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## Guidelines to problems chapter 9

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**Guidelines to problems chapter 9**Nutan S. Mishra Department of mathematics and Statistics University of South Alabama**Important points to remember**• X is a random variable with parameters µ and σ. • In fact X represent the whole population (of some laaarge size) whose average is µ with standard deviation σ • µ and σ are parameters of the population which is represented by X • For a given population values of parameters are always fixed • For a given population values of parameters are often unknown. • So we collect a sample of size n from this population • In fact we could collect hundreds of such samples from the population. • And for each such sample we could compute the sample mean and sample standard deviation s • The value of sample mean is different for different samples • Thus we have a large group of sample means. • All the sample means form a new population represented by • This new population of sample means represented by has mean and standard deviation • The values of the population parameters of the new random variable are given by = µ and = σ/√n**Exercise 9.9**• H0: µ = 20 hours vs H1: µ ≠ 20 hours, this is a two tailed test. • H0: µ = 10 hours vs H1: µ ≠ 10 hours, this a two tailed test. • H0: µ = 3 years vs H1: µ ≠ 3 years, this is a two tailed test • H0: µ = $1000 vs H1: µ < $1000, this is a left tailed test. • H0: µ = 12 minutes vs H1: µ > 12 minutes, this is a right tailed test.**Exercise 9.17**Size of population = 8.1 million X= duration of unemployment for the people of 16 and over ( note that µ and σ are unknown) Size of sample n = 400 sample mean = 16.9 weeks sample s.d. = s=4.2 weeks (note that sample size is large) To test H0: µ = 16.3 weeks Vs H1: µ > 16.3 weeks Given that size of rejection region is α= .02 Note that this is a right tailed test. Since the sample size is large, we choose the z-distribution (standard normal) Thus the test statistic would be z = Since σ is unknown, we replace it by the sample standard deviation s = 4.2 weeks Thus z = = 2.86 and p-value = .0021 Since p-value < α , we make decision to reject H0 at 2% LOS And conclude that at 2% LOS the sample data does not support the H0 hence current duration of unemployment is greater than 16.3 weeks**Exercise 9.42**Population : households in United states. X= amount spent on gifts etc by a house hold during holiday season. In the year 2001 µ = $940 To test if the average amount spent this year is different from 2001. That is to test H0: µ = $940 vs H1 : µ ≠ $940 To test this hypothesis, they recently collected a sample Size of sample n = 324 households. = $1005, s = $330 Size of rejection region α = .01 Note that sample size is large , thus we choose z- distribution also test is two tailed. Thus our test statistic is z = Thus z= = 3.55 p-value = 2*.0002 = .0004 Since p-value << α , we make a decision to reject H0 And conclude that since sample data is not supporting null, the average expenditure by all households in United states on gifts etc, has increased since 2001.**Exercise 9.67**Population: female college basketball players. X= height of a player Assumption: x has normal distribution According to coach : µ = 69.5 inches To test this claim, a sample is collected with n = 25 players = 70.25 inches sample standard deviation = 2.1 inches α= .01 H0: µ = 69.5 inches vs H1: µ ≠ 69.5 inches Note that • Population is normal • Population standard deviation σ is unknown • Sample size is small Under these three conditions the test statistic follows t-distribution with n-1 d.f. Thus the test statistics t = follows t-distribution with 24 d.f. t = = 1.79 This is a two tailed test , p-value = 2* .05 (approximately) = .10 approximately Using flash apps on TI-89 exact p-value = .086 Since p-value >> α , we make a decision not to reject H0 at 1%LOS Conclude that at !% LOS sample data supports null that is average height is 69.5**Exercise 9.99**Population: Affluent Americans (annual income ≥ $75,000) X= # of households having serious problems in paying unexpected bill 0f $5000 According to “Money” Survey in 2002 , population proportion of such households p = .32 Question: has this proportion gone up since 2002? H0: p = .32 vs H1 : p > .32 In a recent survey they collected a sample of n = 1100 households with X = 396 that is sample proportion = 396/1100 = .36 To test the pair of hypotheses we have to choose a distribution first. Before choosing a distribution we need to decide if the sample size is large enough Here np = 1100* 0.32 = 352 and nq = 1100* (1-.32) = 748 Since both np and nq are larger than 5 the sample is large. Since the sample size is large, we choose z- (standard Normal) distribution and the test statistic is z = = = 2.84 hence p-value = .0023 Since p-value << α (which is .025) we make a decision to reject H0 And conclude that at 2.5% LOS data does not support the H0 hence the proportion of such people has gone up since 2002**Exercise 9.99 (b)**Type I error = decide that population proportion has gone up when in fact it has not. P(type I error ) = α = .025