# ECON1203 Hw Solution week09

UNSW ECON1203...

#### Description

BES Tutorial Sample Solutions, S1/13 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 Rejection region: ̅ 50 1.645 . 10⁄√100 Alternatively 10 51.645 50 1.645 ̅ ̅ . √ √100 Since 55 50 5 1.645 . 10⁄√100 Can reject H 0 and conclude that the population mean is greater than 50 .

0.05

50

X

51.645 reject

1

0.05 0 1.645

Z reject

5 (b)

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

Rejection region: ̅

25

̅

̅

.

24

25

5⁄√100

1.28

.

5⁄√100

Alternatively Since

0.0000

25

2

1.28

.

5 √100

24.36

1.28

Can reject H 0 and conclude that the population mean is less than 25 . 0.1 X 24.36 25 reject 2

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 Rejection region: ̅

80

Alternatively: ̅

̅

.

or ̅ Since

̅

.

1.96

.

4⁄√100

√ √ 80.5

80

1.96

80

1.96

80

4⁄√100

1.96

4 √100 4 √100

79.216 80.784

1.25

is not less than ‐1.96 or nor greater than 1.96 we do not reject H 0 and conclude that the population mean is equal to 80. 3

0.025

0.025

79.216

80

80.784

X

reject

reject

0.025

0.025

‐1.96 reject

2

0

1.96

1.25

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  value of a house in the area ̅

\$265,000,

\$50,000, ~

,

We wish to test :

Rejection region:

250,000;

̅

̅ Since

̅

.

265,000

250,000

250,000

50,000⁄√150

or

:

250,000

250,000

50,000⁄√150

.

1.645

3.67

1.645

50,000

256,715.68

√150

.

1.645

Hence we reject H 0 and conclude that the mean house value in the area is more than \$250,000 . (b)

Why is an upper tail test most appropriate in this case?

The nature of the research problem dictates an upper tail test. In this case we will not believe the expert’s claim unless there is ‘significant’ sample evidence to do so. This implies an upper tail test. 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

The p‐value is the probability of obtaining a test statistic more extreme than the realized value, assuming the null hypothesis is true. The lower the p‐value, the greater is the evidence for rejection of the null hypothesis. In this case it is very unlikely to find a sample mean as extreme as \$265,000 given a population mean of \$250,000. (d)

Define the type I and II errors in the context of the part (a) test.

Type I Error: Concluding that housing price is more than \$250,000, while it is really \$250,000. Type II Error: Not being able to reject the claim that housing price is \$250,000, while it is really more. 3. What effect does increasing the sample size have on the outcome of a hypothesis test? Explain your answer using the example of a one-tail test concerning the mean of a normally distributed population with known variance. (It is expected that students will find this question difficult) Suppose an upper tail test Under

:

;

:

: ~

The point

,

~

,

⁄√

on N(0,1) corresponds to the point ̅

distribution of

.

~

0,1

on the

6

The distribution of

is:

 0

 0  z

X  n

But suppose the true  is to the right of . Then the true distribution of is say:  0 

 0  z

 n

The shaded area in the above diagram gives the probability of correctly rejecting H0 (i.e. the power, 1‐ which is greater than ) Now suppose the sample size is increased. As a result: decreases & hence

decreases.

Suppose the new sample size is n1>n.

7

The distribution of

will now look something like:

0  0  z

 n1

X

Note that with a fixed  the rejection region cutoff is now smaller. Again, if the true  is actually to the right of , the probability of rejecting the same incorrect null hypothesis is higher than before. Diagrammatically the true distribution of will be say X  0 

 0  z

 n1

Again the shaded area in the above diagram gives the probability of correctly rejecting H0. Conclusion: The probability of correctly rejecting a false H0. (the power of the test, as discussed in lectures this week) increases as n increases given we keep the Type I error () fixed. 8