Null hypothesis: H0 : The pine trees are randomly distributed north to south (µ = 100) Alternative hypothesis H1 : The pine trees are not randomly distributed north to south (µ <> 100) Z Test of Hypothesis for the Mean Data Null Hypothesis µ= 100 Level of Significance 0.05 Population Standard Deviation 58 Sample Size 584 Sample Mean 99.74 Intermediate Calculations Standard Error of the Mean 2.4 Z Test Statistic -0.11 Two-Tail Test Lower Critical Value -1.96 Upper Critical Value 1.96 p-Value 0.91 Do not reject the null hypothesis As p-value is grater than 0.05 (alpha), we do not reject the null hypothesis. Means, The pine trees are randomly distrib
- 20. the distribution of longleaf pine trees in the Wade Tract. the tract is to examine the average location in the d in this direction, any difference from the middle ract is 99.74. A theoretical calculation based on the and alternative hypotheses in terms of this variable. of the trees into specific statements about the mean mary of what you have found. pine trees are randomly distributed north to south
- 21. 6.68 Who is the author? Statistics can help decide the authorship of literary works. Sonnets by a certain Elizabethan poet are known to contain an average of µ=8.9 new words (words are not used in the poet’s other works). The SD of the number of new word is δ=2.5. Now a manuscript with debating whether it is the poet’s work. The n =10.2 words not used in the poet’s known works. We expect poems by another author to contain more new words, so to see H0:µ = 8.9 Ha: µ > 8.9 Give the z test statistics and its P-value. What do you conclude about the authorship of the new poems? Solution: Z Test of Hypothesis for the Mean Data Null Hypothesis µ= 8.9 Level of Significance 0.05 Population Standard Deviation 2.5 Sample Size 1 Sample Mean 10.2 Intermediate Calculations Standard Error of the Mean 2.5 Z Test Statistic 0.52 Upper-Tail Test Upper Critical Value 1.6449 p-Value 0.3015 Do not reject the null hypothesis As we are not rejecting the null hypothesis, it is saying that µ=8.9, so the authorship of the new poems is Elizabethan
- 22. s (words are not used in the poet’s whether it is the poet’s work. The new sonnets contain an average of contain more new words, so to see if we have evidence that the new sonnets are not by our poet we test new poems? of the new poems is Elizabethan poet

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