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

x

x

XX

XX

XX

Exercise 9.9a. H0: µ = 20 hours vs H1: µ ≠ 20 hours, this

is a two tailed test.b. H0: µ = 10 hours vs H1: µ ≠ 10 hours, this a

two tailed test.c. H0: µ = 3 years vs H1: µ ≠ 3 years, this is

a two tailed testd. H0: µ = $1000 vs H1: µ < $1000, this is a

left tailed test.e. H0: µ = 12 minutes vs H1: µ > 12 minutes,

this is a right tailed test.

Exercise 9.17Size of population = 8.1 millionX= 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 weeksGiven that size of rejection region is α= .02Note 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

x

n

x

/

200/2.4

3.169.16

Exercise 9.42Population : 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.

x

n

x

/

324/330

9401005

Exercise 9.67Population: 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 1. Population is normal

2. Population standard deviation σ is unknown

3. 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

x

ns

x

/

25/1.2

5.6925.70

Exercise 9.99Population: Affluent Americans (annual income ≥ $75,000)X= # of households having serious problems in paying unexpected bill 0f $5000According to “Money” Survey in 2002 , population proportion of such households p = .32Question: has this proportion gone up since 2002?H0: p = .32 vs H1 : p > .32In 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) = 748Since 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 H0And conclude that at 2.5% LOS data does not support the H0 hence the proportion of such

people has gone up since 2002

p̂

npq

pp ˆ

110068.*32.

32.36.

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