WEEK 9 TUTORIAL EXERCISES (To be discussed in the week starting May 6) 1. Perform the following hypothesis tests of the population mean. In each case, illustrate the rejection regions on both the Z and ̅ distributions, and calculate the p-value (prob-value) of the test. (a) H0:μ=50,H1:μ>50,n=100, ̅ =55,σ=10,α=0.05 Rejectionregion: ̅ 50 1.645 . 10⁄√100 Alternatively 10 51.645 50 1.645 ̅ ̅ . √ √100 Since 55 50 5 1.645 . 10⁄√100 Canreject H 0 andconcludethatthepopulationmeanisgreaterthan 50 .

0.05

50

51.645 reject

X

1

0.05 0 1.645 reject Z

5 (b)

0.0000

H0: μ = 25, H1: μ < 25, n = 100, ̅ = 24, σ = 5, α = 0.1 ̅ 25 .

Rejectionregion:

Alternatively ̅ Since ̅

5⁄√100

1.28

.

√

25

1.28

5 √100

24.36

24

25

Canreject H 0 andconcludethatthepopulationmeanislessthan 25 . 0.1 X 24.36 25 reject 2

5⁄√100

2

.

1.28

0.1 ‐1.28 0 Z

reject 2 0.0228

(c) H0: μ = 80, H1: μ ≠ 80, n = 100, ̅ = 80.5, σ = 4, α = 0.05 Rejectionregion: ̅

80

.

Alternatively: ̅ or ̅ Since ̅ ̅

4⁄√100

1.96

1.96

.

√ √ 80.5

80 80

1.96 1.96

4 √100 4 √100

79.216 80.784

.

80

is not less than ‐1.96 or nor greater than 1.96 we do not reject H 0 and concludethatthepopulationmeanisequalto80. 3

4⁄√100

1.25

0.025

0.025

79.216 reject

80

80.784

X

reject

0.025

0

1.25

0.025

‐1.96 reject 2

1.96 2

Z

reject 0.1056 0.2112

4

2. A real estate expert claims the current mean value of houses in a particular area is more than $250,000. A random sample of 150 recent sales prices in the area yields a sample mean of $265,000. It is known that house values in the area are approximately normally distributed with a standard deviation of $50,000. (a)Perform an upper tail test of the null hypothesis that the population mean house value in the area is $250,000. Use a 5% level of significance and state the rejection (critical) region in terms of both ̅ and z. Let X valueofahouseinthearea ̅ $265,000, $50,000, ~ , Wewishtotest Rejectionregion: : 250,000; : 250,000

̅

250,000

.

or ̅ Since ̅

.

50,000⁄√150

1.645

√

250,000

1.645

50,000 √150

256,715.68

265,000

250,000

Hence we reject H 0 and conclude that the mean house value in the area is morethan $250,000 . (b) Why is an upper tail test most appropriate in this case?

50,000⁄√150

3.67

.

1.645

Thenatureoftheresearchproblemdictatesanuppertailtest.Inthiscasewe willnotbelievetheexpert’sclaimunlessthereis‘significant’sampleevidence todoso.Thisimpliesanuppertailtest. 5

(c)

What is the p-value associated with the test statistic used in the part (a) test? Interpret this value. 3.67 0.5 0.4999 0.0001

Thep‐valueistheprobabilityofobtainingateststatisticmoreextremethan therealizedvalue,assumingthenullhypothesisistrue.Thelowerthep‐value, thegreateristheevidenceforrejectionofthenullhypothesis.Inthiscaseitis veryunlikelytofindasamplemeanasextremeas$265,000givenapopulation meanof$250,000. (d) Define the type I and II errors in the context of the part (a) test.

TypeIError:Concludingthathousingpriceismorethan$250,000,whileitis really$250,000. TypeIIError:Notbeingabletorejecttheclaimthathousingpriceis$250,000, whileitisreallymore. 3. What effect does increasing the sample size have on the outcome of a hypothesis test? Explain your answer using the example of a...